Haritha

Doppler Effect In Sound

Effect Of Relative Motion Between A Source Of Souno And A Listener When a train approaches a station sounding its horn, the pitch of the sound seems to be higher to a listener standing on the platform.

Again, when the train passes the platform and moves away from the station, the same sound seems to be of a lower pitch to the listener.

• On the other hand, if the source of sound is at rest and the listener approaches it or moves away from it, the pitch of the sound appears to be higher or lower, respectively. These phenomena are known as the Doppler effect.
• We know that the pitch of sound is determined by its frequency. So, an apparent change in the frequency of sound is caused by the relative motion of the source of sound and the listener and this is known as the Doppler effect.

Doppler effect: The apparent change in the pitch of a note due to the relative motion of the source of sound and the listener is called the Doppler effect.

Doppler shift: Let n be the actual frequency of a source of sound and n’ be the apparent frequency of it due to the relative motion of the source and the listener. The apparent change in frequency, i.e., (n’ -n) is called the Doppler shift. The apparent increase or decrease of the pitch of sound refers to positive or negative Doppler shifts, respectively.

Experimental demonstration: A source of sound S which can emit continuous sound of the same frequency (for example, a whistle or a bell driven by a battery) is taken. It is tied to one end of a strong thread of length about 2 m to 3 m and whirled at a high speed along a circle ABCDA with P as the centre.

• Under this condition, a listener at O can easily recognise the successive increase and decrease of the pitch of the sound.
• This increase or decrease of the pitch is due to the Doppler effect. When the source S along the arc ABC approaches the listener the pitch of the sound is increased. When along CDA the source recedes from the listener, the pitch of the sound is decreased.

Doppler Effect In Sound Calculation Of Apparent Frequency And Doppler Shift

Let the velocity of sound in the air be V and the actual frequency of the source of sound be n.

So, the wavelength of the sound wave emitted from the source in air, \(\lambda=\frac{V}{n}\)…(1)

Effect of the motion of the listener: Suppose, the listener O is approaching a stationary source of sound S with velocity u₀. Since the source is at rest, n number of waves extend over the distance V in the direction of the listener, i.e., n number of waves extend over the distance traversed by the sound wave in unit time.

• So, the length of the sound wave that reaches the ears of the listener at O is given by \(\frac{V}{n}\). Thus, the wavelength of the sound remains unchanged. So it may be said that, n ∝ V.
• In this case, the velocity of sound relative to the listener is not V, rather the apparent velocity of sound relative to u₀ (the velocity of the listener) increases and becomes V’ = V+ u₀. So, the frequency of the sound heard by the listener also increases.

Effect of the motion of the source: Suppose, the source S is approaching the listener O at rest with velocity u_s.

• Since the listener is at rest, the velocity of sound relative to him is, V i.e., in this case, the velocity of sound remains unchanged. So, it may be said that, \(n \propto \frac{1}{\lambda}\).
• Now in this case, n number of waves do not extend over the distance V in the direction of the listener; rather due to the velocity of the source (u_s), the distance covered by n number of waves = V- u_s.
• Thus the apparent wavelength of the sound wave decreases and becomes \(\lambda^{\prime}=\frac{V-U_s}{n}\). Hence, the frequency of the sound heard by the listener increases.

Effect of the motion of both the listener and the Source: If the listener and the source both are in motion, the apparent frequency of the sound heard by the listener is given by,

⇒ \(n^{\prime}=\frac{V^{\prime}}{\lambda^{\prime}}=\frac{V+u_o}{\frac{V-u_s}{n}} \text { or, } n^{\prime}=\frac{V+u_o}{V-u_s} \times n\) ….(2)

∴ Doppler shift = \(n^{\prime}-n=\frac{u_s+u_o}{V-u_s} \times n\)…(3)

Frequency And Doppler Shift Special cases:

Source at rest and listener In motion: In this case, u_s = 0. So, from equations (2) and (3),

n’ = \(\frac{V+u_o}{V} \times n \quad \text { and } n^{\prime}-n=\frac{u_o}{V} \times n\)

Listener at rest and source in motion: In this case, u₀ = 0. So, from the equations (2) and (3),

n’ = \(\frac{V}{V-u_s} \times n \quad \text { and } n^{\prime}-n=\frac{u_s}{V-u_s} \times n\)

Frequency And Doppler Shift Discussions:

1. It is very important to use positive and negative signs properly while putting the values of u₀ and u_s in the equations (2) and (3). The rule, that is followed, is:

1. If the listener moves towards the source, u₀ is positive.
2. If the source moves towards the listener, u_s is positive.
   • Conversely:
     1. If the listener moves away from the source, u₀ is negative.
     2. If the source moves away from the listener, u_s is negative.

In general, it may be said that if u₀ and u_s are inclined at an angle θ with the direction of motion of sound, u₀cosθ and u_scosθ are to be placed in equations (2) and (3).

2. It is evident that if there Is no relative motion between the source and the listener, then u_s = -u₀. In that case, no Doppler effect takes place [equations (2) and (3)]. If the distance between the source and the listener decreases with respect to time, the apparent pitch of the sound increases and vice versa.

3. While calculating apparent frequency and Doppler shift it has been assumed that the velocities of both the source and the listener are less than that of sound in air, i.e., u_s < V and u₀ < V. If the velocity of either the source or the listener exceeds the velocity of sound (i.e., it becomes supersonic), the nature of Doppler effect becomes entirely different.

4. Effect of wind: Let the velocity of wind be v. If the wind blows in the direction of motion of sound, v is positive. Then the apparent velocity of sound relative to the listener at rest = V+ v. So, in the calculation of the Doppler effect, the velocity of sound V is to be replaced by V+ v. In that case, equations (2) and (3) become,

n’ = \(\frac{V+v+u_o}{V+v-u_s} \times n\)…(4)

and \(n^{\prime}-n=\frac{u_s+u_o}{V+v-u_s} \times n\)…(5)

It is obvious that if the wind blows in the opposite direction, -v is to be placed instead of v in equations (4) and (5).

5. Doppler effect in case of echo: Let the source of sound S be moving towards the stationary reflector R with velocity u_s. The velocity of the listener O is u₀ in the same direction.

In this case for the echo produced by the reflector R, S’ is the apparent source of sound which is the image of the principal source S. The listener O is approaching the apparent source S’ and the apparent source S’ is also approaching the listener O. So, u₀ and u_s are both positive.

∴ Apparent frequency n’ = \(\frac{V+u_o}{V-u_s} \times n\)…(6)

Here, both the source S and the observer O are moving towards the reflector. If any of them moves in the opposite direction, i.e., recedes from the reflector, u_s or u₀ in equation (6) is replaced by -u_s or -u₀.

Again, as a special case, if the positions of the source of sound and the listener always remain the same (for example, the horn of a car and a passenger of the same car), then u_s = u₀.

Then according to equation (6), \(n^{\prime}=\frac{V+u_o}{V-u_o} \times n\)

Formation of beats due to original sound and its echo: If the difference between the actual frequency and the apparent, frequency due to the Doppler effect (i.e., n-n’ or n’-n) is less than 10 Hz, then the original sound and its echo are superposed and beats are formed.

Doppler Effect In Sound Frequency And Doppler Shift Numerical Examples

Example 1. The frequency of the whistle of a train Is 512 Hz. The train crosses a station at a speed of 72km · h^-1. Calculate the frequency of the sound heard by a listener, standing on the platform, before and after the train crosses the station. Neglect the effect of wind. The velocity of sound is 336 m · s^-1.
Solution:

Velocity of the train = 72 km · h^-1

= \(\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 20 m · s^-1

When the train approaches the station, the distance covered by 512 sound waves =336-20 =316 m

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{316}{512} \mathrm{~m}\)

∴ Apparent frequency due to the Doppler effect

= \(\frac{\text { velocity of sound }(V)}{\lambda^{\prime}}\)

= \(\frac{336}{\frac{316}{512}}=\frac{336 \times 512}{316}=544.4 \mathrm{~Hz}\)

Again, when the train recedes from the station, the distance covered by 512 sound waves = 336 + 20 = 356 m

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{356}{512} \mathrm{~m}\)

∴ Apparent frequency = \(\frac{V}{\lambda^{\prime}}=\frac{336}{\frac{356}{512}}=\frac{336 \times 512}{356}=483.2 \mathrm{~Hz}\)

Alternative Method:

According to the question, the listener is at rest and the source is in motion. So, the apparent frequency to the listener,

n’ = \(\frac{V}{V-u_s} \times n\)

In the first case, V = \(336 \mathrm{~m} \cdot \mathrm{s}^{-1}, n=512 \mathrm{~Hz}\)

and \(u_s=72 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{72 \times 1000}{60 \times 60}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(n^{\prime}=\frac{336}{336-20} \times 512=\frac{336}{316} \times 512=544.4 \mathrm{~Hz}\)

In the second case, V = \(336 \mathrm{~m} \cdot \mathrm{s}^{-1}, n=512 \mathrm{~Hz}\)

and \(u_s=-20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(n^{\prime}=\frac{336}{336-(-20)} \times 512=\frac{336 \times 512}{356}=483.2 \mathrm{~Hz}\)

Example 2. A sound of frequency 512 Hz Is emitted from a stationary source. A train running at a speed of 72 km · h^-1 passes the source. What will be the frequency of the sound heard by a passenger of the train before and after passing the source? Neglect the effect of wind. The velocity of sound is 336 m · s^-1.
Solution:

Velocity of the train = 72 km · h^-1

= \(\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

When the train approaches the source, the velocity of sound relative to the passenger,

V’ = 336 + 20 = 356 m · s^-1

Since the source is at rest, the wavelength of sound remains the same.

∴ Apparent frequency due to the Doppler effect

= \(\frac{V^{\prime}}{\lambda}=\frac{V^{\prime}}{\frac{V}{n}}=\frac{V^{\prime} n}{V}=\frac{356 \times 512}{336}=542.5 \mathrm{~Hz}\)

Again when the train recedes from the source, the velocity of sound relative to the passenger V’ = 336 – 20 = 316 m · s^-1

∴ Apparent frequency = \(\frac{V^{\prime} n}{V}=\frac{316 \times 512}{336}=481.5 \mathrm{~Hz}\)

Example 3. When a train approaches a listener, the apparent frequency of the whistle is 100Hz, while the frequency appears to be 50 Hz when the train recedes. Calculate the frequency when the listener is in the train.
Solution:

If the listener is in the train, he will listen to the actual frequency of the whistle (n).

If V is the velocity of sound, u_s is the velocity of the source and u₀ is the velocity of the listener,

the apparent frequency, \(n^{\prime}=\frac{V+u_o}{V-u_s} \times n\)

In the given problem, in both cases, the listener is at rest. So, u₀ = 0

In the first case, the motion of the train is towards the listener. So, u_s is positive.

Again in the second case, the train recedes from the listener. So, u_s is negative.

Therefore, in the two cases we have, \(100=\frac{V+0}{V-u_s} \times n\)

or, \(\frac{V}{V-u_s} \times n=100\)….(1)

and \(50=\frac{V+0}{V-\left(-u_s\right)} \times n\)

or, \(\frac{V}{V+u_s} \times n=50\)….(2)

From equation (1), \(\frac{n}{100}=\frac{V-u_s}{V}=1-\frac{u_s}{V} \quad \text { or, } \frac{u_s}{V}=1-\frac{n}{100}=\frac{100-n}{100}\)

From equation (2), \(\frac{n}{50}=\frac{V+u_s}{V}=1+\frac{u_s}{V} \quad \text { or, } \frac{u_s}{V}=\frac{n}{50}-1=\frac{n-50}{50}\)

∴ \(\frac{100-n}{100}=\frac{n-50}{50} \quad \text { or, } 100-n=2 n-100\)

or, \(3 n=200 \quad \text { or, } n=66 \frac{2}{3} \mathrm{~Hz}\)

Example 4. Two engines pass each other in opposite directions. One of them blows a whistle of frequency 540 Hz. Find the frequencies heard by a passenger sitting on the other engine before and after passing each other. The velocity of both engines = 72 km · h^-1; velocity of sound =340 m · s^-1.
Solution:

Velocity of the engine,

u = \(72 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Velocity of sound, V = 340 m · s^-1;

Actual frequency, n = 540 Hz

At the time of approaching each other, the velocity of the whistle, u_s = +20 m · s^-1

The velocity of the passenger on the other engine, u₀ = +20 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340+20}{340-20} \times 540=\frac{360}{320} \times 540=607.5 \mathrm{~Hz}\)

Again at the time of receding from each other, u_s = -20 m · s^-1 and u₀ = -20 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340-20}{340-(-20)} \times 540=\frac{320}{360} \times 540=480 \mathrm{~Hz}\)

Example 5. A car travelling at a speed of 36 km · h^-1 sounds its horn of frequency 500 Hz. It is heard by the driver of another car which Is travelling behind the first car In the same direction with a velocity of 20 m · s^-1. Another sound Is heard by the driver of the second car after reflection from a bridge ahead. What will be the frequencies of the two sounds heard by the driver of the second car? Sound travels in air with a speed of 340m · s^-1.
Solution:

The driver of the second car O is the listener. The first car S and its image S’ due to reflection on the bridge ahead are two sources of sound.

The direction of motion of O is towards S and S’.

So velocity of the listener, u₀ = + 20 m · s^-1

Here, S is receding from O. So, the velocity of the source S,

u₀ = -36 km · h^-1

= \(-\frac{36 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)=\(-10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Apparent frequency to the listener due to S,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340+20}{340-(-10)} \times 500\)

= \(\frac{360}{350} \times 500=514.3 \mathrm{~Hz}\)

On the other hand, S’ is approaching O.

So, velocity of the source S’, u_s = +10 m · s^-1

∴ Apparent frequency to the listener due to S’,

n’ = \(\frac{V+u_o}{V-u_s} \times n=\frac{340+20}{340-10} \times 500\)

= \(\frac{360}{330} \times 500=545.5 \mathrm{~Hz}\)

So the driver of the second car will hear the sounds of frequencies 514.3 Hz and 545.5 Hz.

Example 6. A whistle, emitting a sound of frequency 440 Hz, is tied to a thread of length 1.5 m and rotated with an angular velocity of 20 rad · s^-1 on a horizontal plane, Calculate the range of frequency of the sound heard Velocity of sound in air =330 m · s^-1
Solution:

Linear velocity of the whistle (u)

= angular velocity x radius of the circular path

= 20 x 1.5 = 30 m · s^-1

The velocity of the whistle at that position of the circular path where the whistle approaches the listener.

u_s = +30m · s^-1

∴ Apparent frequency, \(n_1 =\frac{V}{V-u_s} \times n=\frac{330}{330-30} \times 440\)

= \(\frac{330}{300} \times 440=484 \mathrm{~Hz}\)

Again, the velocity of the whistle at that position of the circular path where the whistle recedes from the listener, u’s =-30m-s_1

Apparent frequency, \(n_2 =\frac{V}{V-u_s^{\prime}} \times n=\frac{330}{330-(-30)} \times 440\)

= \(\frac{330}{360} \times 440=403.3 \mathrm{~Hz}\)

So, the range of frequency is 403.3 Hz to 484 Hz.

Example 7. Each of the two persons has a whistle of frequency 500 Hz. One person is at rest at a particular place and the second person recedes from him with a velocity of 1.8 m · s^-1. If both of them blow whistles, how many beats will be heard by each of them? The velocity of sound = 330 m · s^-1.
Solution:

Both of them listen to the sound of frequency 500 Hz for their own whistles.

In the first case, let us suppose that the first person is listening to the sound coming from the whistle of the second person.

Here the velocity of the listener, u₀ = 0

Since the source is receding, the velocity of the source, u_s = -1.8 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V}{V-u_s} \times n=\frac{330}{330-(-1.8)} \times 500=497.29 \mathrm{~Hz}\)

∴ Number of beats per second = n-n’ = 500-497.29

= 2.71 ≈ 3

In the second case, let us suppose that the second person is listening to the sound coming from the whistle of the first person.

Here the velocity of the source, u_s = 0

Since the listener is receding,

The velocity of the listener, u₀ = -1.8 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V} \times n=\frac{330-1.8}{330} \times 500=497.27 \mathrm{~Hz}\)

∴ Number of beats per second = n-n’ =500-497.27

= 2.73 ≈ 3

Example 8. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. What is the ratio of the velocity of train B to that of train A?
Solution:

Let, u_A be the velocity of train A.

∴ Frequency of siren relative to the passenger of train A \(n_1=n\left(\frac{V+u_A}{V}\right)\)

∴ \(5\left(\frac{V+u_A}{V}\right)=5.5\)….(1)

Similarly, the frequency of the siren relative to the passenger of train B,

⇒ \(n_2=n\left(\frac{V+u_B}{V}\right) \quad\left[u_B=\text { velocity of train } B\right]\)

∴ \(5\left(\frac{V+u_B}{V}\right)=6\)…(2)

From equation (1), 5.5 V- 5 V = 5 u_A

∴ V = \(\frac{5}{0.5} u_A\)….(3)

From equation (2), 6V-5V = 5u_B

∴ V = 5u_B……(4)

Comparing equations (3) and (4), \(\frac{5}{0.5} u_A=5 u_B\)

∴ \(\frac{u_B}{u_A}=\frac{1}{0.5}=\frac{2}{1}\)

∴ \(u_B: u_A=2: 1\)

Example 9. A railway track and a road are mutually perpendicular. A train Is approaching the railway crossing at a speed of 80 km/h. When the train h at a distance 1 km from the crossing it blows a whistle of frequency 400 Hz. Which frequency of sound will he hear from a man on a road at a distance of 600 m from the crossing? velocity of sound =330 m · s^-1
Solution:

Train (S) is moving with a velocity 80 km/h along SA. At the time of blowing the whistle, the distances of the train and a man (O) from the crossing (A) are 1 km and 600 m respectively.

∴  SA = 1 km = 1000 m; OA = 600 m

If u_S be the velocity of the train, the component along SO is u_S cosθ.

∴ \(\cos \theta=\frac{S A}{S O}=\frac{S A}{\sqrt{S A^2+A O^2}}=\frac{100}{\sqrt{(1000)^2+(600)^2}}=0.8575\)

∴ \(u_S \cos \theta=80 \times \frac{5}{18} \times 0.8575=19.05 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Apparent frequency of sound, n’ = \(\frac{V}{V-u_S \cos \theta} n\)

Here, V = velocity of sound = 330 m · s^-1 and actual frequency, n = 400 Hz

∴ n’ = \(\frac{330}{330-19.05} \times 400=424.5 \mathrm{~Hz}\)

Doppler Effect In Sound and Light

Like sound, the Doppler effect also takes place in the case of light waves. When a source of light and an observer are in relative motion, an apparent change in the frequency, i.e., the wavelength of light is perceived in the eyes of the observer.

It means that the colour of the light is found to change in the eyes of the observer. This Doppler effect of light is of two kinds.

Redshift: When the source and the observer recede from each other, the wavelength of the light apparently increases. This apparent increase of wavelength due to the Doppler effect is called redshift. This is a displacement of the Lines in the spectra towards the red end of the visible spectrum (i.e., towards a longer wavelength).

Blueshift: When the source and the observer approach each other, the wavelength of the light apparently decreases. This apparent decrease of wavelength is called blue shift. This is a displacement of the lines in the spectra towards the blue end of the visible spectrum (i.e., towards a shorter wavelength).

Calculation of apparent wavelength and Doppler shift: Let the velocity of light in a vacuum be c, the original frequency of a monochromatic light be n and the original wavelength of the light, \(\lambda=\frac{c}{n}\).

Now suppose that the source of this monochromatic light and an observer approach each other with relative velocity u. In that case, n number of waves produced per second do not occupy the distance c, rather the distance occupied by those n number of waves = c- u

So, the apparent wavelength is \(\lambda^{\prime}=\frac{c-u}{n}=\frac{c}{n}\left(1-\frac{u}{c}\right) \quad \text { or, } \lambda^{\prime}=\left(1-\frac{u}{c}\right) \lambda\)…(1)

∴ Doppler shift of wavelength = \(\lambda^{\prime}-\lambda=-\frac{u}{c} \lambda\)….(2)

Doppler Effect Of Light Special cases:

1. When the source approaches the observer, u is positive in equations (1) and (2). So, λ’ is less than λ, i.e., the wavelength apparently decreases. It is the phenomenon of blue shift.
2. When the source recedes from the observer, u is negative in equations (1) and (2). So, λ’ is greater thanλ, i.e., the wavelength apparently increases. It is the phenomenon of redshift.

Doppler Effect Of Light Discussions:

1. Velocity of light in vacuum, c = 3 x 10⁸ m · s^-1. If the relative velocity of the source and the observer is very small, i.e., u<<c or, \(\frac{u}{c} \ll 1\), then according to equation (1), λ’ ≈ λ. In this case, the Doppler effect is practically absent. So when a train or motor car lighting their headlights approaches an observer, red or blue shift becomes negligible, i.e., no apparent change in the colour of light takes place.

2. On the other hand, if the relative velocity of the source and the observer is nearly equal to the velocity of light, i.e., u<<c or, \(\frac{u}{c} \approx 1\) then according to equation (1), λ’  ≈ 0. But in practical cases the wavelength never becomes zero or nearly zero. So it is evident that if u ≈ c, equation (1) is no longer applicable.

In that case, the corrected form of equation (1) will be \(\lambda^{\prime}=\lambda \sqrt{\frac{c-u}{c+u}}\) and the associated frequency will be, \(n^{\prime}=n \sqrt{\frac{c+u}{c-u}}\).

The theory of relativity is essential to reach these equations which is out of the current syllabus.

3. The Doppler effect of light differs remarkably from that of sound. The relative velocity of sound increases or decreases when the listener is in motion with respect to the source of sound.

• On the other hand, the velocity of light, as proposed in the theory of relativity, is always equal to c, whatever be the relative velocity of the observer and the source.
• Hence Doppler effect of light takes place for the relative velocity u of the source and the observer. Here, it is not necessary to consider the velocities of the source of light and the observer separately.

Application of Doppler effect of light:

1. Velocity of Slots: The velocity of a distant star relative to the earth can be determined by measuring the Doppler shift of the light that comes to the Earth from the star. Generally, for a star or galaxy of stars, this shift is a red shift from which it is understood that these stars are receding gradually from the earth (velocity of separation is 10 km · s^-1 to 300km ·  s^-1). This observation supports the theory of an expanding universe.

2. Binary Star: For a star, if both red shift and blue shift are observed, it may be concluded that the star is actually a binary star. The two stars are revolving around a common axis.

3. Quasar: The redshift of these stars is very high. Calculations show that these stars are receding from the earth at a tremendously high speed (in some cases tire speed becomes 80% of the speed of light).

4. Rotation of the sun about its own axis: As the sun rotates about its axis, one side faces the earth while the other side goes behind it. So, a blue shift for one side and a rod shift for the other side are observed. From the calculations of these shifts, it is found that the speed of rotation of the sun is about 2 km · s^-1.

5. Doppler radar: It is used to measure the speed of an aeroplane flying at a high speed. The waves emitted from a Doppler radar are reflected from a flying aeroplane. By measuring the Doppler shift of the reflected radar waves, the speed of the aeroplane can be determined.

6. Measurement of high temperature: A blue shift is observed in the light emitted by a hot substance when the molecules of the substance move towards the observer. Whereas a red shift is observed when the molecules move away from the observer.

Measuring this shift, the expression of rms velocity of die molecules is determined. Next from the expression of rms velocity = \(\sqrt{\frac{3 R T}{M}}\), the temperature T is obtained (M = Mass of 1 mol of the substance).

Doppler Effect In Sound Doppler and Light Conclusion

The apparent change in the pitch of a note due to relative motion between a source of sound and a listener is called the Doppler effect.

If n is the actual frequency of a source of sound and n’ is the apparent frequency of it due to the relative motion of the source and the listener, the apparent change in frequency (n’ -n) is called Doppler shift.

• No Doppler effect can be noticed by an observer in the absence of any relative motion between the source and observer.
• When the source and the observer recede from each other, an apparent increase in the wavelength of light Is observed. This apparent increase of wavelength due to the Doppler effect is called redshift. This is a displacement of the lines in a spectrum towards the red end (i.e., towards a longer wavelength).
• When the source and the observer approach each other, an apparent decrease in the wavelength of light is observed. This apparent decrease of wavelength is called blue shift. This is a displacement of the lines in a spectrum towards the blue end (i.e., towards a shorter wavelength).

Doppler Effect In Sound Doppler Effect Of Light Useful Relations For Solving Numerical Examples

If the listener and the source are both in motion, the apparent frequency of the sound heard by the listener is,

n’ = \(\frac{V+u_o}{V-u_s} n\)

where V = velocity of sound in air, n = actual frequency of source of sound, u_s = velocity of the source of sound.

Doppler shift, n’ -n = \(\frac{u_s+u_o}{V-u_s} n\)

Sign convention:

1. If the listener moves towards the source, u₀ is positive,
2. If the source moves towards the listener, u_s is positive.
   • Conversely,
     1. If the listener moves away from the source, u₀ is negative,
     2. If the source moves away from the listener, u_s is negative.

If the velocity of wind is v and if the wind blows in the direction of motion of sound, then n’ = \(\frac{V+v+u_o}{V+v-u_s} n\)

If the velocity of wind is v and if the wind blows in the opposite direction of motion of sound, then n’ = \(\frac{V-v+u_o}{V-v-u_s} n\)

A car moving towards a stationary reflector with velocity u₀ blows a horn of frequency n. The frequency of echo heard by the passenger of the car will be

n’ = \(\frac{V+u_0}{V-u_0} n\)

A stationary car blows a horn of frequency n. If a reflecting surface moves towards the car with velocity u_s, the frequency of echo heard by the passenger of the car is

n’ = \(\frac{V+u_s}{V-u_s} n\)

Some Important Cases of the Doppler Effect

Doppler Effect In Sound Doppler Effect Of Light Very Short Answer Type Questions

Question 1. Which property of sound undergoes an apparent change due to the Doppler effect?
Answer: Pitch

Question 2. When a source moves at a speed greater than that of sound, will the Doppler effect hold?
Answer: No

Question 3. Will there be a Doppler effect for sound when the source and listener move at a right angle to the line joining them?
Answer: No

Question 4. When a source of sound approaches a stationary listener, the sound appears to be _______ to the listener.
Answer: Sharper

Question 5. When a listener approaches a source of sound, the sound appears to be ______ to the listener.
Answer: Sharper

Question 6. The apparent change of frequency of sound due to the Doppler effect is called Doppler ______
Answer: Shift

Question 7. If there is a relative motion between a source of light and an observer, a change of colour of the light appears in the eyes of the observer. What is the name of this phenomenon?
Answer: Doppler effect of light

Question 8. When a source of light and an observer recede from each other, the apparent change in the wavelength of light is called _______ shift.
Answer: Red

Doppler Effect In Sound Doppler Effect Of Light Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The intensity of sound waves changes when the listener moves towards or away from the stationary source.

Statement 2: The motion of the listener causes the apparent change in wavelength.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 2.

Statement 1: When there is no relative velocity between the source and observer, then the observed frequency is the same as emitted.

Statement 2: The velocity of sound when there is no relative velocity between the source and observer is zero.

Answer: 3. Statement 1 is true, statement 2 is false.

Doppler Effect In Sound Doppler Effect Of Light Match Column 1 With Column 2

Question 1. Source has frequency f. Source and observer both have the same speed. The apparent frequency observed by the observer matches the following:

Answer: 1. C, 2. A, 3. B, 4. C

Doppler Effect In Sound Doppler Effect Of Light Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A source 5 of an acoustic wave of frequency v₀ = 1700 Hz and a receiver R are located at the same point. At the instant t = 0, the source starts from rest to move away from the receiver with a constant acceleration ω. The velocity of sound in air is v = 340 m · s^-1.

1. If ω = 10 m · s^-2, the apparent frequency that will be recorded by the stationary receiver at t = 10 s will be

1. 1700 Hz
2. 1.35 Hz
3. 850 Hz
4. 1.27 Hz

Answer: 2. 1.35 Hz

2. if ω = 0 for t> 10 s, the apparent frequency recorded by the receiver at t = 15 s will be

1. 1700Hz
2. 1310Hz
3. 850Hz
4. 1.23kHz

Answer: 2. 1310Hz

3. If ω = 10 m · s^-2, the apparent frequency that will be recorded by the stationary receiver just at the instant when the source is exactly 1 km away from the receiver will be

1. 1700 Hz
2. 1310 Hz
3. 850 Hz
4. 1.26 kHz

Answer: 4. 1.26 kHz

Question 2. A small source of sound vibrating at a frequency 500 Hz is rotated in a circle of radius (100/π)cm at a constant angular speed of 5.0 revolutions per second, The speed of sound in air is 330 m · s^-1, An observer (A) is situated at a great distance on a straight line perpendicular to the plane of the circle, through its centre. Another observer (B) is at rest at a great distance from the centre of the circle but nearly in the same plane. After some time the source of sound comes to rest after reaching the centre of the circle, At that time, another observer (C) moves towards the source with a constant speed of 20 m · s^-1, along the radial line to the centre.

1. The apparent frequency of the source heard by A will be

1. Greater than 500 Hz
2. Smaller than 500 Hz
3. Always 500 Hz
4. Greater for half the circle and smaller during the other half

Answer: 3. Always 500 Hz

2. The minimum and the maximum values of the apparent frequency heard by B will be

1. 455 Hz and 535 Hz
2. 485 Hz and 515 Hz
3. 485 Hz and 500 Hz
4. 500 Hz and 515 Hz

Answer: 2. 485 Hz and 515 Hz

3. The change in the frequency of the source heard by C will be

1. 6%
2. 3%
3. 2%
4. 9%

Answer: 1. 6%

Doppler Effect In Sound Doppler Effect Of Light Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. The frequency of the sound of a car horn as perceived by an observer towards whom the car is moving differs from the frequency of the horn by 2.5%. Assuming that the velocity of sound in air is 320 m · s^-1, find the velocity (in m · s^-1) of the car.
Answer: 8

Question 2. A man is watching two trains, one leaving and the other coming in with equal speeds of 4m · s^-1. If they sound their whistles, each of frequency 240 Hz, find the number of beats heard by the man (velocity of sound in air =320 m · s^-1).
Answer: 6

Question 3. The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of sound in air is 300 m · s^-1, find the velocity (in m · s^-1) of the source.
Answer: 3

Question 4. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f₀. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m· s^-1.
Answer: 7

Doppler Effect In Sound Doppler Effect Of Light Short Answer Type Questions

Question 1. Show that, the change in frequency of sound during the motion of the source towards the audience is more than that when the audience moves towards the source with the same velocity.
Answer:

Natural frequency of the source = n;

velocity of sound = V;

velocity of the source = v and

velocity of the audience = u.

So, apparent change in frequency = Doppler shift (Δn)

= \(\frac{\nu+u}{V-\nu} n\)

When the source of sound remains stationary, v = 0.

When the audience moves towards the source, \(\Delta n_1=\frac{u}{V} n\)

When the audience is stationary, u = 0.

when the source moves towards the audience with the same velocity (i.e., v= u), \(\Delta n_2=\frac{u}{V-u} n\)

Clearly, Δn₂ > Δn₁

Question 2. A car is moving with a speed of 72 km · h^-1 towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is 340 m · s^-1, the difference of the two frequencies, the driver hears is

1. 50 Hz
2. 85 Hz
3. 100 Hz
4. 150 Hz

Answer:

Velocity of the car, u₀ = 72 km · h^-1 = 20 m · s^-1

When the car is approaching the source, the apparent frequency is,

∴ \(n_1=\left(\frac{V+u_o}{V}\right) n=\left(\frac{340+20}{340}\right) 850=900 \mathrm{~Hz}\)

When the car moves away from the source, then appar¬ent frequency is,

∴ \(n_2=\left(\frac{V-u_o}{V}\right) n=\left(\frac{340-20}{340}\right) 850=800 \mathrm{~Hz}\)

∴ n₁ -n₂ = 900 – 800 = 100 Hz

The option 3 is correct.

Question 3. A train is moving with a uniform speed of 33 m/s and an observer is approaching the train with the same speed. If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s, then the apparent frequency of the sound that the observer hears is

1. 1220 Hz
2. 1099 Hz
3. 1110 Hz
4. 1200 Hz

Answer:

Here, the velocity of the source, u_s = 33 m/s

The velocity of the observer, u₀ = 33 m/s,

Now, velocity of sound, v = 333 m/s

As the source and the observer are approaching each other,, the apparent frequency,

n’ = \(\frac{\nu+u_0}{\nu-u_2} \cdot n=\frac{333+33}{333-333} \times 1000=1220 \mathrm{~Hz}\)

The option 1 is correct.

Question 4. A train is moving on a straight track with a speed 20 m · s^-1. It is blowing its whistle at the frequency of 1000Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m · s^-1) close to

1. 6%
2. 12%
3. 18%
4. 24%

Answer:

Velocity of the listener, v₀ = 0 ;

The velocity of the train, v_s = 20 m/s

So, apparent frequency when the train moves towards the listener,

⇒ \(n_1=\frac{v}{v-\nu_s} n=\frac{320}{320-20} \times 1000=1066.7 \mathrm{~Hz}\)

Again, apparent frequency when the train moves away from the listener,

⇒ \(n_2=\frac{v}{\nu-\left(-v_s\right)}=\frac{320}{320+20} \times 1000=941.2 \mathrm{~Hz}\)

Hence, the percentage change in the apparent frequency

= \(\frac{n_1-n_2}{n} \times 100=\frac{1066.7-941.2}{1000} \times 100\)

= 12.55% ≈ 12%

The option 2 is correct.

Question 5. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at a frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 x 10⁸ m · s^-1)

1. 10.1 GHz
2. 12.1GHz
3. 17.3 GHz
4. 15.3 GHz

Answer:

Speed of the observer = \(\frac{c}{2}\)

∴ Doppler effect in relevant with the theory of relativity as follows:

v’ = \(\nu \sqrt{\frac{1+\beta}{1-\beta}}\)

β = \(\frac{\text { speed of observer }}{\text { speed of light }}=\frac{1}{2}\)

∴ \(\nu^{\prime} =10 \sqrt{\frac{1+\frac{1}{2}}{1-\frac{1}{2}}}=10 \sqrt{3}=10 \times 1.73\)

= 17.3 GHz

The option 3 is correct.

Question 6. A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km · h^-1. He finds that traffic has eased and a car moving ahead of him at 18 km · h^-1 is honking at a frequency of 1392 Hz. If the speed of sound is 343 m · s^-1, the frequency of the honk as heard by him will be

1. 1332 Hz
2. 1372 Hz
3. 1412 Hz
4. 1454 Hz

Answer:

Speed of motorcyclist, u₀ = 36 km · h^-1 = 10 m · s^-1

Speed of car, u_s = 18 km · h^-1 = 5 m · s^-1

Now, the fundamental frequency of honk, v = 1392 Hz

∴ \(\lambda^{\prime}=\frac{V+u_s}{n}=\frac{343+5}{1392}=0.25 \mathrm{~m}\)

Hence, the frequency of honking as heard by the cyclist

= \(\frac{V+V_0}{\lambda^{\prime}}=\frac{343+10}{0.25}=1412 \mathrm{~Hz}\)

The option 3 is correct.

Question 7. A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15m · s^-1. Then the frequency of sound that the observer hears in the echo reflected from the cliff is (take the velocity of sound in air = 330 m · s^-1

1. 800 Hz
2. 838 Hz
3. 885 Hz
4. 765 Hz

Answer:

Apparent frequency of echo

n’ = \(n\left(\frac{v}{v-v_s}\right)=800\left(\frac{330}{330-15}\right)\)

= \(\frac{800 \times 330}{315}=838 \mathrm{~Hz}\)

Question 8. Due to the Doppler effect, the shift in wavelength observed is 0.1 Å, for a star producing a wavelength of 6000 Å. The velocity of recession of the star will be

1. 20 km s^-1
2. 2.5 km s^-1
3. 10 km s^-1
4. 5 km s^-1

Answer:

Doppler shift, \(\Delta \lambda=\lambda \frac{v}{c}\)

∴ v = \(c \frac{\Delta \lambda}{\lambda}=\left(3 \times 10^8\right) \times \frac{0.1}{6000}\)

= \(5 \times 10^3 \mathrm{~m} / \mathrm{s}=5 \mathrm{~km} / \mathrm{s}\)

The option 4 is correct.

Question 9. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.

1. What is the frequency of the whistle for a platform observer when the train

1. Approaches the platform with a speed of 10 m/s,
2. Recedes from the platform with a speed of 10 m/s.

2. What is the speed of sound in each case if the speed of sound in still air is 340 m/s.

Answer:

1. Here, n = 400Hz; v = 340m/s

When a train approaches the platform, v_s = 10m/s

So, required frequency, = \(\frac{v}{v-v_s} \times n=\frac{340 \times 400}{340-10}=412.12 \mathrm{~Hz}\)

When the train recedes from the platform, v_s = 10 m/s

Hence, required frequency = \(\frac{v}{v+v_s} \times n=\frac{340 \times 400}{340+10}=388.6 \mathrm{~Hz}\)

2. The speed of sound in both the cases is same.

Question 10. Once Amit was going to his house. He was listening to music on his mobile with earphones while crossing the railway line and he did not hear the sound of an approaching train though the train was blowing the horn. A person nearby ran towards him and pushed away just as the train reached there. Amit realised his mistake and thanked the person.

1. Describe the value possessed by the person.
2. Name the phenomenon of change in frequency of sound when there is relative motion between the observer and source of sound.

Answer:

1. The values possessed by the person are—
   1. Presence of mind,
   2. General awareness,
   3. Good understanding,
   4. Prompt decision-making ability,
   5. Concern for other people’s safety and well-being,
   6. Helping and caring nature.
2. The phenomenon is Doppler’s effect on sound.

 Nature Of Vibration – Free Or Natural Vibration

A vibrating body always moves to and fro about an equilibrium position. At the instant when it is at the equilibrium position, there are no forces acting on the body. However, the body does not stop because of its inertia of motion.

As soon as it crosses the equilibrium position, a force acts on it. This force is always directed toward the equilibrium position and is called the restoring force.

If no force other than the restoring force acts on the body, or the effect of other forces is negligible, the body can vibrate without interruption, i.e., its vibration is a free or natural vibration. The amplitude of this vibration remains unchanged with the passage of time.

Free Or Natural Vibration Definition: If the effect of forces other than the restoring force is negligible on a vibrating body, its motion is called a free or natural vibration.

A body undergoing free vibration has a definite frequency, i.e., the body executes a fixed number of vibrations in unit time. It is called its natural frequency. The natural frequency (n₀) depends on the density, shape elasticity, etc. of the vibrating body. For example:

A simple pendulum: \(n_0=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{g}{L}}\); so if g is constant, the natural frequency of a simple pendulum depends on its length L.

An elastic spring: \(n_0=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\) = here the spring constant k is related to the elasticity of the material of the spring.

It may be said that the natural frequency of the spring depends on the mass (m) hanging at the end of the spring and the elasticity of the material of the spring. Every vibrating body such as a tuning fork, the string of a musical instrument, etc. has its characteristic natural frequency.

Nature Of Vibration – Damped Vibration

Damped Vibration Definition: If resistive forces act on a vibrating body in addition to the restoring force, its amplitude gradually diminishes. After some time, the body comes to rest at its position of equilibrium. This type of vibration is called damped vibration. The resistive effect is called damping.

• In fact, free vibration does not exist in real life. All vibrating bodies ultimately come to rest after some time, i.e., all vibrations are damped. Different types of resistive forces act on different types of vibrating bodies.
• For a simple pendulum, the resistive force is the viscous force of air; in a moving coil galvanometer, the resistive force is electromagnetic damping. The vibrating body has to work against these resistive forces. So its energy decreases. When the energy of the body becomes zero, it comes to rest.

Graphical representation of free and damped vibration: The characteristics of free and damped vibrations can be understood easily with the help of displacement time graphs. If the vibration of a body is free, it will vibrate forever with its amplitude unchanged. In real life, the amplitude gradually diminishes and ultimately the body comes to rest.

Beneficial examples of damped vibration: Damping plays a beneficial role in our modern-day life. One such application of damped oscillation is the car suspension system. It makes use of damping to make our ride less bumpy and more comfortable by counteracting and hence reducing the vibrations of the car when it is on the road for optimal passenger comfort, the system is critically damped or slightly underdamped.

Different types of damped motion:

1. If the damping is very weak, the body vibrates with almost its natural frequency. For example, if the bob of the simple pendulum is heavy enough, then the damping becomes insignificant. So the pendulum continues to oscillate for a long time. The time period and frequency of such a pendulum are almost equal to those of a free pendulum.

2. If the damping is stronger, the vibration of a body does not continue for a long time. For example, if a light piece of wood is used as the bob of a simple pendulum and is allowed to oscillate, it comes to rest after a few oscillations. In this case the time period becomes very large, i.e., the frequency of vibration is much less than the natural frequency.

3. In case of very strong damping, the vibrating body comes back to its position of equilibrium from its displaced position and stops there. So the body cannot move past its position of equilibrium. The body does not vibrate at all. This is called overdamped motion or aperiodic motion.

4. There is a particular state of damping, between small damping and overdamping, for which the body returns to its equilibrium position in the least time, but cannot travel past its position of equilibrium. This is called critical damping, In practical cases, if we want to stop the vibration of a body quickly, its damping is kept close to the state of critical damping.

5. If the damping is less than critical damping, the body oscillates with decreasing amplitude. This is known as underdamping.

Decrement of amplitude In damped vibration: From the graph for damped vibration we get, A₁ = initial amplitude of vibration, A₂ = amplitude of vibration after one complete oscillation, A₃ = amplitude of vibration after two complete oscillations.

Obviously, A₁ > A₂ > A₃

Two important characteristics of such a damped vibration are:

1. The amplitude of vibration decreases in a constant ratio for each complete vibration.

That is, \(\frac{A_1}{A_2}\)=\(\frac{A_2}{A_3}\)=\(\frac{A_3}{A_4}\)=\(\cdots\) = constant. This constant is called the decrement.

2. From the very initiation of motion, damping comes into play. Therefore, even the first amplitude A₁ of damped vibration is less than the amplitude A₀ of free vibration.

Equation of motion: A particle under damped har-monic vibration in one dimension is subject to two types of forces:

A restoring force: F₁ = -kx, where k is constant.

One or more resistive forces: Each resistive force is proportional to the velocity \(\left(v=\frac{d x}{d t}\right)\)of the particle, and acts in a direction opposite to that of the instantaneous velocity.

So, the resultant of the resistive forces is, \(F_2=-k^{\prime} v=-k^{\prime} \frac{d x}{d t}\), where k’ is another constant.

The acceleration of the particle is, \(a=\frac{d v}{d t}=\frac{d^2 x}{d t^2}\). Thus from the relation F = ma, we get, ma = F₁ + F₂

or, \(m \frac{d^2 x}{d t^2}=-k x-k^{\prime} \frac{d x}{d t}\)

or, \(m \frac{d^2 x}{d t^2}+k^{\prime} \frac{d x}{d t}+k x=0\)

or, \(\frac{d^2 x}{d t^2}+2 b \frac{d x}{d t}+\omega^2 x=0\)

Were, \(\omega=+\sqrt{\frac{k}{m}} and b=\frac{k^{\prime}}{2 m}\),

The equation (1) is known as the equation of motion of a damped SHM.

Nature Of Vibration – Damped Vibration Numerical Examples

Example 1. A second pendulum is shifted 4 cm away from its equilibrium position and then released. After 2 s  the pendulum is 3 cm away from its position of equilibrium. What will be the position of the pendulum after another 2s?
Solution:

Time period of a second pendulum is 2s. In the case of this damped vibration, let the change in the time period be negligible. For the given pendulum the initial amplitude of vibration A₁ = 4 cm. After one complete oscillation, the amplitude of vibration A₂ = 3 cm.

So, if the amplitude of vibration after two complete oscillations is A₃, then the decrement is:

⇒ \(\frac{A_1}{A_2}=\frac{A_2}{A_3} \text { or, } A_3=\frac{\left(A_2\right)^2}{A_1}=\frac{(3)^2}{4}=2.25 \mathrm{~cm} \text {. }\)

Example 2. After 100 complete oscillations, a pendulum’s amplitude becomes 1/3 rd of its initial value. What will be its amplitude after 200 complete oscillations? Express it as a fraction of the initial amplitude.
Solution:

The pendulum has a damped vibration. So, the amplitude decreases at the same rate. Since, after 100 complete oscillations the amplitude of vibration becomes 1/3 rd of the initial amplitude, after 200 complete oscillations, the amplitude will be 1/3 x 1/3 = 1/9 th of the initial amplitude.

Nature Of Vibration – Forced Vibration

Practically all vibrations are damped vibration. The vibrating body works against different resistive forces. So its energy diminishes and the amplitude gradually decreases.

• To maintain a steady vibration, energy from an external source is needed. If energy is supplied from an external source in such a way that the rate of supply of energy exactly balances the rate of loss of energy, then the amplitude of the body remains constant.
• The value of the amplitude is similar to that of the free vibration of the body. This type of vibration is called a forced vibration. For example, if we do not wind a pendulum clock, it will stop after a while due to damping.
• When we wind the clock, we compress a spring within the clock which stores potential energy and supplies that energy continuously. The pendulum oscillates continuously with constant amplitude and time period.

External means are required not only for maintaining the vibration but also to vibrate a body that is initially at rest.

Examples of forced vibration:

1. If we strike a prong of a tuning fork, the intensity of the emitted sound is not very high so it cannot be heard from a distance. Now, if the handle of the vibrating tuning fork is made to touch the surface of a table, the tuning fork sets the table surface into forced vibration. Sound is emitted from the table also. Hence, the sound is amplified.

• It must be kept in mind that, according to the principle of conservation of energy, the total amount of energy cannot be increased. When the handle of the tuning fork is pressed against a table surface, a part of the energy from the tuning fork is transferred to the table.
• As the damping of the table is higher than that of the tuning fork, the energy transferred to the table by the tuning fork decays at a faster rate. Hence, the vibration of the table stops earlier and the intensity of the sound produced due to vibration of the table is comparatively higher.
• In this example of forced vibration, the upper surface of the table is made to vibrate by the tuning fork. The vibration of the tuning fork is the driving vibration and the vibration of the upper surface of the table is the driven vibration.

2. A thread is tied loosely between P and Q. Two pendulums C and D, having different lengths, are sus¬pended from two points between P and Q. If the pen¬dulum C is made to oscillate, it will continue its oscillation with its natural frequency. As the thread PQ is tied loosely, energy will be transferred from pendulum C to pendulum D through the thread.

• As a result, the pendulum D will begin to oscillate. As the lengths of the pendulums are different, their natural frequencies are not the same. It is found that the pen¬dulum D initially tries to oscillate at its natural frequency.
• But its vibration is damped quickly. Then pendulum D begins to oscillate at the natural frequency of C. In this example, pendulum C provides the driving vibration and the vibration of pendulum D is the driven vibration.
• In these two examples, it is to be noted that the external driving forces are neither steady nor momentary. Rather, the force originating from the vibrating body is a periodic force. In fact, the spring of a clock exerts a periodic force on the pendulum.

From the above discussions forced vibration can be defined in the following way.

Forced Vibration Definition: If an external periodic force is applied to a freely vibrating body, the body tries to maintain its vibrations at its own frequency; but after some time, the body begins to vibrate with the frequency of the applied periodic force. Such a vibration of a body is called a forced vibration.

Equation Of motion: A particle under forced harmonic vibration in one dimension is subject to three types of forces:

A restoring force: F₁ = -kx, where k is constant.

Resistive forces: The resultant of all the resistive forces acting on a particle is, \(F_2=-k^{\prime} v=-k^{\prime} \frac{d x}{d t}\)

where v is the velocity of the particle and k’ is a constant.

Externally impressed periodic force: This is of the form, \(F_3=F_0 \sin \omega t\)

where ω is the circular frequency and F₀ is the amplitude of the periodic force; both ω and F₀ are constants.

The acceleration of the particle is, a = \(\frac{d v}{d t}=\frac{d^2 x}{d t^2}\). Then from the relation F= ma, we get, ma = F₁ + F₂ + F₃

or, \(m \frac{d^2 x}{d t^2}=-k x-k^{\prime} \frac{d x}{d t}+F_0 \sin \omega t\)

or, \(m \frac{d^2 x}{d t^2}+k^{\prime} \frac{d x}{d t}+k x=F_0 \sin \omega t\)

or \(\frac{d^2 x}{d t^2}+2 b \frac{d x}{d t}+\omega_0^2 x=a_0 \sin \omega t\) where, \(\omega_0= \pm \sqrt{\frac{k}{m}}, b=\frac{k^{\prime}}{2 m}\) and \(a_0=\frac{F_0}{m}\).

The equation (1) is known as the equation of motion of a forced SHM. It is to be noted that, the natural frequency ω₀ of the particle vibration is, in general, different from the frequency co of the external periodic force. In the special case, when cω = ω₀, the phenomenon of resonance comes into play.

Comparison Between Free Vibration And Forced Vibration

Nature Of Vibration – Resonance Or Resonant Vibration

Usually, when a body executes a forced vibration, its amplitude and velocity remain small. In most cases, the frequency of the applied periodic force does not come close to the natural frequency of the vibrating body.

• However, when the frequency of the applied periodic force becomes equal to the natural frequency of the body, the amplitude and the velocity of the body become very large. This phenomenon is called resonance.
• It is important to keep in mind that some objects are very resonant at a particular frequency while others barely resonate. For example, a temple bell will only ring at a fixed frequency, which is its resonant frequency but a lump of jelly will vibrate at many different frequencies but will not resonate at all.

Resonance Or Resonant Vibration Definition: When the frequency of the applied periodic force matches the natural frequency of a vibrating body, its amplitude rapidly increases to a large value. This phenomenon is called resonance.

This vibration is also known as sympathetic vibration. Resonant vibration is of two types.

Amplitude resonance: In this case, the amplitude of vibration becomes maximum. At amplitude resonance die frequency of applied force is slightly less than the natural frequency i.e., \(\omega=\sqrt{\omega_0^2-2 b^2}\)

Velocity resonance: In this case, the velocity of the vibrating body becomes maximum. Velocity resonance occurs when the natural frequency is exactly equal to the frequency of applied force i.e., ω = ω₀.

Examples of resonance:

1. A thread is tied loosely between P and Q, and four simple pendulums C, D, E, and F are suspended from the thread. Pendulums C and D have the same length.

• So they have the same natural frequency. But the lengths of pendulums E and F are different from those of C and D. So they have different natural frequencies. Now if pendulum C is set into vibration, it executes SHM.
• Hence, a periodic force acts on the pendulums D, E, and F through the thread PQ. After a while, it is observed that, though E and p are set in forced vibration, their amplitudes and velocities of vibration are not large: but pendulum D vibrates with a large amplitude. This is because pendulum D resonates with pendulum C.
• As soon as pendulum D is set into resonant vibration, it applies, in mm, periodic forces on pendulums C F., and F through die thread PQ. This does not affect the vibrations of E and F very much, but pendulum C experiences resonance. In this wave, energy is alternately transferred from C to D and D to C.
• As a result, it is observed diet. when the amplitude of the vibration of pendulum C becomes very large, pendulum D almost comes to a stop; in die next moment the amplitude of pendulum D starts to increase while that of pendulum C gradually decreases.

2. Resonant air column: The length of an air column contained in a tube determines die natural frequency of the column. If a vibrating tuning fork is held at the mouth of such a tube, forced vibration is set up in the air column.

• If the frequency of the tuning fork is the same as the natural frequency of the air column, then resonance occurs in the air column and a loud sound is heard. Using this phenomenon, sometimes a tuning fork is mounted on a hollow box.
• The box is so shaped that its frequency becomes equal to the natural frequency of the tuning fork. So, when the tuning fork is struck a resonance is produced and a loud sound is heard.

3. Hollow box in musical instruments: In musical instruments, such as violin, esraj, sitar, etc., the strings are stretched on a wooden hollow box or cavity The vibration of a string induces forced vibration on the wooden box and on a large mass of air inside it. This increases die intensity of die emitted sound. If resonance occurs die emitted sound is further intensified.

4. In musical instruments like violin, esraj, sitar, etc., there are a number of strings in addition to the principal string, which are adjusted or tuned to predefined notes of different frequencies.

When the principal string is played to produce a certain note, resonant vibrations may occur in some other string. This contributes to the increase in both the loudness and quality of the musical sound.

5. Helmholtz’s resonator: To detect the presence of tones of different frequencies in a note, the German scientist Helmholtz devised a resonator. It consists of a brass shell of nearly spherical shape with two openings a and b of different diameters.

• The natural frequency of die air inside the spherical shell depends on the size of the shell. The larger opening a, called hole or neck, is turned towards the source of sound while we place our ears in front of the smaller opening b, called pip.
• Now, suppose that in a note there is a tone whose frequency is equal to the natural frequency of the resonator. In that case, resonance is produced in the air inside the resonator and the tone will be heard distinctly. With the help of different resonators of known frequencies, we can detect different tones in a note.

Comparison between Forced Vibration and Resonance

 

Superposition Of Waves

Principle Of Superposition Of Waves

The simultaneous progress of more than one wave through a region of space produces the phenomenon known as the superposition of waves. During superposition, each of the waves travels independently, i.e., a wave retains its individual properties though it overlaps on other waves. For example, when different musical instruments are played at a time, the notes from each instrument can be detected separately.

If more than one waves are incident simultaneously on a particle in a medium, the particle would have different displacements for each of the separate waves. But these displacements of the particle occur at the same time, i.e., a resultant displacement of the particle occurs for all the waves.

Since displacement is a vector quantity, the resultant displacement is the vector sum of all the individual displacements. This is known as the principle of superposition of waves.

Superposition Of Two Single Waves Or Two Wave Pulses:

• For example, let two wave pulses P and Q of equal displacement and of the same sign (both upward) travel with equal speed along the length of a string but in opposite directions the two wave pulses at a later instant before superposition. The result is when the two pulses superpose each other at a time t.
• It is found from that at time t, a wave pulse R of displacement equal to the sum of the displacements of wave pulses P and Q is formed. Then, after that, the wave pulses P and Q retaining their original shapes move along the string as shown.

• Now, let the two wave pulses P and Q having equal displacement but of opposite signs are move along the length of a string with equal speed in opposite directions and show that after a short time, the two pulses get closer to each other.
• At the time t’, the two wave pulses superpose and produce a resultant wave pulse R of zero displacements. After that, the two wave pulses retain their original shapes and continue to move along the length of the string as shown.

Superposition Of Waves Statement: The resultant displacement of a particle In a medium due to more than one wave Is equal to the vector sum of the different displacements produced by the Individual waves separately.

Let n number of waves traveling in a medium superpose on each other. If \(\overrightarrow{y_1}, \overrightarrow{y_2}, \overrightarrow{y_3}, \cdots \cdots \vec{y}_n\) are the displacements at a point due to n waves, then the resultant displacement will be \(\vec{y}=\overrightarrow{y_1}+\overrightarrow{y_2}+\overrightarrow{y_3}+\cdots \overrightarrow{y_n}\)

If the displacements due to the two wave pulses are equal and in the same direction, i.e., \(\left|\overrightarrow{y_1}\right|=\left|\overrightarrow{y_2}\right|=A\), then from the superposition principle, the magnitude of the resultant displacement will be \(\left|\overrightarrow{y_1}\right|=A+A=2 A\), as shown.

If the displacements are equal but in opposite directions, then the magnitude of the resultant displacement will be \(|\vec{y}|=A+(-A)=A-A=0\), as shown.

• If different displacements are collinear, it is sufficient to take their algebraic sum, i.e., to determine the resultant displacement, two like vectors are added while two unlike vectors are subtracted. The principle of superposition is applicable to all types of waves, say, electromagnetic waves, sound waves, etc.
• The best example of the superposition of waves is the melody of musical instruments. Another classic example is the throwing of more than one stone in a lake. In fact, most of the sounds we produce while speaking is a superposition. In the case of light waves, this is also valid. Any light in nature we see is in general a superposition.

The Superposition Of Similar Waves Gives Rise To The Following Important Phenomena:

Stationary Waves: The superposition of two identical but oppositely directed progressive waves produces stationary waves.

Beats: Two progressive waves of the same amplitude and velocity, but of slightly different frequencies, produce beats on superposition.

Interference: Two identical progressive waves, on superposition with a constant phase difference, produce interference.

The first two of these three phenomena will be discussed in this chapter, with special emphasis on sound waves.

Superposition Waves Numerical Examples

Example 1. The displacement of a periodically vibrating particle is y = 4cos(\(\frac{1}{2}\)t) sin (1000t). Calculate the number of harmonic waves that are superposed.
Solution:

y = \(4 \cos ^2\left(\frac{1}{2} t\right) \sin (1000 t)\)

= \(2 \cdot 2 \cos ^2\left(\frac{1}{2} t\right) \cdot \sin (1000 t)\)

= 2(1 + cos t) sin (1000t)

= 2sin(1000t) + 2 sin (1000t) cost

= 2 sin(1000t) + sin(1000t+ t) + sin(1000t- t)

= 2 sin( 1000t) + 1 sin(1001t)+ 1 sin(999t)

= y₁+y₂+y₃

Here each of y₁, y₂, and y₃ is in the form of A sinωt. Thus, each of them represents a harmonic wave.

Hence, the number of superposed harmonic waves = 3.

Example 2. The displacements of a particle at the position x = 0 in a medium due to two different progressive waves are y₁ = sinπrt and y₂ = sin2πt, respectively. How many times would the particle come to rest in every second?
Solution:

According to the principle of superposition, the resultant displacement of the particle is

y = y₁ +y₂= sin4πt+sin2πt

= \(2 \sin \frac{4 \pi t+2 \pi t}{2} \cos \frac{4 \pi t-2 \pi t}{2}=2 \sin 3 \pi t \cdot \cos \pi t\)

The particle comes to rest (y = 0) when either sin3πt = 0 or cosπt = 0.

When sin3πt = 0, we have t = 0, \(\frac{1}{3}\)s, \(\frac{2}{3}\)s (t< 1s)

Again, when cosπt = 0, we have t = \(\frac{1}{2}\)s (t<1s)

∴ y = 0, when t = 0, \(\frac{1}{3}\)s, \(\frac{2}{3}\)s, \(\frac{1}{2}\)s

In every second, the particle comes to rest 4 times.

Superposition Waves Stationary Or Standing Waves

Stationary Or Standing Waves Definition: When two progressive waves of the same amplitude, frequency, and velocity traveling in opposite directions superpose in a region of space, the resultant wave is confined to that region and cannot progress through the medium. Such a type of wave is called a stationary wave or standing wave.

• A stationary wave keeps on repeating itself in a region and there is no transfer of energy along the medium in either direction. In can be explained as below.
• Let a thin metallic wire AB be stretched between two rigid supports. The wire is struck at any arbitrary point in a normal direction. Two separate waves are produced and they travel towards the two ends of the wire.
• After getting reflected from the end support, each wave starts moving towards the opposite end. As a result, they superpose in the region between A and B. The resultant wave is confined in the region AB and cannot travel like a progressive wave.

Stationary waves in a stretched string are produced in this way. These waves are the sources of notes emitted from stringed instruments like sitar, violin, etc. Stationary waves are also produced in the air columns in instruments like flute, organ, etc.

Nodes And antinodes: A stationary wave remains confined in a region and cannot progress through the medium. As a result, the waveform is such that in a few positions (like A, D, F, H, B), the particles in the medium remain stationary at all times, i.e., the wave amplitude at these positions is always zero.

These positions are called nodes. On the other hand, the particles in a few positions (like C, E, G, I) continue to vibrate with maximum amplitude—these positions are the antinodes.

Nodes And Antinodes Definition: In a stationary wave, the positions where the particles of the medium always remain at rest are called nodes, and the positions where the particles vibrate with the maximum amplitude are called antinodes.

For vibrations of a stretched string, nodes and antinodes are formed at points on the string; they are nodal points and antinodal points, respectively. Similarly, during vibrations of stretched membranes, in instruments like tabla, drum, etc., nodal lines and antinodal lines are formed, while nodal surfaces and antinodal surfaces are formed in vibrating air columns.

Loop: The region between two consecutive nodes in a stationary wave is called a loop. For example, each of AD, DF, FH, and HB is a loop. If we consider any single loop, it is evident that all the particles are either in the equilibrium position or above or below that position at any instant. As the vibrations are usually very fast, the whole loop is visible.

Now, we consider two adjacent loops. At any instant, if the particles in loop AD are below the equilibrium position along line AB, the particles in loop DF would be above line AB. So, the particles in adjacent loops of a stationary wave are in opposite phases, i.e., the phase difference is 180°.

Wavelength Of A Stationary Wave: The two antinodal points C and G are in the same phase because

1. When the displacement of the particle at C is maximum, the particle at G also goes to its maximum displaced position and
2. At every instant, the particles at C and G are on the same side relative to their equilibrium positions. Since C and G are consecutive points lying in the same phase, the distance between C and G is known as the wavelength (λ).

It is to be noted that there is another antinodal point E between C and G, but E is in the opposite phase with respect to C and G. It is observed that the distance between the nodal points A and F is also equal to CG.

Wavelength Of A Stationary Wave Definition: The distance between three consecutive nodes or three consecutive antinodes, is the wavelength (λ) of a stationary wave.

So, the length of a loop = distance between two consecutive nodes (say, AD) = \(\frac{\lambda}{2}\);

The distance between a node and the adjacent antinode (say, AC) = half of the length of a loop = \(\frac{\lambda}{4}\).

Resonant Frequency: Let us consider a string of a guitar, is stretched between its two ends. Suppose a continuous sinusoidal wave of a certain frequency is propagating along the string to the right. When the wave reaches the right end, it will reflect and begin to move towards the left.

• While moving towards the left end of the sting it must superpose with the wave that is still travelling towards the right. Similarly, after reaching the left end, the left-going wave reflects and begins to travel to the right, which results in a superposition with the left and right-going waves.
• In other words, within a very short time, we may find many overlapping traveling waves, interfering with each other.
• For certain frequencies, the interference produces a standing wave pattern accurately with nodes and antinodes. Such a standing wave is said to be produced at resonance and the string is said to be a resonator at this certain resonant frequency. If the string is oscillated at some frequency other than its resonant frequency, a standing wave is not formed.

Explanation Of The Formation Of Stationary Waves By Graphical Method: Two identical, but oppositely directed progressive waves superpose in a region of an elastic medium to produce stationary waves. The formation of these stationary waves can be explained graphically.

Consider a progressive wave (wave 1, denoted by a blue line) is moving towards right through a medium. Another progressive wave (wave 2, denoted by a broken blue line) of the same amplitude, frequency, and velocity is moving toward the left. The relations T = \(\frac{1}{n}\) and λ = \(\frac{V}{n}\) confirm that their time periods and wavelengths are also equal. When the two waves superpose in the region AE of the medium, the following cases can occur:

1. At the beginning of a period (time, t = 0), the two waves are in opposite phases. So, the resultant displacement is zero for every particle in the medium, i.e., every particle is in its equilibrium position. The graph of the resultant wave is the straight line AE.
2. During the time t = \(\frac{1}{4}\), the 1st wave covers a distance towards the right; the 2nd wave also travels the same distance towards the left. So, at this instant, the two waves are in the same phase.
   • The resultant displacements of the points A, C, and E become maximum, but the points B and D remain in equilibrium position. The graph of the resultant wave is shown as a red line.
3. By the time t = \(\frac{T}{2}\), both the waves have progressed further towards their direction of propagation through a distance \(\frac{\lambda}{4}\); so they are again in opposite phases. At this instant, every particle in the medium returns to its equilibrium position. So, the graph of the resultant wave becomes a straight line again.
4. At time t = \(\frac{3T}{4}\), the resultant displacement of every particle is equal but opposite to that at time t = \(\frac{T}{4}\). Here again, the points B and D remain in their equilibrium positions and the resultant displacements of the points A, C, and E become maximum. The graph of the resultant wave is again shown as a red line.
5. Finally, at time t = T, each of the progressive waves completes a period and returns to its position as that at time t = 0. As a result, each particle in this medium is in its equilibrium position again. The graph of the resultant wave again becomes a straight line.

The above discussion, on the superposition of two identical but oppositely directed progressive waves, leads to the following inferences:

1. The displacements of particles at B and D are zero at all times, i.e., they are always at rest. These points are called the nodal points. On the other hand, the particles at A, C, and E vibrate with maximum amplitudes on the two sides of the equilibrium and are known as the antinodal points.
2. The nodes and the antinodes do not travel through the medium. So the resultant wave is a stationary or standing wave.
3. The two ends B and D of the portion BD remain stationary and the intermediate points vibrate periodically across the position of equilibrium. So, a loop is formed in the portion BD of the stationary wave.

Mathematical Representation Of A Stationary Wave

Suppose each of two progressive waves has amplitude = A, frequency = n, velocity = V, time period = T = \(\frac{1}{n}\) and wavelength = \(\lambda=\frac{V}{n}\). If they approach each other from two opposite directions along the x-axis, their equations are

\(\left.\begin{array}{rl}
y_1 & =A \sin (\omega t-k x) \\
\text { and } \quad y_2 & =A \sin (\omega t+k x)
\end{array}\right\}\)…(1)

Where, ω = 2πn and \(k=\frac{2 \pi}{\lambda}\)

According to the principle of superposition of waves, the resultant displacement of a particle in the medium is

y = \(y_1+y_2\)

= \(A[\sin (\omega t-k x)+\sin (\omega t+k x)]\)

= \(2 A \sin \frac{(\omega t-k x)+(\omega t+k x)}{2} \cos \frac{(\omega t+k x)-(\omega t-k x)}{2}\)

= \(2 A \sin \omega t \cos k x=A^{\prime} \sin \omega t\)…(2)

Where, \(A^{\prime}=2 A \cos k x=2 A \cos \left(\frac{2 \pi}{\lambda} x\right)\)…(3)

Equation (2) represents a stationary wave. It cannot represent a progressive wave because a term like (ωt± kx) is not present in the argument of its trigonometric functions. The frequency of this motion is n = ω/2π, and the amplitude is \(A^{\prime}=2 A \cos \left(\frac{2 \pi}{\lambda} x\right)\), which varies harmonically and depends on the value of x.

Change Of Amplitude With Position: The amplitude will be zero at points, where

cos k x = \(\cos \left(\frac{2 \pi}{\lambda} x\right)=0=\cos \left(n+\frac{1}{2}\right) \pi\)

where n = 0, 1, 2 ……

∴ \(\frac{2 \pi x}{\lambda}=\left(n+\frac{1}{2}\right) \pi\)

or, x = \((2 n+1) \frac{\lambda}{4}\)

or, x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \cdots\)

These points where amplitude is zero, are called nodes. Clearly, the separation between two consecutive nodes is λ/2.

The amplitude will have a maximum value of 2A at points, where,

cos kx = \(\cos \left(\frac{2 \pi}{\lambda} x\right)= \pm 1=\cos n \pi\) [n = 0,1,2, •••]

These points of maximum amplitude are called antinodes. Clearly, the antinodes are also separated by λ/2 and are located halfway between pairs of nodes.

It is evident from equation (2) that every particle in the medium executes simple harmonic motion. The frequency of this motion is n = \(\frac{\omega}{2 \pi}\) and \(A^{\prime}=2 A \cos \left(\frac{2 \pi}{\lambda} x\right) .\)

The main features of the resultant displacement are, however, inherent in equation (3). Here, the quantity A’ represents the amplitudes of the particles at different positions (i.e., different values of x) of the medium. Evidently, the amplitude is different for different values of x.

For Example,

1. At the points x = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}, \cdots, \frac{n \lambda}{2}\), the amplitude A’ = ±2A = maximum.These are the antinodal points. The distance between two consecutive antinodes = \(\frac{\lambda}{2}\).
2. At the points x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \cdots,(2 n+1) \frac{\lambda}{4}\), the amplitude A’ = 0. These are the nodal points. Here again, the distance between two consecutive nodes = \(\frac{2 \lambda}{4}=\frac{\lambda}{2}\)

The values of x also show that an antinode is present between two consecutive nodes.

Characteristics Of Stationary Waves

1. Two identical but oppositely directed progressive waves superpose to form a stationary wave.
2. Along a stationary wave, the particles at different points vibrate with different amplitudes. The points at which the amplitudes of vibration are always zero are called nodes the points with the maximum amplitudes, of vibration are called antinodes.
3. The distance between two consecutive nodes or two consecutive antinode is \(\frac{\lambda}{2}\). (λ = wavelength of the stationary wave.)
4. Nodes and mummies are fixed points; they do not change their positions with time. So. stationary waves do not travel through the medium. They remain confined to a region.
5. All die particles in a loop, between two nodes, are displaced in the same direction relative to their equilibrium positions. So, these particles are in the same phase.
6. Particles in adjacent loops are displaced in opposite directions at any instant; so they are in opposite phases.
7. All die particles come to rest simultaneously twice in each period and they cross the equilibrium position simultaneously twice in each period.
8. At the instant when all the particles are at their equilibrium positions, die potential energy of the stationary wave becomes zero, but the kinetic energy becomes maximum. On die other hand, in the maximum displaced positions, the kinetic energy becomes zero, while the potential energy becomes maximum. The total energy of the stationary wave is always conserved.
9. The changes in density and pressure are maximum at nodal points but zero at antinodal points.

Difference Between Progressive And Stationary Waves:

Superposition Of Waves Stationary Or Standing Waves Numerical Examples

Example 1. A sound wave of frequency 80 Hz gets reflected normally from a large wall. Estimate the distance of the first node and the first antinode from the wall. Given, the velocity of sound in air = 320 m · s^-1.
Solution:

Here, the superposition of the incident and the reflected waves produces a stationary wave. The air particles adjacent to the wall cannot vibrate; so a node is developed at the wall.

Wavelength of the stationary wave, \(\lambda=\frac{V}{n}=\frac{320}{80}=4 \mathrm{~m}\)

∴ Distance of the 1st node from the wall = distance between two consecutive nodes = \(\frac{\lambda}{2}=\frac{4}{2}=2 \mathrm{~m}\)

The 1st antinode is at the midpoint between these two nodes; so its distance from the wall = 1 m.

Example 2. A wave represented by the equation y=A cos(kx-ωt) superposes on another wave to produce a stationary wave with a node at x = 0. What is the equation of this second wave?
Solution:

As a stationary wave is produced, the second wave will have the same amplitude, frequency, and velocity, but it will be oppositely directed. So its general equation will be, y’ = Acos(kx + ωt+ θ), where θ = initial phase

According to the principle of superposition, the equation of the resultant stationary wave is y_r = y+y’ = A [cos(kx-ωt) + cos(kx+ ωt+ θ)]

y_r = 0 at the point x = 0 due to the presence of a node

So, 0 = A[cos(-ωt) + cos(ωt+ θ)]

or, cosωt + cos (ωt+θ) = 0

or, cosωt = -cos(ωt+θ) = cos(ωt+θ +π)

Hence, ωt = ωt+ θ + Kπ or, θ = -π

So, the required equation is

y’ = A cos(kx + ωt-π) or, y’ = -A cos(kx + ωt).

Example 3. The equation of the vibration of a wire is y = \(5 \cos \frac{\pi x}{3} \sin 40 \pi t,\), where x and y are given in cm and t is given in s. Calculate the

1. Amplitudes and velocities of the two waves which on superposition, form the above-mentioned vibration
2. Distance between the two closest points of the wire that are always at rest;
3. The velocity of a particle at x = 1.5 cm at the instant t = \(\frac{9}{8}\)s.

Answer:

y = \(5 \cos \frac{\pi x}{3} \sin 40 \pi t=\frac{5}{2} \cdot 2 \sin 40 \pi t \cos \frac{\pi x}{3}\)

= \(\frac{5}{2}\left[\sin \left(40 \pi t+\frac{\pi x}{3}\right)+\sin \left(40 \pi t-\frac{\pi x}{3}\right)\right]\)

= \(\frac{5}{2} \sin 40 \pi\left(t+\frac{x}{120}\right)+\frac{5}{2} \sin 40 \pi\left(t-\frac{x}{120}\right)\)

= \(y_1+y_2\)

So, the resultant vibration is produced due to the superposition of two waves y₁ and y₂. Comparing these two waves with the general equation, y = \(A \sin \omega\left(t-\frac{x}{V}\right)\)

1. Amplitude, A₁ = A₂ = \(\frac{5}{2}\) = 2.5 cm; wave velocity, V₁ = 120 cm · s^-1 (towards negative x – direction) and V₂ = 120 cm · s^-1 (towards positive x-direction).

2. Wavelength, \(\lambda=\frac{V}{n}=\frac{V}{\omega / 2 \pi}=\frac{2 \pi V}{\omega}\)

The closest points that are always at rest denote two consecutive nodes. So, the distance between them = \(\frac{\lambda}{2}=\frac{\pi V}{\omega}=\frac{\pi \times 120}{40 \pi}=3 \mathrm{~cm} .\)

3. Particle velocity,

v = \(\frac{d y}{d t}=5 \cos \frac{\pi x}{3} \cdot 40 \pi \cos 40 \pi t\)

= \(200 \pi \cos \frac{\pi x}{3} \cos 40 \pi t\)

So, at x = \(1.5 \mathrm{~cm} and t=\frac{9}{8} \mathrm{~s}\),

v = \(200 \pi \cos \frac{\pi \times 1.5}{3} \cos \left(40 \pi \times \frac{9}{8}\right)\)

= \(200 \pi \cos \frac{\pi}{2} \cos 45 \pi=0 .\)

 

Superposition Of Waves Sonometer

A sonometer is an instrument designed to verify the laws of transverse vibration in a stretched string and to determine the emitted frequency.

Sonometer Description: A hollow box H is kept on a table. A uniform, thin metal wire whose one end is fixed with the rigid support R is stretched over two fixed bridges A and C, kept on the box H. This wire is passed through a small fixed pulley P and suspended beside the table.

The movable bridge B can be put anywhere between A and C. Different known masses can be suspended at the free end Q of the wire to produce different tensions. More than one wire can be set similarly parallel to one another. In general, wires of different materials and of different cross sections are used.

Sonometer Working Principle: A wire of the sonometer is forced to vibrate in the region between the two bridges A and B. As a result, two nodes are generated at the ends of A and B. So, the effective length of the vibrating wire is equal to the distance between A and B.

• The handle of a vibrating tuning fork of known frequency is touched with the hollow box H. So, the wire AB vibrates at that instant due to forced vibration. Now, the length of the wire is adjusted by moving the bridge B until a resonance is achieved.
• This means that the tones emitted from the tuning fork and that from the wire are in unison and a loud sound is heard. In this situation, the frequencies become equal and the frequency of the vibrating wire is obtained from the known frequency of the tuning fork.

Verification Of The Laws Of Transverse Vibration In A String:

Law Of Length: To verify this law, only one wire of the sonometer is used and the mass suspended at the free end is kept unaltered. As a result, the mass per unit length (m) and the tension (T) of the wire remain constant. Now, a tuning fork of a known frequency (n₁) is vibrated and the bridge B of the sonometer is adjusted until the wire AB resonates. Let the length of the wire in this case be l₁. The wire is then similarly brought to resonance by using a few other tuning forks of different frequencies n₂, n₃, •••, etc. If the corresponding lengths of the wire AB at resonance are \(l_2, l_3, \cdots, \text { etc. }\) respectively, it is observed that \(n_1 l_1=n_2 l_2=n_3 l_3=\cdots\)

i.e., \(n \propto \frac{1}{l}\), when T and m are constants.

Law Of Tension: In this case, only one wire of the sonometer is used, so that the mass per unit length (m) of the wire remains constant. Moreover, if bridge B is kept at a fixed position, the length (l) of the wire does not change.

Now the wire AB is brought to resonance with different tuning forks of known frequencies n₁, n₂, n₃,….., respectively. These resonances are generated by increasing or decreasing the mass M suspended at the free end of the wire. Let the values of M at the resonances be M₁, M₂, ……, respectively. So, the corresponding values of the tension in the wire are T₁ = M₁g, T₂ = M₂g,…… Here, it is observed that,

∴ \(\frac{T_1}{T_2}=\frac{n_1^2}{n_2^2} \text { or, } \frac{n_1}{n_2}=\sqrt{\frac{T_1}{T_2}}\)

i.e., n ∝ √T, when l and m are constants.

Law Of Mass: A few wires made of different materials and having different cross sections are taken. These are known as experimental wires. The mass per unit length of the wires are different; let the values be m₁, m₂, m₃, …….., respectively Initially the 1st wire is used in the sonometer.

• Bridge B is kept at a definite position and a particular mass is suspended from its free end. The position of B and the value of the mass are not changed throughout the experiment. So the length l and the tension T remain constant.
• Now, another wire called the reference wire is set at a parallel position, and a fixed mass is hung from its free end. So the tension in the reference wire also remains fixed. Then the 1st experimental wire and the reference wire are vibrated simultaneously and resonance is obtained by adjusting the position of the movable bridge (say B’) of the reference wire.
• Let n₁ = frequency of vibration of both the wires and l₁ = length of the reference wire at resonance. The experiment is repeated by replacing the 1st experimental wire with the 2nd, 3rd,… and so on. Now, from the law of length, it is evident for the reference wire that, n₁l₁ = n₂l₂ = n₃l₃ =  …….

If the experimentally obtained values are analyzed, it is observed that \(\frac{m_1}{l_1^2}=\frac{m_2}{l_2^2}=\frac{m_3}{l_3^2}=\cdots\)

So, \(\frac{l_1^2}{l_2}=\frac{m_1}{m_2} or, \frac{l_1}{l_2}=\sqrt{\frac{m_1}{m_2}} or, \frac{n_2}{n_1}=\sqrt{\frac{m_1}{m_2}}\)

i.e., \(n_1 \sqrt{m_1}=n_2 \sqrt{m_2}=n_3 \sqrt{m_3}=\cdots\)

So, \(n \propto \frac{1}{\sqrt{m}}\), when l and T are constants.

Here, it is to be noted that the values m₁, m₂, …… correspond to the experimental wires, and l₁, l₂, ….. correspond to the reference wire. In each case, the length of the experimental wires l = constant.

However, the frequencies n₁, n₂,……. are the frequencies at resonance of the experimental wire as well as of the reference wire.

Determination of the frequency of a tuning fork by a Sonometer: The arm of a tuning fork vibrating with an unknown frequency is loosely held in contact with the hollow box of the Sonometer.

• Now the position of bridge B under the sonometer wire is adjusted until resonance is obtained. This resonance denotes that the tuning fork and the sonometer wire are vibrating with the same frequency, i.e., they are in unison.
• For the detection of the resonance, a V-shaped thin piece of paper is kept inverted at the center of the wire AB. At resonance, the amplitude of vibration at the central antinode is very large.

As a result, the wire throws away the piece of paper. If the wire vibrates in a single loop, it emits the fundamental tone. Then its frequency is,

n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{2 l} \sqrt{\frac{M g}{m}}\)

Here, l = length of the portion of the wire AB at resonance, M = mass suspended at the free end, and m = mass per unit length of the wire.

The values of l, M, and m are measured by usual methods. By using these values of l, M, and m, the value of the frequency (n) can be determined. The calculated value of n is equal to the frequency of the tuning fork.

Superposition Of Waves Sonometer Numerical Examples

Example 1. A sonometer wire emits a tone of frequency 150 Hz. Find out the frequency of the fundamental tone emitted by the wire if the tension is increased in the ratio 9:16 and the length is doubled.
Solution:

The mass per unit length m remains the same for a single wire.

So from the relation n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}\), we get

∴ \(\frac{n_1}{n_2}=\frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}\)

or, \(n_2=n_1 \cdot \frac{l_1}{l_2} \sqrt{\frac{T_2}{T_1}}=150 \times \frac{1}{2} \times \sqrt{\frac{16}{9}}=100 \mathrm{~Hz}\).

Example 2. The fundamental frequency of a 100 cm long sonometer wire is 330 Hz. Find out the velocity of the transverse wave in the wire and the wavelength of the resulting sound waves in air. Given, the velocity of sound in air = 330 m · s^-1.
Solution:

For fundamental tone in a wire, two nodes are produced at the two ends, So, the length of the wire = distance between two consecutive nodes = \(\frac{\lambda}{2}\)

Here, λ = length of the transverse wave,

According to the question, \(\frac{\lambda}{2}=100 \mathrm{~cm}=1 \mathrm{~m}\)

So, λ = 2

∴ Velocity of transverse wave in the wire, V = frequency x wavelength = 330 x 2 = 660 m · s^-1

The frequency of the resulting sound waves in air will also be 330 Hz.

The velocity of sound waves in air = 330 m · s^-1.

∴ The wavelength of the resulting sound waves in air

= \(\frac{\text { velocity }}{\text { frequency }}=\frac{330}{330}=1 \mathrm{~m}\).

Example 3. A wire kept between two bridges 25 cm apart, Is stretched through a linear expansion of 0.04 cm. Find the fundamental frequency of vibration of the wire. Given, the density and Young’s modulus of the material of the wire are 10 g · cm^-3 and 9 x 10¹¹ dyn · cm^-2, respectively.
Solution:

Longitudinal strain = \(\frac{\text { linear expansion }}{\text { original length }}=\frac{x}{l}\)

Longitudinal stress = \(\frac{\text { tension in the wire }}{\text { area of cross-section }}=\frac{T}{\alpha}\)

Y = \(\frac{\text { longitudinal stress }}{\text { longitudinal strain }}=\frac{\frac{T}{a}}{\frac{x}{l}}=\frac{T l}{\alpha x}\)

or, T = \(\frac{Y \alpha x}{l}\)

Mass per unit length of the wire, m= density x area of cross-section = \(\rho \alpha\)

∴ Fundamental frequency,

n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{2 l} \sqrt{\frac{Y \alpha x}{l} \cdot \frac{1}{\rho \alpha}}=\frac{1}{2 l} \sqrt{\frac{Y x}{l \rho}}\)

= \(\frac{1}{2 \times 25} \times \sqrt{\frac{\left(9 \times 10^{11}\right) \times 0.04}{25 \times 10}}=240 \mathrm{~Hz}\).

Example 4. The linear density of mass of a metal wire is 9.8 g · m^-1. It is kept between two rigid supports, 1 m apart, with a tension of 98 N. The midpoint of the wire is kept between the poles of an electromagnet driven by an alternating current. Find out the frequency of this alternating current that will produce resonance in the wire.
Solution:

Mass per unit length of the wire,

m = 9.8 g · m^-1 = 9.8 x 10^-3 kg · m^-1; length of the wire, l = 1m; tension in the wire, T = 98 N

∴ Fundamental frequency,

n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{2 \times 1} \sqrt{\frac{98}{9.8 \times 10^{-3}}}=50 \mathrm{~Hz}\)

For this reason, the frequency of the alternating current is also 50Hz.

Example 5. A 1m long wire Is clamped at both ends. Find out the positions of two bridges that will divide the wire in three parts such that the fundamental frequencies are in the ratio 1:2:3.
Solution:

Here, n₁: n₂: n₃ = 1:2:3

As \(n \propto \frac{1}{l}\), we have \(l_1: l_2: l_3=1: \frac{1}{2}: \frac{1}{3}\)

So, \(l_2=\frac{l_1}{2}\) and \(l_3=\frac{l_1}{3}\)

Now, \(l_1+l_2+l_3=1\) ; so, \(l_1+\frac{l_1}{2}+\frac{l_1}{3}=1\)

or, \(\frac{11}{6} l_1=1$ or, $l_1=\frac{6}{11} \mathrm{~m}\)

Then, \(l_2=\frac{l_1}{2}=\frac{3}{11} \mathrm{~m} and l_3=\frac{l_1}{3}=\frac{2}{11} \mathrm{~m}\)

So, the first bridge is at a distance of \(\frac{6}{11}\)m from one end and the second bridge is at a distance of \(\frac{2}{11}\)m from the other end.

Example 6. The linear, density of a wire is 0.05 g · cm^-1. The wire is stretched with a tension of 4.5 x 10⁷ dyn between two rigid supports. A driving frequency of 420 Hz resonates with the wire. At the next higher frequency of 490 Hz, another resonance is observed. Find out the length of the wire.
Solution:

Let 420 Hz = frequency of the p -th harmonic.

So, 490 Hz = frequency of the (p + 1) – th harmonic.

Then, 420 = \(\frac{p}{2 l} \sqrt{\frac{T}{m}}\) and \(490=\frac{p+1}{2 l} \sqrt{\frac{T}{m}}\)

∴ \(\frac{420}{490}=\frac{p}{p+1}\) or, p=6

So, we get, 420 = \(\frac{6}{2 l} \sqrt{\frac{T}{m}}\)

or, \(l=\frac{6}{2 \times 420} \sqrt{\frac{4.5 \times 10^7}{0.05}}=214.3 \mathrm{~cm}\)

Example 7. One end of a wire of radius is sealed with the end of another wire of radius 2r. This is used as a sonometer wire with tension T, keeping the junction at the mid-point of the vibrating length. If a stationary wave having a node at the junction is generated, find out the ratio between the number of loops in the two portions.
Solution:

Let ρ₁ and ρ₂ be the densities of the materials of the two wires respectively.

Mass per unit length of the first wire, m₁ = πr²ρ₁

Mass per unit length of the second wire, \(m_2=\pi(2 r)^2 \rho_2=4 \pi r^2 \rho_2\)

∴ \(\frac{m_1}{m_2}\) =\(\frac{\rho_1}{4 \rho_2}\)

The junction is at the mid-point of the vibrating length 2l (say).

So, the vibrating length of each wire = l.

The two wires are vibrating simultaneously; so the frequency of vibration is the same. Let it be n.

Again, a node is generated at the junction; so an integral number of loops is produced in each wire. If the number of loops in the two wires are p and q, respectively, then

n = \(\frac{p}{2 l} l \sqrt{\frac{T}{m_1}} \text { and } n=\frac{q}{2 l} \sqrt{\frac{T}{m_2}}\)

∴ \(\frac{p}{2 l} \sqrt{\frac{T}{m_1}}=\frac{q}{2 l} \sqrt{\frac{T}{m_2}}\)

or, \(\frac{p}{q}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{\rho_1}{4 \rho_2}}=\frac{1}{2} \sqrt{\frac{\rho_1}{\rho_2}}\)

If the two wires are of the same material, then \(\rho_1=\rho_2\)

So, \(\frac{p}{q}=\frac{1}{2}\).

Superposition Of Waves Longitudinal Vibration In A String

If a string is vibrated by stretching it along its length, longitudinal waves are generated. After reflection from the end-supports, incident waves, and reflected waves superpose. So, longitudinal stationary waves are produced.

The velocity of these waves is equal to the velocity of longitudinal sound waves propagating through a string and is given by, \(V_l=\sqrt{\frac{Y}{\rho}}\)

where Y and ρ are Young’s modulus and the density of the material of the string, respectively.

Now let l = length of the string, α = area of cross-section, T = tension in the string, and x = linear expansion due to tension.

So, mass per unit length of the wire, m = pa; longitudinal strain = \(\frac{x}{l}\); longitudinal stress = \(\frac{T}{\alpha}\)

∴ Y = \(\frac{\frac{T}{\alpha}}{\frac{x}{l}}=\frac{T l}{\alpha x} \text { and } \rho=\frac{m}{\alpha}\)

∴ \(V_l=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{T l}{\alpha x} \cdot \frac{\alpha}{m}}=\sqrt{\frac{T l}{m x}}\)

We know that the velocity of transverse waves in a stretched string is \(
V_t=\sqrt{\frac{T}{m}}\)

∴ \(\frac{V_t}{V_l}=\sqrt{\frac{x}{l}}\)

For elastic metal wires, the linear expansions is much less than the original length, i.e., x<<1, so V_t <<V_l.

So, the longitudinal wave velocity in a stretched string is far greater than the transverse wave velocity.

For example, we consider a stretched string made of steel. The velocity of sound waves in steel is about 5000 m · s^-1.

This is the velocity of longitudinal waves in a steel string. If the longitudinal stationary waves in such a stretched steel string produce a fundamental tone of frequency 250 Hz, then the wavelength, \(\lambda=\frac{5000}{250}=20 \mathrm{~m}\).

So, the length of the wire, l = \(\frac{\lambda}{2}=10 \mathrm{~m}\). Clearly, it is practically impossible to construct any musical instrument with such a long string. For this reason, the longitudinal vibrations in a stretched string have no practical importance.

Superposition Of Waves Velocity Of Sound Numerical Examples

Example 1. A tuning fork of frequency 384 Hz produces the 1st and the 2nd resonances with air columns of a pipe closed at one end at lengths 22 cm and 67 cm, respectively. Find out the velocity of sound in air, and the end error for the open end of the tube.
Solution:

Velocity of sound in air, V = \(2 n\left(l_2-l_1\right)=2 \times 384 \times(67-22)=2 \times 384 \times 45\)

= 34560 \mathrm{~cm} \cdot \mathrm{s}^{-1}[/latex].

End error, c = \(\frac{1}{2}\left(l_2-3 l_1\right)=\frac{1}{2}(67-3 \times 22)\)

= \(\frac{1}{2} \times 1=0.5 \mathrm{~cm}\) .

Example 2. A 200 cm long vertical tube is filled with water, and a vibrating tuning fork of frequency 256 Hz is held over the open upper end of the tube. Then water is allowed to escape gradually through the lower end. Find out the positions of the water surface at the 1st and 2nd resonances. Neglect the end error. The velocity of sound in air = 320 m s^-1.
Solution:

The 1st resonance corresponds to the fundamental tone for the tube closed at one end. If l₁ is the length of the air column at the 1st resonance, then n = \(\frac{V}{4 l_1}\)

or, \(l_1 =\frac{V}{4 n}=\frac{320 \times 100}{4 \times 256}=31.25 \mathrm{~cm}\)

The 2nd resonance corresponds to the 1st overtone, which is the 3rd harmonic. If l₂ is the corresponding length of the air column then,

n = \(3 \cdot \frac{V}{4 I_2}\)

or, \(l_2=3 \cdot \frac{V}{4 n}=3 l_1=3 \times 31.25=93.75 \mathrm{~cm}\)

So, the height of the water surface from the bottom of the tube is respectively, (200-31.25) or 168.75 cm and (200- 93.75) or 106.25 cm.

Superposition Of Waves Conclusion

The resultant displacement of a particle in a medium due to more than one wave is equal to the vector sum of the different displacements produced by the individual waves separately. This is the principle of superposition of waves.

• When two progressive waves of the same amplitude, frequency, and velocity, traveling in opposite directions, superpose in a region of space, the resultant wave is confined to that region, and cannot progress through the medium. Such a type of wave is called a stationary wave or standing wave.
• In a stationary wave, the positions where the particles of the medium always remain at rest are called nodes and the positions where the particles vibrate with maximum amplitude, are called antinodes.
• The distance between two successive nodes or two successive antinodes is equal to half the wavelength of a stationary wave.
• The vibrating region between two successive nodes is a loop of a stationary wave. All the particles in a loop lie in the same phase and those in adjacent loops belong to opposite phases.

Laws of transverse vibrations in a stretched string: Let Z be the length of a stretched string, m be the mass per unit length of the string and T be the tension in the string. If the frequency of transverse vibrations in the string is n, then the string is n, then

1. \(n \propto \frac{1}{l}\), when T and m are constants.
2. \(n \propto \sqrt{T}\), when l  and m are constants.
3. \(n \propto \frac{1}{\sqrt{m}}\), when l  and T are constant.

The fundamental tone is emitted when a stretched string vibrates in a single loop. All the odd and even harmonics may also be emitted from the string depending on the number of loops formed during vibrations.

• A sonometer is a suitable instrument to study the vibrations in a stretched string.
• The longitudinal wave velocity in a stretched string is many times higher than the transverse wave velocity. The longitudinal waves in a string have no practical importance.
• The stationary waves in a stretched string are transverse stationary waves, whereas those in an air column are longitudinal stationary waves.
• A closed pipe (a pipe dosed at one end and open at the other) can emit its fundamental tone and only the odd harmonics. But an open pipe (a pipe with both ends open) can emit its fundamental tone and both odd and even harmonics. So an open pipe emits a musical sound of higher quality.
• The fundamental frequency of an open pipe is twice that of a closed pipe of equal length.
• The periodic rise and fall of the loudness of the resultant produced fay the superposition of two 4- If two progressive waves approach each other from two progressive sound waves of equal amplitude but of slightly different frequencies, is called beats.

The difference between the frequencies of rise two component waves of a beat is called the beat frequency. Beats are distinctly audible when the two superposing waves have a frequency difference of 10 Hz or less.

Superposition Of Waves Useful Relations For Solving Numerical Problems

If two progressive waves approach each other from two opposite directions along the x-axis, their equations are \(y_1=A \sin (\omega t-k x) \text { and } y_2=A \sin (\omega t+k x)\)

where for the two progressive waves, amplitude = A, frequency = n, velocity = V, time period = T = \(\frac{1}{n}\) and wavelength = \(\lambda=\frac{V}{n}\)

They superpose to form a stationary wave: y = y₁ + y₂ = 2A coskx sinωt = A’sinωt,

where \(A^{\prime}=2 A \cos k x=2 A \cos \left(\frac{2 \pi}{\lambda} x\right)\)

1. At x  = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}, \cdots[latex], the amplitude is  A’ = ±2A = maximum. These antinodes of the stationary’ wave.
2. At x = [latex]\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \cdots\), the amplitude is A’ = 0 = minimum. These points are nodes.

For the transverse vibrations in a sealed string of length l, mass per unit length m, and tension T, the frequency is, \(n_p=\frac{p}{2 l} \sqrt{\frac{T}{m}}\) where p = a number of loops.

Let the length of a pipe be l and the velocity of sound be V. For dosed and open pipes, the wavelength and the frequency of longitudinal stationary sound waves are given in the following table:

The end error occurring at each open end of a dosed or an open pipe is c ≈ 0.6r, where r = radius of the pipe.

Then, the effective length of a dosed pipe of length l is l+ c; the effective length of an open pipe of length l is l+2c.

So, the fundamental frequencies are,

For a closed pipe, \(n_0=\frac{V}{4(l+c)}\)

For an open pipe, \(n_0=\frac{V}{2(l+2 c)}\)

In case of beat production, two sound weaves have the amplitude A and the same initial phase but have slightly different frequencies n₁ and n₂, (where n₁ > n₂). Their equations are \(y_1=A \sin 2 \pi n_1 t \text { and } y_2=A \sin 2 \pi n_2 t \text {. }\)

The equation of the resultant wave formed on superposition is, \(y=y_1+y_2=A^{\prime} \sin 2 \pi n t\)

where n = \(\frac{n_1+n_2}{2}\)

and A’ = \(2 A \cos \left\{\pi\left(n_1-n_2\right) t\right\}\)

Beat frequency = number of beats heard per second = n₁-n₂ = the difference between the frequencies of the two-component sound waves

Superposition Of Waves Very Short Answer Type Questions

Question 1. What type of wave is formed when two identical but oppositely directed progressive waves superpose?
Answer: Stationary wave

Question 2. Name the type of wave which does not transmit energy from one place to another.
Answer: Stationary wave

Question 3. If λ is the wavelength of a stationary wave, what would be the distance between two consecutive nodes?
Answer: \(\frac{\lambda}{2}\)

Question 4. If λ is the wavelength of a stationary wave, what would be the distance between a node and the adjacent antinode?
Answer: \(\frac{\lambda}{4}\)

Question 5. How many times in each period all the particles in the medium for a stationary wave, come to rest simultaneously?
Answer: Twice

Question 6. The phase is always the same for all particles between two consecutive nodes of a stationary wave. Is the statement true or false?
Answer: True

Question 7. At the instant when all the particles are at their equilibrium positions, the potential energy of a stationary wave becomes maximum. Is the statement true or false?
Answer: False

Question 8. The fundamental frequency of transverse vibration in a taut string is 200 Hz. What will be the fundamental frequency if the length of the string is doubled, with its tension unchanged?
Answer: 100 Hz

Question 9. The fundamental frequency of transverse vibration in a taut string is 200 Hz. Keeping the length of the string unaltered, if its tension is doubled, what will be the fundamental frequency?
Answer: 200√2 Hz

Question 10. How does the frequency of the fundamental tone of transverse vibration in a stretched string change when a comparatively thicker string of the same material is used?
Answer: Decreases

Question 11. At the ends of an organ pipe open at both ends, what do we always get nodes or antinodes?
Answer: Antinode

Question 12. If the fundamental frequency emitted by a pipe closed at one end is 200 Hz, what is the frequency of the first overtone?
Answer: 600 Hz

Question 13. If the fundamental frequency emitted by a pipe open at both ends is 200 Hz, what is the frequency of the first overtone?
Answer: 400 Hz

Question 14. If the fundamental frequency of a closed pipe is 200 Hz, what would be the fundamental frequency of an open pipe of equal length?
Answer: 400 Hz

Question 15. Which harmonics are present in the note produced from a pipe closed at one end?
Answer: Fundamental tone and its odd harmonics

Question 16. What will be the beat frequency when two tuning forks of frequencies 256 Hz and 260 Hz are vibrated simultaneously?
Answer: 4

Question 17. As in sound, can beats be observed by two light sources?
Answer: No

Question 18. The superposition of two progressive sound waves of equal speed and amplitude but of slightly different frequencies produces ______
Answer: Beats

Question 19. Beats are heard when two tuning forks of frequencies 256 Hz and 260 Hz are vibrated simultaneously. If some wax is dropped at one of the prongs of the first fork, how will the beat frequency change?
Answer: Increase

Superposition Of Waves Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 Is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 Is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: When two vibrating tuning forks have f₁ = 300 Hz and f₂ = 350 Hz and are held close to each other, beats cannot be heard.

Statement 2: The principle of superposition is valid only when f₁ – f₂ < 10 Hz.

Answer: 3. Statement 1 is true, and statement 2 is false.

Question 2.

Statement 1: When a wave goes from one medium to another, the average power transmitted by the wave may change.

Statement 2: Due to changes in medium, amplitude, speed, wavelength, and frequency of wave may change.

Answer: 3. Statement 1 Is true, and statement 2 is false.

Question 3.

Statement 1: For a closed pipe, the 1st resonance length is 60 cm. The 2nd resonance position will be obtained at 120 cm.

Statement 2: In a closed pipe, n₂ = 3n₁, where n₁ = frequency of the fundamental tone and n₂ = frequency of the 1st overtone.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: If two waves of some amplitude, produce a resultant wave of the same amplitude, then the phase difference between them will be 120°.

Statement 2: The velocity of sound is directly proportional to the square of its absolute temperature.

Answer: 3. Statement 1 Is true, and statement 2 is false.

Superposition Of Waves Match Column 1 With Column 2

Question 1. Three successive resonance frequencies in an organ pipe are 1310, 1834, and 2358 Hz. The velocity of sound in air is 340 m · s^-1.

Answer: 1. C, 2. A, 3. D, 4. B

Question 2. A string fixed at both ends is vibrating in resonance. In Column 1 some statements are given which can match with one or more entries in Column 2.

Answer: 1. A, B, D, 2. A, B, D 3. C, 4. A, B, C, D

Superposition Of Waves Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A closed-air column 32 cm long is in resonance with a tuning fork. Another open-air column of length 66 cm is in resonance with another tuning fork. The two forks produce 8 beats per second when sounded together.

1. The speed of sound in air

1. 33792 cm · s^-1
2. 35790 cm · s^-1
3. 31890 cm · s^-1
4. 40980 cm · s^-1

Answer: 1. 33792 cm · s^-1

2. The frequencies of the forks

1. 230 Hz, 290 Hz
2. 250 Hz, 300 Hz
3. 264 Hz, 256 Hz
4. 150 Hz, 300 Hz

Answer: 3. 264 Hz, 256 Hz

Question 2. Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below 1250 Hz. The length of the pipe is 85 cm. The velocity of sound is 340 m · s^-1. Consider the following two cases:

1. The pipe is closed from one end

1. 2
2. 4
3. 8
4. 6

Answer: 4. 6

2. The pipe is opened from both ends

1. 3
2. 7
3. 6
4. 9

Answer: 3. 6

Question 3. In the arrangement shown a mass can be hung from a string with a linear mass density of 2 x 10^-3 kg ·m^-1 that passes over a light pulley. The string is connected to a vibrator of frequency 700 Hz and the length of the string between the vibrator and the pulley is 1 m.

1. If the standing waves are to be observed, the largest mass that can be hung is

1. 16 kg
2. 24 kg
3. 32 kg
4. 400 kg

Answer: 4. 400 kg

2. If the mass suspended is 16 kg, then the number of loops formed in the string is

1. 1
2. 3
3. 5
4. 8

Answer: 3. 5

3. The string is set into vibrations and represented by the equation y = \(6 \sin \left(\frac{\pi x}{10}\right) \cos \left(14 \times 10^3 \pi t\right)\), where x and y are in cm and t is in s. The maximum displacement at x = 5 m from the vibrator is

1. 6 cm
2. 3 cm
3. 5 cm
4. 2 cm

Answer: 1. 6 cm

Superposition Of Waves Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. A closed organ pipe and an open organ pipe of the same length produce 2 beats when they are set into vibration simultaneously in their fundamental mode. The length of the open organ pipe is now halved and that of the closed organ pipe is doubled. What will be the number of beats produced?
Answer: 7

Question 2. The displacement y of a particle executing periodic motion is given by \(y =4 \cos ^2\left(\frac{1}{2} t\right) \sin (1000 t)\) This expression may be considered as a result of the superposition of how many simple harmonic motions?
Answer: 3

3. A sound wave starting from source S, follows two paths AOB and ACB to reach the detector D. ABC is an equilateral triangle of side l and there is silence at point D. If the maximum wavelength is nl, find the value of n.
Answer: 8

Question 4. A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. The amplitude at the center of the string is 4 mm. Find the distance between the two points (in m) having an amplitude 2 mm.
Answer: 1

Wave Motion Introduction

When a small piece of stone is thrown into a pond, the water surface gets disturbed. The ripples formed due to this disturbance do not remain confined to the region where the stone hits the water’s surface, but it spreads in all directions along the surface.

• As the disturbance passes, the water particles start oscillating, i.e., moving up and down about their mean positions but are not displaced along the water surface. The pattern that moves along the pond due to such movement of individual particles of the medium, is called a wave.
• This wave motion transfers energy from one point to another, but no mass transport is associated with this transfer.

Wave: A wave is a disturbance that travels through a medium, whereas the particles constituting the medium do not travel.

• We are familiar with different types of waves; like sound waves, light waves, radio waves, etc. Among these, sound wave is a mechanical wave and the other two are electromagnetic waves.
• In this chapter, we are mainly concerned with mechanical waves. Mechanical waves can propagate only through a material medium, whereas electromagnetic waves do not require any material medium for propagation, i.e., they can propagate through a vacuum.

Wave Motion Mechanical Waves

The origin and propagation of mechanical waves depend on three properties of materials

1. Elasticity
2. Interia and
3. Cohesion
4. Elasticity: If any part of a material or a medium is displaced from its equilibrium position, stress is developed in it due to its elasticity. This stress tries to bring that part back to its equilibrium position.
5. Inertia: When a particle returns to its equilibrium position, it has a motion due to inertia. So the particle cannot come to rest immediately on reaching the equilibrium position. Due to the inertia of motion, it moves to the opposite side crossing the equilibrium position.
   • These two incidents— coming back to the equilibrium position due to elasticity and moving to the opposite side due to inertia of motion—happen alternately in any part of a material or a medium, causing that part to vibrate.
6. Cohesion: The adjacent molecules of a material medium attract each other. This phenomenon is called cohesion. If any part of a material medium begins to vibrate, the adjacent part is then forced to vibrate due to cohesion.
   • In this way, vibration propagates from one layer to the next. This type of propagating vibration is known as a mechanical wave.

Mechanical Waves Definition: The disturbance which travels through a material medium, due to collective vibration of the particles of the medium, is known as a mechanical wave.

Characteristics Of A Mechanical Wave:

1. For the propagation of a mechanical wave, a material medium is necessary. This wave cannot propagate through a vacuum.
2. Each particle of the medium forces the adjacent particles to vibrate, but the particle itself is not carried away from its equilibrium position; rather, the wave advances through the medium.
3. The vibrating particles of any material medium have their own mechanical energies and potential energy. This mechanical energy is forced vibration, i.e., a wave in motion transfers mechanical energy through the medium. So, this type not wave is called a mechanical wave.

Wave Motion Transverse And Longitudinal Waves

Simple Harmonic Wave: If the motion of the particles of a medium is sample harmonic, then the corresponding wave is known as a simple harmonic wave.

There Are Two Types Of Simple Harmonic Waves

1. Transverse wave and
2. Longitudinal waves

Transverse Wave Definition: A wave that propagates in a direction, perpendicular to the direction of motion of the vibrating particles of the medium, is called a transverse wave.

Examples Of Transverse Waves:

1. If a small stone is thrown into a pond, it generates waves that spread out horizontally in all directions. Now, if a small piece of cork is floated on the water surface, it oscillates vertically about its mean position. So the wave motion is directed along the horizontal direction, but the direction of motion of the cork and the vibrating water panicles are vertical. This is an example of a transverse wave.
2. A rubber rope AB is taken. The end B is fixed to a rigid support and the other end A is held in such a way that the rope is in a stretched condition. Now the end A is oscillated vertically A wave is found to be generated along the rope and it propagates towards the end B.

It is evident that the direction of vibration of each particle of the rope is perpendicular to the direction of propagation of the wave through the rope.

• The particle whose position at some instant is P occupies the position Q at another instant. Since PQ and AB are perpendicular to each other, the wave is obviously a transverse wave.
• A transverse wave can be described generally through. The straight line OBDF is the equilibrium position of the rope. Suppose at an instant, the particle at O is passing through its mean position in its course of vibration from an upward to a downward direction.
• The directions of motion of the different particles of the medium at that instant have been shown by arrows. A and E are two points having a maximum displacement in the positive direction.
• These are called crests. Again C is a point having maximum displacement in the negative direction. This is called a trough.

It is evident that at the said instant, points A and E are situated on the same side of and at the same distance from the mean position. Their velocities are also the same. In short, the conditions of motion at points A and E are the same.

• So, these two points are in the same phase. Similarly, the points O and D, or B and F are in the same phase.
• On the other hand, the conditions of motion at points A and C are opposite. So, these two points are in opposite phases.
• Electromagnetic waves like light waves, radio waves, etc., are transverse waves. The two quantities responsible for the production of electromagnetic waves are the electric field vector and magnetic field vector, which are always perpendicular to the direction of propagation of the wave.
• Electromagnetic waves can propagate through solid, liquid, and gaseous media and even through a vacuum.

Longitudinal Wave Definition: A wave which propagates of motion of the vibrating particles of t a longitudinal wave.

Example Of Longitudinal Wave: A thin and long spring AB is taken whose spring constant is very small. The end B is fixed to a rigid support and the other end A is held in such a way that the spring is in stretched condition.

• Now the end A is made to oscillate back and forth so that a wave moves along its length to end B.
• Some coils of the spring come close to each other creating compression and some other coils move away from each other creating rarefaction.
• Compressions and rarefactions are formed alternately along the length of the spring and they propagate towards the end B.
• Since the wave motion is directed from A to B, and during compressions and rarefactions the coils of the spring oscillate parallel to the length of the spring, this wave is a longitudinal wave.

Actually, a longitudinal wave travels through a medium in the form of periodic compressions and rarefactions.

• Let the point O be the equilibrium position of a particle in a material medium. The medium may be imagined to consist of many layers of equal thickness.
• A few layers on the right side of O have been shown. Now, some energy from outside is supplied to the layer at O so that the layer oscillates along the line AB.
• When the layer moves from A to B due to oscillation, it exerts pressure on the layers in front of it. So, those layers get compressed due to the property of compressibility of solid, liquid and gaseous media.
• Thus compression takes place in the region CD of the medium due to the motion of the layer at point O from A to B.
• The opposite incident happens at the time of motion of the layer from B to A, i.e., the layers of the region CD get rarefied due to the decrease in pressure.
• By that time, the previous compression reaches the region DE by compressing the next layers leaving a rarefaction behind.

So, a complete oscillation (ABA) generates compression and a rarefaction. These compressions and rarefactions are not confined to a region, but move through the medium due to the property of compressibility, thereby producing a wave.

• Again, if we think of any array of layers, it is found that the layers alternately get compressed and rarefied parallel to the direction of wave motion.
• It is evident that density and pressure in the medium increase in the zones of compression and decrease in the zones of rarefaction.
• It is the process of propagation of sound wave in air. It is clear that a sound wave is a longitudinal mechanical wave.

Nature Of The Medium: All longitudinal waves are mechanical waves. This type of waves cannot propagate without any medium. Longitudinal waves can propagate through any solid, liquid or gaseous medium.

• These media revert to their original conditions after the external driving oscillation (like ABA) is withdrawn.
• Till that instant, compression or rarefaction passes to the next parallel layers continuously in the direction of the force applied.
• Transverse elastic waves can be formed only in solids, not in liquids and gases. This is because liquids and gases have negligible compressibility and hence cannot sustain shearing stress as they have no definite shape.
• A solid has a definite shape and it opposes any force exerted to change its shape, i.e., a solid substance can sustain shearing stress.
• So, if one of its layers oscillates, its adjacent layer is also forced to oscillate in the same direction.
• Light waves, radio waves, etc., are not mechanical waves. These are electromagnetic waves. The elasticity of the medium has no relation with the electric and magnetic fields.
• So electromagnetic waves can propagate through vacuum, as well as through solid, liquid, or gas.

Actually, the waves set up on the surface of the water are not elastic waves; they are mechanical waves produced due to Earth’s gravity.

Difference Between Transverse And Longitudinal Waves:

Wave Motion Other Classifications Of Waves

Considering the nature of vibration, we classify waves as mechanical waves, electromagnetic waves, etc. Again, considering the direction of motion of the vibrating particles of the medium and that of the wave, we classify them as transverse waves and longitudinal waves.

In addition, waves can also be categorized on the basis of their different properties. A few of them are discussed below.

On the basis of the direction of energy transmission: The wave that transmits energy in a single direction is called a one-dimensional wave, For example, a transverse wave formed in a stretched string and a longitudinal wave formed In an elastic spring are one-dimensional waves.

• The wave which transmits energy along a plane is called a two-dimensional wave. The wave formed on the surface of water, when a stone is thrown on it, is two-dimensional.
• The wave which transmits energy in all directions is called a three-dimensional wave. Sound waves, light waves, radio waves, etc., are three-dimensional waves.

On the basis of characteristics of particle vibration: If the particles of a medium vibrate simply harmonically, the corresponding wave is called a simple harmonic wave. Practically most waves are produced by complex vibrations. But any complex wave may be described as a superposition of a number of simple harmonic waves.

On the basis of the limit of wave motion: If any wave advances through a medium continuously with a definite velocity, the wave is called a travelling or progressive wave. If the wave is not damped, it can propagate up to infinity. On the other hand, if the wave does not advance, but remains confined in a region, it is called a standing or stationary wave.

Wave Motion Some Physical Terms Related To Waves

Waves Phase: The quantity from which the motion of a wave can be known completely is called the phase of the wave.

Displacement, velocity, acceleration, etc., of a vibrating particle can be obtained from it. The particles at O and D are in the same phase. The particles at A and E, or the particles at C and E, are also in the same phase. Particles at A and C and particles at C and D are in opposite phases.

Complete Wave: The wave in between two consecutive particles, having the same phase at an instant, is known as a complete wave. Show respectively how a complete transverse wave is formed In between two consecutive crests, and how a complete longitudinal wave is formed by the combination of a compression and, a rarefaction.

Wavelength: The length of a complete wave, i.e., the distance between two consecutive particles having the same phase at an instant, is called tire wavelength (λ). OD or AE is the wavelength of the transverse wave. Again CE or DF is the wavelength of the longitudinal wave. Generally, it can be said that

The wavelength of a transverse wave = distance between any two consecutive crests or consecutive troughs.

Wavelength of a longitudinal wave = total length of a pair of successive compression and rarefaction.

In general wavelength of a wave is the distance between two consecutive points in the same phase of motion at the same instance of time.

Time period: The time required to form a complete wave, i.e., time taken by a wave to cover a distance between two consecutive particles having the same phase of vibration, is called the time period (T) of the wave.

Frequency: Frequency (n) of a wave is the number of complete waves formed in unit time.

Amplitude: The amplitude of a wave is the maximum displacement of any particle producing the wave, from its mean position. The distance of the points A, C, or E from the straight line OBDF is the amplitude of the wave.

Wave Velocity: The distance traveled by a wave in a unit of time is called its wave velocity (V). Energy is transmitted through the medium by the wave with this velocity. It may be noted that wave velocity is different from particle velocity. Particle velocity is the velocity of the particles of the medium which execute simple harmonic motions about their mean positions.

Wavefront: All the particles on a surface normal to the direction of propagation of a wave have the same phase. A surface of this type is called a wavefront. In other words, the wavefront is the locus of all points having the same phase of motion at the same instance of time.

This surface is perpendicular to the direction of propagation of wave at any point. The plane A is a wavefront because the phase of all the particles lying on plane A is the same. Similarly, the plane B is another wavefront. Clearly, planes A and B are parallel.

Wavefront Definition: Any surface, that is normal to the direction of propagation of a wave, is known as a wavefront. The particles lying on a wavefront have the same phase.

Wave Ray: A normal drawn on a wavefront is called a ray. The energy of a wave is transferred from one part of the medium to another along the ray.

Relations Among The Physical Quantities

Relation Between Time Period And Frequency: If T is the time period of a wave, one complete wave is formed in time T. Therefore, the number of complete waves formed in unit time is \(\frac{1}{T}\).

So, according to the definition of frequency,

n = \(\frac{1}{T} \quad \text { or, } T=\frac{1}{n}\)…(1)

Relation Between Wavelength And Wave Number: The number of complete waves in a length λ is 1.

So, the number of complete waves in unit length is \(\frac{1}{\lambda}\). This quantity multiplied by 2π is known as the wave number k, i.e.,

k = \(2 \pi \cdot \frac{1}{\lambda}=\frac{2 \pi}{\lambda} \quad \text { or, } \lambda=\frac{2 \pi}{k}\)…(2)

Relation Among Wave Velocity, Frequency, And Wavelength: Let the frequency of a wave be n, wavelength λ, and wave velocity V. According to the definition of frequency, the number of complete waves formed in unit time is n. Again, the wave covers a distance of V in unit time. So, the length of n number of complete waves is V.

So, the length of one complete wave = \(\frac{V}{n}\);

Then, according to the definition of wavelength, \(\lambda=\frac{V}{n} \quad \text { or, } V=n \lambda\)…(3)

i.e., wave velocity = frequency x wavelength

Again if T is the time period of a wave, n = \(\frac{1}{T}\)

∴ V = \(n \lambda=\frac{\lambda}{T} \quad \text { or, } \lambda=V T\)….(4)

From equation (3), n = \(\frac{V}{\lambda}\). If more than one wave move through a medium with the same velocity, V will be a constant. In that case, \(n \propto \frac{1}{\lambda},\) i.e., frequency and wavelength will be inversely proportional to each other. So, greater the wavelength of a wave, the smaller its frequency, and vice versa.

Wave Motion Numerical Examples

Example 1. The wave generated on a water surface advances 1 m in Is. If the wavelength is 20 cm, how many waves are produced per second?
Solution:

Wavelength, λ = 20 cm;

wave velocity, V = 1 m · s^-1 = 100 cm · s^-1

If n is the frequency, V = nλ

or \(n=\frac{V}{\lambda}=\frac{100}{20}=5 \mathrm{~s}^{-1}\)

So, 5 waves are produced per second.

Example 2. A radio center broadcasts radio waves of length 300 m. What is the frequency of this wave? Given, the velocity of light = 3 x 10⁵ km · s^-1.
Solution:

Wavelength of radio wave, λ = 300 m.

Radio waves are electromagnetic waves like light and both of them have the same velocity.

So, the velocity of radio waves,

V = \(3 \times 10^5 \mathrm{~km} \cdot \mathrm{s}^{-1}=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Frequency, n = \(\frac{V}{\lambda}=\frac{3 \times 10^8}{300}=10^6 \mathrm{~Hz}=1 \mathrm{MHz}\)

Example 3. The frequency of a tuning fork is 400 Hz and the velocity of sound in air is 320 m · s^-1. Find how far would the sound travel when the fork just completes 30 vibrations.
Solution:

Here V = 320 m · s^-1; n = 400

We know, V = nλ

∴ 320 = 400 x λ

or, \(\lambda=\frac{320}{400}=\frac{4}{5} \mathrm{~m}\)

So, when the fork completes 1 vibration, sound travels \(\frac{4}{5}\)m.

∴ When the fork completes 30 vibrations, sound travels \(\frac{4}{5}\)x30 = 24m

Example 4. A light pointer attached to one arm of a tuning fork touches a plate. The tuning fork is made to vibrate and the plate is allowed to fall freely downwards simultaneously. The tuning fork completes 8 vibrations when the pointer shows a downward displacement of 10 cm of the plate. What is the frequency of the tuning fork?
Solution:

Suppose a time t is taken by the plate to move 10 cm downwards.

So, from the relation h = \(\frac{1}{2}\)gt², we have

10 = \(\frac{1}{2}\) x 980 x t²

or, \(t^2=\frac{1}{49} \quad \text { or, } t=\frac{1}{7} \mathrm{~s}\)

The tuning fork completes 8 vibrations in that time,

So, frequency of the tuning fork = \(\frac{8}{1 / 7}=56 \mathrm{~Hz}\)

Example 5. What is the length of a compression in the sound wave, produced by a tuning fork of frequency 440 Hz? Given, the velocity of sound in air is 330 m · s^-1.
Solution:

Frequency of the sound wave (n) = frequency of the tuning fork = 440 s^-1

Velocity of sound (V) = 330 m · s^-1

∴ Wavelength, \(\lambda=\frac{V}{n}=\frac{330}{440}=\frac{3}{4} \mathrm{~m}\)

∴ Length of a compression = \(\frac{\text { wavelength }}{2}=\frac{3}{4 \times 2}\)

= 0.375 m = 37.5 cm.

Example 6. A rod hanging from a spring is dipped partly in water. The rod vibrates 180 times per minute. As a result, waves are formed on water and have 6 consecutive crests within a distance of 30 cm. Find the velocity of the wave in water.
Solution:

Here, frequency n = \(\frac{180}{60}\) = 3 Hz

There are 5 waves within 6 consecutive crests.

So, the length of these 5 waves = 30 cm

∴ Wavelength, \(\lambda=\frac{30}{5}=6 \mathrm{~cm}\)

∴ Velocity of the wave, V= nλ = 3 x 6 = 10 cm · s^-1.

Example 7. The frequency of a tuning fork is 280 advances 80 m In a medium while the tuning fork executes 70 complete oscillations. Determine the velocity of sound in that medium.
Solution:

Wavelength, \(\lambda=\frac{80}{70}\) = \(\frac{8}{7}\) cm

So, the velocity of sound in the medium,

V = \(n \lambda=280 \times \frac{8}{7}=320 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 8. The frequency of a tuning fork is 512 Hz. When the tuning fork makes 30 vibrations, the emitted sound travels 20 m in the air. Determine the wavelength and the velocity of sound waves in air.
Solution:

Wavelength, \(\lambda=\frac{20}{30}\) = 0.667 m

∴ Velocity of sound in air,

V = nλ = 512 x = 341.33 m · s^-1

Example 9. When two vibrating tuning forks of frequencies 50 Hz and 100 Hz touch the surface of water, they produce waves of wavelengths 0.6 cm and 0.36 cm, respectively. Compare the velocities of the two surface waves.
Solution:

If V₁ and V₂ are die velocities of die two surface waves, then

⇒ \(\frac{V_1}{V_2}=\frac{n_1 \lambda_1}{n_2 \lambda_2}=\frac{50 \times 0.6}{100 \times 0.36}=\frac{5}{6}\)

i.e., \(V_1: V_2=5: 6 .\)

 

Wave Motion Equation Of A Travelling Or Progressive Wave Numerical Examples

Example 1. The equation of a progressive wave is y = 15sin(660πt-0.02πx) cm. Calculate the frequency and the velocity of the wave.
Solution:

The general equation of a progressive wave is,

y = \(A \sin \omega\left(t-\frac{x}{V}\right)\)….(1)

The equation of the given wave is

y = \(15 \sin (660 \pi t-0.02 \pi x)=15 \sin 660 \pi\left(t-\frac{0.02 x}{660}\right)\)

= \(15 \sin 660 \pi\left(t-\frac{x}{660 / 0.02}\right) \mathrm{cm}\)…(2)

Comparing equations (1) and (2) we have ω = 660π

⇔ Frequency, n = \(\frac{\omega}{2 \pi}=\frac{660 \pi}{2 \pi}=330 \mathrm{~s}^{-1}=330 \mathrm{~Hz} \text {; }\)

Wave velocity, V = \(\frac{660}{0.02}=\frac{660 \times 100}{2}=33000 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

= \(330 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

Example 2. A wave vibrating along the y-axis propagates along the negative direction of x-axis. The values of its amplitude/frequency and wavelength are 10cm, 500 Hz, and 100 cm, respectively. Write down the equation of the progressive wave.
Solution:

Amplitude, A = 10 cm; frequency n = 500 Hz and wavelength, λ = 100 cm

∴ Wave velocity, V = nλ = 500 x 100 = 5 x 10⁴ cm · s^-1

Angular frequency, ω = 2πn = 2π x 500 = 1000π Hz

Since the wave propagates along the negative x-direction, the equation of the progressive waves,

y = \(A \sin O\left(t+\frac{x}{V}\right)=10 \sin 1000 x\left(t+\frac{x}{5 \times 10^4}\right) \mathrm{cm}\)

Example 3. The equation of a progressive wave is y = 20sinπ(4t – 0.01x)cm. Determine the amplitude, frequency, wavelength, and velocity of the wave.
Solution:

The general equation of a progressive wave is y = \(A \sin \frac{2 \pi}{d}(V t-x)\)…(1)

The given equation is y = \(20 \sin \pi(4 t-0.01 x)\)

= \(20 \sin \pi\left(4 t-\frac{x}{100}\right)=20 \sin \frac{\pi}{100}(400 t-x)\)

= \(20 \sin \frac{2 \pi}{200}(400 t-x) \mathrm{cm}\)…(2)

Comparing equations (1) and (2) we have,

Amplitude, A = 20 cm; wavelength, λ = 200 cm;

Wave velocity, V = 400 cm · s^-1.

So, frequency, n = \(\frac{V}{\lambda}=\frac{400}{200}=2 \mathrm{~Hz}\)

Example 4. The equation of a progressive wave is y = \(10 \sin 2 \pi\left(\frac{t}{0.005}-\frac{x}{10}\right)\)cm; here t and x are given in CGS units. Determine the amplitude, wavelength, and velocity of the wave.
Solution:

y = \(10 \sin 2 \pi\left(\frac{t}{0.005}-\frac{x}{10}\right)\)

= \(10 \sin \frac{2 \pi}{10}\left(\frac{t}{0.0005}-x\right) \mathrm{cm}\)…(1)

Comparing equation (1) with the general equation of a progressive wave, y = \(A \sin \frac{2 \pi}{\lambda}(V t-x)\), we have,

Amplitude, A = 10 cm; wavelength, λ = 10 cm

Velocity of wave, V = \(\frac{1}{0.0005}=2000 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Example 5. A wave traveling along the X-axis is described by the equation y(x, t) = 0.005cos(αx—βt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s respectively, what are the values of α and β?
Solution:

= \(\frac{2 \pi}{\lambda}=\frac{2 \pi}{0.08}=25 \pi \mathrm{m}^{-1} ; \beta=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{s}^{-1}\)

[In this question β and α are used instead of the usual symbols ω and k respectively].

Example 6. The equation of a progressive wave is given by y = 10sinπ(t-0.002x) cm. Determine

1. Maximum displacement
2. Maximum velocity and
3. Maximum acceleration of the particles of the medium,

Solution: Given the equation,

y = 10sinπ(t-0.002x) cm….(1)

General equation of a progressive wave,

y = \(A \sin \omega\left(t-\frac{x}{V}\right)\)…(2)

Comparing equations (1) and (2) we have,

A = 10 cm; \(\omega=\pi \mathrm{Hz} ; V=\frac{1}{0.002}=500 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

1. Maximum displacement of the particles of the medium = amplitude of the wave (A) =10 cm.

2. Velocity of a particle, \(v=\frac{\partial y}{\partial t}=\omega A \cos \omega\left(t-\frac{x}{V}\right) \mathrm{cm} \cdot \mathrm{s}^{-1}\)

∴ Maximum velocity of the particle = ωA = π x 10 = 31.4 cm · s^-1.

3. Acceleration of a particle,

a = \(\frac{\partial v}{\partial t}=-\omega^2 A \sin \omega\left(t-\frac{x}{V}\right) \mathrm{cm} \cdot \mathrm{s}^{-2}\)

∴ Maximum acceleration of the particle = ω²A = π² x 10 = 98.7 cm · s^-2.

Example 7. The equation y = \(y_0 \sin 2 \pi\left(n t-\frac{x}{\lambda}\right)\) expresses a transverse wave. If the maximum particle velocity is equal to four times the wave velocity, show that \(\lambda=\frac{1}{2} \pi y_0\)
Solution:

Given equation,

y = \(y_0 \sin 2 \pi\left(n t-\frac{x}{\lambda}\right)=y_0 \sin \frac{2 \pi}{\lambda}(\lambda n t-x)\)

Comparing this equation with the general equation of a progressive wave, y = \(A \sin \frac{2 \pi}{\lambda}(V t-x)\), we have,

A = y₀; V= λn

Again, particle velocity is given by v = \(\frac{\partial y}{\partial t}=\frac{2 \pi V}{\lambda} A \cos \frac{2 \pi}{\lambda}(V t-x)\)

So, maximum particle velocity, \(v_0=\frac{2 \pi V A}{\lambda}\)

According to the equation \(\frac{v_0}{V}=4 \text { or, } \frac{2 \pi A}{\lambda}=4 \text { or, } \lambda=\frac{2 \pi A}{4}=\frac{1}{2} \pi y_0\)

Example 8. A wave equation that gives a displacement along y direction Is given by y = 10^-4 sin(60t+ 2x) where x and y are in meters and f is time in seconds. Determine the wavelength, frequency, and velocity of the wave.
Solution:

Given equation is y = \(10^{-4} \sin (60 t+2 x)\)

= \(10^{-4} \sin 2(30 t+x)\)

= \(10^{-4} \sin \frac{2 \pi}{\pi}(30 t+x)\)

Comparing the given equation with the general equation of a progressive wave, y = \(A \sin \frac{2 \pi}{\lambda}(V t+x)\), we have,

Wavelength, λ =π= 3.14 m;

Wave velocity, V = 30 m · s^-1; (in the direction of negative x-axis)

Frequecy, n = \(\frac{V}{\lambda}=\frac{30}{3.14}=9.55 \mathrm{~Hz} .\)

Example 9. A wave having a frequency 200 Hz is advancing with a velocity 40 m · s^-1. What is the phase difference of two particles separated by 5 cm in the direction of wave motion?
Solution:

Here n = 200 Hz; V = 40 m · s^-1 = 4000 cm · s^-1

So, \(\lambda=\frac{V}{n}=\frac{4000}{200}=20 \mathrm{~cm}\)

∴ Phase difference = \(\frac{2 \pi}{\lambda}\) x path difference

= \(\frac{2 \pi}{20} \times 5=\frac{\pi}{2} \mathrm{rad}=90^{\circ} .\)

Example 10. \(y_1=0.1 \sin \frac{\pi}{2}(200 t-x) \mathrm{cm}\) and \(y_2=0.2 \sin \frac{\pi}{2}(200 t-x+5) \mathrm{cm}\) are two wave equations. Show that the phase difference of the two waves does not change. What is the value of that phase difference?
Solution:

Comparing the given equations of the two waves with the general equation of a progressive wave,

y = \(A \sin \frac{2 \pi}{\lambda}(V t-x)\),

The wavelength (λ) and the wave velocity (V) of the given two waves are equal. So, the two waves can propagate with any special phase (crest or trough) maintaining equal distance with respect to each other. So, the phase difference of the two waves does not change.

Phase of the first wave \(\left(\theta_1\right)=\frac{\pi}{2}(200 t-x)\)

Phase of the second wave \(\left(\theta_2\right)=\frac{\pi}{2}(200 t-x+5)\)

So, phase difference = θ₂-θ₁

= \(\frac{\pi}{2} \cdot 5=\frac{5 \pi}{2} \mathrm{rad}=\frac{5}{2} \times 180^{\circ}=450^{\circ}\)

Since 450° = 360° + 90°, the two angles 450° and 90° are equivalent, i.e., the required phase difference = 90°.

Example 11. The equation of a progressive wave is y = \(0.1 \sin \frac{\pi}{2}(200 t-x) \mathrm{cm}\)

1. What is the phase of the wave at the point x = 2 at time t =0?
2. What is the phase difference between two points separated by 8 cm?
3. How does the phase at any point change in 0.005 s?

Solution:

The phase of the given progressive wave,

θ = \(\frac{\pi}{2}(200 t-x)\) ….(1)

1. Putting x = 2, t = 0 in equation (1),

θ = \(\frac{\pi}{2}(-2)=-\pi \mathrm{rad}=-180^{\circ}\)

-180° and + 180°—these two angles are equivalent. So, θ = 180°.

2. For two points separated by 8 cm,

⇒ \(\theta_1=\frac{\pi}{2}\left(200 t-x_1\right), \theta_2=\frac{\pi}{2}\left(200 t-x_2\right)\)

∴ Phase difference = θ₂ – θ₁

= \(\frac{\pi}{2}\left(x_2-x_1\right)=\frac{\pi}{2} \times 8=4 \pi \mathrm{rad}\)

But the angle 4π rad is equivalent to 0°.

So, phase difference = 0°.

2. The required phase change

= \(\frac{\pi}{2}(200 \times 0.005)=\frac{\pi}{2} \times 1=\frac{\pi}{2} \mathrm{rad}=90^{\circ}\)

Example 12. The amplitude of a wave propagating in the positive x-direction is given by y = \(\frac{1}{1+x^2}\) at time t = 0 and by y = \(\frac{1}{1+(x-1)^2}\) at t = 2 s where x and y are in meters, The shape of the wave does not change during the propagation. What is the velocity of the wave?
Solution:

Considering the point x = 0 at t = 0, amplitude, y = \(\frac{1}{1+0^2}=1 \mathrm{~m}\)

Now, if the wave propagates up to the point x = x₁ at t = 2s, then

1 = \(\frac{1}{1+\left(x_1-1\right)^2} \quad \text { or } 1+\left(x_1-1\right)^2=1 \)

or, \(\left(x_1-1\right)^2=0 \quad \text { or, } x_1-1=0 \quad \text { or, } x_1=1 \mathrm{~m},\)

i.e., in time 2s tire phase of amplitude 1m has advanced from the point x = 0 to x = 1m.

So, the velocity of the wave = \(\frac{1}{2}\) = 0.52 m · s^-1

Example 13. A plane progressive wave of frequency 25 Has, amplitude 2,5 x 10^-5 m, and initial phase zero propagates along the negative x-direction with a velocity of 300 m · s^-1. At any Instance, what Is the phase difference between the oscillations at two points 6 m apart along the line of propagation of the wave? Also, determine the corresponding amplitude difference.
Solution:

Here, n = 25 Hz, A = 2.5 x 10^-5 m, V= 300 m · s^-1

∴ \(\lambda=\frac{V}{n}=\frac{300}{25}=12 \mathrm{~m}\)

So, two points 6 rn apart have a path difference of \(\frac{\lambda}{2}\).

∴ Phase difference = \(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2}=\pi \mathrm{rad}=180^{\circ}\)

So, the two points are in the opposite phase. For example, if a crest is formed at a point, a trough would be formed 6m apart. However, the magnitudes of the amplitude would be the same.

∴ Amplitude difference = 0.

Example 14. The bulk modulus and the density of steel are 80 and 8 times, respectively, of those of water. Calculate the speed of sound in steel. Given, the speed of sound in water = 1493 m · s^-1.
Solution:

The velocity of sound in an elastic medium is

V = \(\sqrt{\frac{E}{\rho}}\)

Where E = modulus of elasticity of the medium ρ = density of the medium

∴ Velocity of sound in water, \(V_w=\sqrt{\frac{E_w}{\rho_w}}\)

Velocity of sound in steel, \(V_w=\sqrt{\frac{E_w}{\rho_w}}\)

∴ \(\frac{V_s}{V_w}=\sqrt{\frac{E_s}{\rho_s} \cdot \frac{\rho_w}{E_w}}=\sqrt{\frac{E_s}{E_w} \times \frac{\rho_w}{\rho_s}}=\sqrt{80 \times \frac{1}{8}}=\sqrt{10}=3.162\)

∴ \(V_s=3.162 V_w=3.162 \times 1493=4721 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Example 15. A wave is represented by y = \(20 \cdot \sqrt{3} \sin \left(\frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\right) \mathrm{cm}\), Calculate the displacement of a particle at the position x = \(\frac{lambda}{6}\) on the wave at time t = \(\frac{T}{3}\)
Solution:

Given equation, \(y=20 \cdot \sqrt{3} \sin \left(\frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\right)\)

Putting x = \(\frac{\lambda}{6}$ and $t=\frac{T}{3}\), we get

Displacement, y = \(20 \cdot \sqrt{3} \sin \left(\frac{2 \pi T}{3 T}-\frac{2 \pi \lambda}{6 \lambda}\right)=20 \cdot \sqrt{3} \sin \left(\frac{2 \pi}{3}-\frac{\pi}{3}\right)\)

= \(20 \cdot \sqrt{3} \sin \frac{\pi}{3}=20 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2}=30 \mathrm{~cm}\)

Example 16. The equations of two waves are, \(y_1=0.30 \sin (314 t-1.57 x)\),
\(y_2=0.10 \sin (314 t-1.57 x+1.57)\) Determine the phase difference and the ratio of the intensities of the two waves.
Solution:

Phase of the first wave, θ₁ = 314t – 1.57x

Phase of the second wave, θ₂ = 314t – 1.57x +1.57

∴ The phase difference of the two waves

= \(\theta_2-\theta_1=1.57 \mathrm{rad}=1.57 \times \frac{180}{\pi}=90^{\circ}\)

We know that the intensity of a wave is proportional to the square of its amplitude.

∴ Ratio of the amplitudes of the two waves = \(\frac{A_1}{A_2}=\frac{0.30}{0.10}=\frac{3}{1}\)

∴ Ratio of the intensities of the two waves = \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{1}\)

Wave Motion Velocity Of Sound In Solid And Liquid Media

Sound propagates through solids and liquids as longitudinal elastic wave tike in gaseous media. But the velocities of sound in solid and liquid are different. If E is the modulus of elasticity of the medium and ρ is its density, Newton’s formula for the velocity of sound is given by,

c = \(\sqrt{\frac{E}{D}}\)…(1)

• The density of any solid or liquid is greater than that of a gas. Bin the modulus of elasticity of a solid or a liquid is much greater than the Thar of a gas.
• So, the velocity of sound in a solid or B. Liquid is greater than that in a gas. For example, the psVjcities of sound in iron and in water are about 15 times and 4.5 times that in air, respectively.
• The velocity of sound in iron is greater than that in air—it can be easily understood from a simple experiment. If one end of a long iron pipe is struck heavily, the sound of striking is heard twice at the other end of the pipe.
• Sound is heard for the first time due to its propagation through the iron and for the second time due to its propagation through the air in the pipe. An iron pipe nearly 100 m in length should be taken to hear the two sounds distinctly.

There is one important difference between solids and liquids as media of propagation of sound. The modulus of elasticity for the solid is, Young’s modulus (Y) while it is the bulk modulus (k) in the case of the liquid. So the equation (1)

For solid, c = \(\sqrt{\frac{Y}{\rho}}\)….(2)

and for liquid, c = \(\sqrt{\frac{k}{\rho}}\) …(3)

In case of steel, Y = 2 x 10¹¹ N · m^-2; ρ = 7850 kg · m^-3;

From equation (2) we get,

c = \(\sqrt{\frac{2 \times 10^{11}}{7850}} \approx 5048 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

In case of water, k = 2.1 x 10⁹ N · m^-2; ρ = 1000 kg · m^-3

From equation (3) we get, c = \(\sqrt{\frac{2.1 \times 10^9}{1000}}=1449 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Velocity Of Sound In Solid And Liquid Media Numerical Examples

Example 1. Two explosions are made simultaneously from a ship, one above the surface of the water and another just below it. Sound is heard in a hydrophone placed below water from another ship 5 km apart 11s earlier than the sound reaching the deck of the ship. What is the velocity of sound in water? Given that the velocity of sound in air = 348 m · s^-1.
Solution:

Distance between the two ships = 5 km = 5000 m

So time taken by the sound to come through air = \(\frac{5000}{348} \mathrm{~s}\)

According to the question, time is taken by the sound to come through water = \(\frac{5000}{348}-11=\frac{5000-3828}{348}=\frac{1172}{348} \mathrm{~s}\)

Therefore, the velocity of sound in water

= \(\frac{5000}{\frac{1172}{348}}=1485 \mathrm{~m} \cdot \mathrm{s}^{-1}\) (approx.)

Example 2. If one end of a long steel pipe is struck, the sound is heard twice at an interval of 3 s at the other end. What is the length of the pipe? (The velocities of sound in air = 350m · s^-1, in steel = 5000 m · s^-1)
Solution:

Let the length of the pipe be l m.

So, time is taken by the sound to come through the air in the pipe = \(\frac{l}{350}\)s

Again, time taken by die sound to come through steel =\(\frac{l}{5000}\)s

According to the question, \(\frac{l}{350}-\frac{l}{5000}=3 \quad \text { or, } l\left(\frac{1}{350}-\frac{1}{5000}\right)=3\)

or, \(l\left(\frac{100-7}{35000}\right)=3\)

or, \(l=\frac{35000 \times 3}{93}=1129 \mathrm{~m}=1.129 \mathrm{~km} .\)

Example 3. Sound moves through a liquid with a velocity 1340 m · s^-1. If the density of the liquid is 0.8 g · cm^-3, determine the compressibility of the liquid.
Solution:

We know, c = \(\sqrt{\frac{k}{\rho}}\); k = bulk modulus of the liquid

∴ k = c²ρ [c = 1340 m · s^-1 = 134 x 10³ cm · s^-1]

Again, compressibility = \(\frac{1}{k}=\frac{1}{c^2 \rho}\)

= \(\frac{1}{\left(134 \times 10^3\right)^2 \times 0.8}=\frac{1}{143648} \times 10^{-5}\)

= \(6.96 \times 10^{-11} \mathrm{~cm}^2 \cdot \mathrm{dyn}^{-1}\)

Wave Motion Musical Sound And Noise

Musical sound: The sound produced due to regular and periodic vibrations, that is pleasing to hear, is called musical sound. While singing, the vocal cords of a singer vibrate with regular periodic motion.

• Similarly when die strings of a sitar are excited to produce a certain note, diey vibrate with regular periodic motion. The sound of clapping is usually unpleasant, but rhythmic clapping of hands may produce a pleasant sound.
• This is due to the regular vibration of the source. Similarly pattern of raindrops on tin roofs or the clickers’ sound of train wheels may be sonorous.

Noise: The sound produced by an irregular or short-lived vibration is called noise. The explosion of a cracker, the firing of a gun, the horn of a car, etc., are noises.

Tone: A musical sound of a single frequency is called a tone. When the source of sound vibrates in SHM, it produces a tone. The two prongs of a tuning fork vibrate in SHM and produce the sound of a single frequency. Thus, the sound of a tuning fork is a pure tone.

Tone note: A musical sound due to a mixture of more than one frequency is called a note. The sound produced by different musical instruments consists, in general, of more than one tone. If a tone is compared with monochromatic light such as red light, a note can be compared with compound light, i.e., white light. The tones within a note are classified on the basis of their frequencies.

• Fundamental Tone: The tone with the lowest frequency present in a note is called the fundamental tone.
• Overtones: The tones in a note, other than the fundamental tone, are called overtones.
• Harmonics: Harmonics are tones having frequencies that are integral multiples of that of the fundamental tone. The fundamental tone is also a harmonic.
• Octave: A tone whose frequency is twice that of the fundamental tone is said to be the octave of the fundamental tone.

Suppose, a musical note consists of tones of frequencies 200 Hz, 300 Hz, 400 Hz, 500 Hz, and 600 Hz. So it can be said:

1. Each of these five different frequencies corresponds to a tone,
2. The combination of these five tones is a note,
3. The tone of 200 Hz is die fundamental tone,
4. The tones having frequencies 300 Hz, 400 Hz, 500 Hz, and 600 Hz are overtones,
5. Both of the tones of frequencies 400 Hz and 600 Hz are harmonics, as the frequencies are 2 and 3 umes that of the fundamental tone respectively. The tone of 400 Hz is called the second harmonic as its frequency is 2 times of the fundamental tone. Similarly, the tone of 600 Hz is called the third harmonic. The fundamental tone is called the first harmonic.
6. The tone of 400 Hz is the octave because this frequency is the mice die frequency of the fundamental tone.

Characteristics Of Musical Sound Or Note: There are three characteristics that differentiate musical notes. These are loudness, pitch, and quality.

1. Loudness: Loudness Is related to the sound energy reaching our ears per unit time. If the sound energy reaching our ears in 1 second goes up, we perceive a corresponding increase in the loudness.
2. Pitch: It is that characteristic that differentiates a sharp or shrill sound from a grave one. It increases with the Increase in frequency of the source.
3. Quality Or Timbre: It is that characteristic of a musical note that enables us to distinguish between a note emitted by one musical instrument from a note of the same loudness and pitch emitted by another instrument. It depends on the number of overtones present in a note and their relation with the fundamental tone.

Wave Motion Conclusion

A wave is a disturbance that travels through a medium, but the particles constituting the medium do not travel.

• The disturbance that travels through a material medium due to a collective vibration of the particles of the medium is known as a mechanical wave.
• It is an elastic wave if the particles vibrate due to the elasticity of the medium. If the motion of the particles of a medium is simple harmonic, the corresponding wave is known as a simple harmonic wave.
• A wave, that propagates in a direction perpendicular to the direction of motion of the vibrating particles of the medium, is called a transverse wave.
• A wave, that propagates along the direction of motion of the vibrating particles of the medium, is called a longitudinal wave.
• All longitudinal waves are mechanical waves. Longitudinal waves can be transmitted through any solid, liquid, or gaseous medium. This type of wave cannot propagate without any medium.
• Transverse elastic waves cannot be produced in liquids or gases but can be created in solids.

Some physical quantities related to waves:

1. Phase: The quantity from which the motion of a wave, except the amplitude, can be known completely is called the phase of the wave.

2. Complete Wave: The wave in between two consecutive particles having the same phase at an instant is known as a complete wave.

3. Wavelength: The length of a complete wave, i.e., the distance between two consecutive particles having the same phase at an instant is called the wavelength of the wave.

4. Time period: The time required to create a complete wave, i.e., the time taken by a wave to cover the distance between two consecutive particles having the same phase of vibration is called the time period of the wave.

5. Frequency: The frequency of a wave is the number of complete waves produced in unit time.

6. Amplitude: The amplitude of a wave is the maximum displacement from the mean position of a particle-producing tire wave.

7. Wave velocity: The distance traversed by a wave in unit time is called the wave velocity.

8. Wavefront: A surface normal to the direction of propagation of a wave is known as a wavefront. Particles lying on a wavefront have the same phase.

9. Ray: A normal drawn on a wavefront is called a ray.

The velocity with which sound wave propagates in a material medium depends on two properties of the medium density, and elasticity.

• Newton assumed that the propagation of sound through a gaseous medium is an isothermal process.
• According to Laplace, the propagation of sound through a gaseous medium is an adiabatic process.
• The pressure of a gas has no effect on the velocity of sound.
• The velocity of sound in a gas is directly proportional to the square root of its absolute temperature.
• The velocity of sound in air increases by 0.61 m · s^-1 or 61 cm · s^-1 for 1°C rise in temperature.
• The velocity of sound in moist air is greater than that in dry air.
• The velocity of sound in a gas is inversely proportional to the square root of its density.
• To obtain regular reflection of sound, the reflector must be large but the surface of the reflector need not be very smooth.
• If the reflected sound is heard separately from an original sound, it is called an echo of the original sound.
• The sensation of an inarticulate sound (sound produced by gunshot, clapping, etc.) persists in our ear for about \(\frac{1}{10}\)th of a second. This is known as the persistence of hearing.
• The velocity of sound in air at 0°C is about 330 m · s^-1.

Wave Motion Useful Relations for Solving Numerical Problems

Relation between time period (T) and frequency (n): T = \(\frac{1}{n}\)

Relation between wavelength (λ) and wave number (k): \(\lambda=\frac{2 \pi}{k}\)

Relation between frequency (n) and angular frequency ω:ω = 2πn

Wave velocity (V) = frequency (n) x wavelength (λ)

Equation of a progressive wave moving along the positive and negative direction of the x-axis:

y = \(A \sin \omega\left(t \mp \frac{x}{V}\right)=A \sin (\omega t \mp k x)\)

= \(A \sin \frac{2 \pi}{\lambda}(V t \mp x)=A \sin 2 \pi\left(\frac{t}{T} \mp \frac{x}{\lambda}\right)\)

If The equation of a progressive wave can be expressed using cosine function also.

Physical quantities related to the motion of a particle in a progressive wave:

\(\begin{array}{|l|l|}
\hline \text { Displacement } & y=A \sin (\omega t-k x \pm \phi) \\
\hline \text { Velocity } & \nu=\frac{d y}{d t}=\omega A \cos (\omega t-k x \pm \phi) \\
\hline \text { Max. velocity } & \pm \omega A \\
\hline \text { Acceleration } & a=\frac{d^2 y}{d t^2}=-\omega^2 A \sin (\omega t-k x \pm \phi) \\
\hline \text { Max, acceleration } & \pm \omega^2 A \\
\hline
\end{array}\)

 

If the particle velocity is v and the wave velocity is V, then \(\nu=-V \frac{\partial y}{\partial x}\)

Velocity of Progressive waves in different media:

\(\begin{array}{|l|l|l|}
\hline \begin{array}{l}
\text { Velocity of a long- } \\
\text { tidal wave in a } \\
\text { solid medium }
\end{array} & V=\sqrt{\frac{Y}{\rho}} & \begin{array}{l}
Y=\text { Young’s modulus of } \\
\text { the medium and } \rho= \\
\text { density of the medium }
\end{array} \\
\hline \begin{array}{l}
\text { Velocity of a long- } \\
\text { tidal wave in a } \\
\text { liquid or a gaseous } \\
\text { medium }
\end{array} & V=\sqrt{\frac{E}{\rho}} & \begin{array}{l}
E=\text { bulk modulus of the } \\
\text { medium and } \rho=\text { density } \\
\text { of the medium }
\end{array} \\
\hline
\end{array}\) \(\begin{array}{|c|c|c|}
\hline \begin{array}{l}
\text { Velocity of sound } \\
\text { wave in a gaseous } \\
\text { medium }
\end{array} & V=\sqrt{\frac{\gamma p}{\rho}} & \begin{array}{l}
p=\text { pressure of the gas, } \\
\gamma=c_p / c_\nu=\text { ratio of the } \\
\text { two specific heats of the } \\
\text { gas and } \rho=\text { density of } \\
\text { the gas }
\end{array} \\
\hline \begin{array}{l}
\text { Velocity of a trans- } \\
\text { verse wave in a } \\
\text { stretched string }
\end{array} & V=\sqrt{\frac{T}{m}} & \begin{array}{l}
T=\text { tension in the string } \\
\text { and } m=\text { mass per unit } \\
\text { length of the string }
\end{array} \\
\hline
\end{array}\)

 

If M is the mass of 1 mol of a gas, then c = \(\sqrt{\frac{\gamma R T}{M}}\)

where T = absolute temperature of the gas,

R = universal gas constant

If c₀ and c are the velocities of sound at 0°C and t °C respectively, then c = c₀(1+ 0.001830 t)

If the densities of two different gases at the same temperature and pressure are ρ₁ and ρ₂ and the velocities of sound in the two gases are c₁ and c₂ respectively, then \(\frac{c_1}{c_2}=\sqrt{\frac{\rho_2}{\rho_1}}\)

If the distance of a reflector from the source of sound is D, the time required to hear the echo is t, and the velocity of sound is V, then, 2D = Vt

Wave Motion Very Short Answer Type Questions

Question 1. What kind of energy is transmitted through an elastic wave?
Answer: Mechanical energy

Question 2. Give an example of an elastic wave.
Answer: Soundwave

Question 3. Is It possible for a transverse wave to propagate in a liquid?
Answer: No

Question 4. What kind of wave is an X-ray?
Answer: Electromagnetic

Question 5. If the direction of motion of a wave and the direction of vibration of the particles of a medium are parallel to each other, the wave is called a transverse wave.
Answer: False

Question 6. If the direction of motion of a wave and the direction of vibration of the particles of a medium are perpendicular to each other, the wave is called a ______ wave.
Answer: Transverse

Question 7. During propagation of sound, compressions and rarefactions occur so rapidly that the _______ of the gas cannot remain constant.
Answer: Temperature

Question 8. If λ is the wavelength of a transverse wave, what will be the distance between two consecutive crests?
Answer: λ

Question 9. If λ is the wavelength of a longitudinal wave, what will be the length of a compression?
Answer: \(\frac{\lambda}{2}\)

Question 10. If the velocity of sound is 330 m · s^-1, what will be the wavelength of the sound wave emitted from a tuning fork of frequency 220 Hz?
Answer: 1.5 m

Question 11. The time required to form a complete wave is called the time period of the wave. Is the statement true or false?
Answer: True

Question 12. What is the number of complete waves formed in unit time called?
Answer: Frequency

Question 13. The maximum displacement of a particle on the wave from its mean position is called the _______ of the wave.
Answer: Amplitude

Question 14. The surface of vibration of the particles of a medium having the same phase during the propagation of a wave is called the ________ of the wave.
Answer: Wavefront

Question 15. The frequency of a progressive wave x = velocity of the wave.
Answer: Wavelength

Question 16. The equation of a progressive wave is given by y = Acos\(\frac{2 \pi}{\lambda}\)(Vt-x). Find the frequency of the wave.
Answer: \(\frac{V}{\lambda}\)

Question 17. The equation of a progressive wave is given by y = \(4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the amplitude of the wave.
Answer: 4cm

Question 18. The equation of a progressive wave is given by y = \(4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the frequency of the wave.
Answer: 10 Hz

Question 19. The equation of a progressive wave is given by \(y=4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the velocity of the wave.
Answer: 100 cm · s^-1

Question 20. Why is the sound heard in CO₂ more intense in comparison to the sound heard in the air?
Answer: Density more

Question 21. By what process does the propagation of sound through a gaseous medium take place?
Answer: Adiabatic

Question 22. What is the ratio of the velocities of sound through hydrogen and oxygen at STP?
Answer: 4: 1

Question 23. If the temperature remains constant, what will be the bulk modulus of a gas?
Answer: Equal to the pressure of the gas

Question 24. Why is the velocity of sound greater in deep water than that on its surface?
Answer: Pressure increases with the depth of water, velocity of sound increases with the increase of pressure

Question 25. Why does a violinist return his instrument on entering a warm room?
Answer: Due to an increase in temperature velocity increases

Question 26. If temperature is kept constant, the velocity of sound in a gas is independent of which of the following properties? Pressure, density, or wind flow.
Answer: Pressure

Question 27. Is the velocity of sound in dry air at constant temperature, less or greater than that in moist air?
Answer: Less

Question 28. The equation of a progressive wave is given by y = \(A \cos \omega\left(t-\frac{x}{V}\right)\). What will be the frequency and amplitude of the wave?
Answer: \(\left[\frac{\omega}{2 \pi}, A\right]\)

Question 29. What should be the minimum time interval between an original sound and its echo, so that the echo is heard separately?
Answer: \(\frac{1}{10}\)s

Question 30. The velocity of sound in air is 330 m • s^-1. What should be the minimum distance of a reflector so that a listener can hear the echo distinctly?
Answer: 16.5 m

Question 31. In a big hall, sound persists even after the original sound is stopped. What is the phenomenon called?
Answer: Reverberation

Question 32. If a sound wave enters water from the air, is the angle of refraction less or greater than the angle of incidence?
Answer: Greater

Wave Motion Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: A sound wave is regarded as a pressure wave.

Statement 2: Energy in this type of wave is transported due to the formation of compression and rarefaction in the medium in which pressure difference is created.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: Any function of space and time that satisfies the following equation represents a wave \(\frac{d^2 y}{d x^2}=\frac{1}{V^2} \frac{d^2 y}{d t^2}\)

Statement 2: y s Asin ωt, y = Acosωt do not satisfy the above equations and hence do not represent waves.

Answer: 2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1

Wave Motion Match The Columns

Question 1. A wave is transmitted from a denser to a rarer medium.

Answer: 1. C, 2. A, 3. A, 4. A

Question 2. Three traveling sinusoidal waves on identical strings have the same tension. The mathematical form of the waves are \(y_1=A \sin (3 x-6 t), y_2=A \sin (4 x-8 t) and y_3=A \sin (6 x-12 t)\).

Answer: 1. D, 2. A, 3. B, 4. C

Wave Motion Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. Represents two snaps of a traveling wave on a string of mass per unit length μ = 0.25 kg · m^-1. The first snap is taken at t = 0 and the second is taken at t = 0.053 s.

1. The speed of the wave is

1. \(\frac{20}{3}\) m · s^-1
2. \(\frac{10}{3}\) m · s^-1
3. 20 m · s^-1
4. 10 m · s^-1

Answer: 2. \(\frac{10}{3}\) m · s^-1

2. The frequency of the wave is

1. \(\frac{5}{3}\) Hz
2. \(\frac{10}{3}\) Hz
3. 5 Hz
4. 10 Hz

Answer: 1. \(\frac{5}{3}\) Hz

3. The maximum speed (in m · s^-1) of the particle is

1. \(\frac{5 \pi}{13}\)
2. \(\frac{5 \pi}{13}\)
3. \(\frac{\pi}{30}\)
4. \(\frac{7 \pi}{20}\)

Answer: 3. \(\frac{\pi}{30}\)

Question 2. A sinusoidal wave is propagating in a negative x -x-direction in a string stretched along the x-axis. A particle of string at x = 2 cm is found at its mean position and it is moving in a positive y-direction at t = 1s. The amplitude of the wave, the wavelength, and the angular frequency of the wave are 0.1 m, \(\frac{\pi}{4}\) m and 4π rad • s^-1 respectively,

1. The equation of the wave is

1. y = 0.1sin[47π(t- 1) + 8(x-2)]
2. y = 0.1 sin [(t- 1) – (x-2)]
3. y = 0.1sin [4π(t- 1)-8(x-2)]
4. None of these

Answer: 1. y = 0.1sin[47π(t- 1) + 8(x-2)]

2. The speed of particle at x = 2m and t = 1 s is

1. 0.2π m · s^-1
2. 0.67π m · s^-1
3. 0.4π m · s^-1
4. 0

Answer: 3. 0.4π m · s^-1

3. The instantaneous power (in J · s^-1) transfer through x = 2 m and t = 1.125 s is

1. 10
2. \(\frac{4 \pi}{3}\)
3. \(\frac{2 \pi}{3}\)
4. Zero

Answer: 4. Zero

Question 3. Suppose a musical note consists of tones of frequencies 100 Hz, 200 Hz, 300 Hz, 400 Hz, 500 Hz, and 600 Hz.

1. What should be the frequency of the fundamental tone?

1. 200 Hz
2. 100 Hz
3. 400 Hz
4. 600 Hz

Answer: 2. 100 Hz

2. Among the frequencies the third and fourth harmonics are respectively

1. 300 Hz, 500 Hz
2. 400 Hz, 500 Hz
3. 300 Hz, 400 Hz
4. None of these

Answer: 3. 300 Hz, 400 Hz

3. Among the tones the octave is

1. 200 Hz
2. 300 Hz
3. 400 Hz
4. None

Answer: 1. 200 Hz

Wave Motion Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. A transverse wave propagating along x -axis is represented by (x, t) = 8.0sin (0.5π x – 4πt – \(\frac{\pi}{4}\)) where x is in meters and t is in seconds. Calculate the speed in m µ s^-1 of the wave.
Answer: 8

Question 2. The water waves are traveling along the surface of an ocean at a speed of 2.5 m · s^-1 and splashing up periodically against a pole. Each adjacent crest is 5 m apart. The crest splashes upon reaching the foot of the pole. How much time (in seconds) passes between each successive splashing?

Answer: 2

Kinetic Theory Of Gases

Molecular Concept Of Matter And Its Applications

For centuries, scientists have studied the structure of matter—how it is formed and which are its smallest entities, possessing properties identical to that of the matter itself.

• In the early nineteenth century, Dalton and Avogadro for the first time proposed the theory of molecular structure of matter.
• A molecule is the smallest entity and matter is composed of molecules having all the chemical properties of the matter itself.
• The physical quantities related to molecules are number of molecules in a material body, molecular velocities, intermolecular distances, intermolecular forces, etc.
• These are the internal microscopic properties of a body. Unfortunately, these properties cannot be measured directly through experiments. As a result, we have to start with certain basic assumptions (postulates) about the behaviour of molecules.
• This gives us a picture of how molecules behave in a body. This is known as the molecular model. The next step is the application of Newton’s laws of motion on the molecules.

Now, the experimentally determined properties of a body as a whole, like pressure, temperature, and internal energy, are expected to be intimately related to the molecular model.

So Newton’s Jaws should give us expressions leading to these Bulk properties. This is essentially the object of the kinetic theory of matter.

• In short, the subject of study in which theoretical expressions of the bulk properties of a body are tamed from the application of Newton’s laws of motion on the internal molecular behaviour is called the kinetic theory of matter.
• Naturally, the values from the theoretical expressions of kinetic theory should match the experimental values. The formulas obtained from kinetic theory should be identical to the experimentally obtained thermodynamic formulas. This is actually the pre-condition for the success of kinetic theory.

This condition is beautifully obeyed in case of gases, leading to the very successful and advanced theory of the kinetic theory of gases. But this is not the case with liquids or solids. Partially successful molecular models exist for solids, but almost none so far has been developed for liquids.

Evidence of molecular motion: Molecules cannot be observed directly, but some natural phenomena clearly indicate the existence of molecular motion.

1. Diffusion: Let a gas jar filled with hydrogen gas be held upside down on another gas jar filled with carbon dioxide gas. Now the lids are removed. After an interval of time, it will be observed that the two gases will produce a homogeneous mixture in the two jars, ignoring gravitation.

• This phenomenon is called diffusion of the two gases. Clearly, molecular motion is evident in this phenomenon. Though carbon dioxide is heavier than hydrogen, CO₂ molecules move up and the hydrogen molecules move down to produce the mixture.
• The molecules of a gas randomly move at different velocities in all directions.
• So the molecules of the two gases mix and produce a homogeneous mixture. Diffusion takes place in liquids and solids also.
• If a few granules of copper sulphate are dropped at the bottom of a container filled with water, the blue colour gradually spreads throughout the whole volume of water.
• The density of copper sulphate solution is more than the density of water.
• But the solution moves up ignoring gravitation and after an interval of time, the whole mixture turns blue.
• It is an example of the motion of copper sulphate molecules diffusing into water. In a similar manner, solid phosphorus or boron can be diffused at high temperatures into solid silicon crystals to produce extrinsic semiconductors.
• In general, diffusion can be defined as the phenomenon by virtue of which movement of molecules occurs from a region of higher concentration to a region of lower concentration in a mixture till a homogeneity of concentration is established.

2. Vaporisation and vapour pressure: Liquid molecules are in motion inside the liquid. They move randomly at different velocities. Some molecules rise to the liquid surface with sufficient kinetic energy and overcome the attraction of other molecules inside the liquid.

• As a result, they may escape from the liquid. This phenomenon is called evaporation. Again, kinetic energies of molecules may be increased by applying heat. Hence, more molecules may escape from the liquid and vaporisation may occur.
• If a liquid is enclosed in a container, molecules leaving the liquid move randomly above the liquid surface. They collide with each other and hit the surface again and again. Some molecules may enter the liquid again. This leads to the vapour pressure corresponding to the vapour above the liquid surface.
• At a particular higher temperature, the kinetic energies of the molecules of the liquid become very high. In comparison, the potential energies due to intermolecular attractions become negligible. So the molecules are effectively free and all of them try to come out of the liquid at the same time. This phenomenon is called boiling of the liquid.

3. Expansion of gas: A gas spreads throughout the whole volume of its container. If the volume of the container increases, the gas spreads again to occupy the whole volume. This shows the property of random and unrestricted motion of the gas molecules.

4. Brownian motion: Very small, but still visible particles are often present as impurities in a liquid or in a gas. Observations through microscopes show that these particles move in a very random manner in all possible directions. This is known as Brownian motion.

This phenomenon can be explained by the concept of molecular motion. Molecules inside a liquid or a gas move randomly in all directions and collide time and again with small foreign particles (called Brownian particles). These collisions are directly responsible for the Brownian motion.

Kinetic Theory Of Gases Brownian Motion

British scientist, Robert Brown, first observed this continuous and irregular motion of the particles with a powerful microscope. He put some pollen grains in water.

• These grains, being very light, remained suspended in water. Brown noticed the random, continuous, and to and fro motion of these grains. However he was not able to determine the mechanisms that caused this motion.
• Albert Einstein published a paper in 1905 that explained in precise detail how the motion that Brown had observed was a result of the pollen being moved by individual water molecules.
• Brownian motion of a particle is shown. Colloids in a colloidal solution, very small feathers suspended in air, etc., are examples of randomly moving Brownian particles.

Explanation of the origin of Brownian motion: Just after the discovery of Brownian motion, scientists assumed that the reason of the origin of Brownian motion was a chemical reaction, irregular change of temperature, surface tension of liquid, etc. However from various experiments, it was proved that these explanations were not correct.

• It became possible to explain Brownian motion with the help of kinetic theory. We know that liquid or gas molecules are moving randomly and colliding with floating particles at every instant and from all directions.
• The force exerted on a floating particle of big size in any direction, due to some colliding molecules, is cancelled by the equal and opposite force due to some other molecules.
• As a result, the resultant force acting on the floating particle becomes zero and it has no Brownian motion. But if the floating particle is very tiny in size, the resultant force on it does not become zero as the colliding particles do not exert force equally from all directions.
• Hence, the floating particle moves along the direction of the resultant force and Brownian motion is observed. As molecular thrusts are random, the magnitude of the resultant force is not equal always.
• Also, the resultant force does not act always in the same direction. So the floating particle moves in a random manner in all possible directions.

Characteristics of Brownian motion:

1. The motion is perpetual, spontaneous, random and continuous. Itoo Brownian particles, even at close proximity, do not have identical motion.
2. The motion does not depend on the motion of the container.
3. The velocities of the particles increase with rise in temperature.
4. Smaller particles have higher average velocities.
5. The velocities of the particles become higher in liquids with lower viscosity.
6. The motion depends only on the mass and the size of the particle and not on the material it is made of.

Kinetic Theory Of Gases – Basic Assumptions Of Kinetic Theory Of Gases

A gas is made of atoms and molecules. The three variables—volume, pressure, and temperature, are all consequences of the motion of the molecules.

• The kinetic theory of gases relates the motion of the molecules to the volume, pressure, and temperature of the gas. Actually, the kinetic theory of gases explains the macroscopic properties of the gases, though it is a microscopic mode.
• Rudolf Clausius and James Clark Maxwell developed the kinetic theory of gases and explained gas laws in terms of motion of the gas molecules.
• In order, to formulate the kinetic theory of gases some simplifying assumptions are made about the behavior of the molecules of the ideal gas. The assumptions are

1. A gas is composed of a large number of molecules. For a particular gas, the molecules are identical. But they are different for different gases.
2. Every gas molecule behaves as a point mass. So the sum of their volumes is negligible compared to the volume of the container. The intermolecular space in the container is an empty space.
3. The molecules are in continuous and random motion in all possible directions. The value of molecular velocities varies from zero to infinity.
4. During random motion, the molecules collide with each other and with the walls of the container. These collisions are perfectly elastic. This means that the velocity changes due to collision, but the net momentum and kinetic energy remain unchanged.
5. The average distance between the molecules is sufficiently large, so that the attractive or repulsive forces between them are negligible except during collision. As a result,
   • Molecular motion is unrestricted and the gas spreads throughout the inner volume of its container
   • The potential energy of a molecule is negligible, the total energy comes from its kinetic energy only
   • Between two collisions, a molecule moves with uniform velocity (according to Newton’s first law of motion). The straight line path between two successive collisions is called a free path.
6. Every collision is instantaneous the time of a collision is negligible compared to the time taken by a molecule to describe a free path.
7. The gas is homogeneous and isotropic. This means that the properties of the gas in any small portion are identical to those in any other equivalent portion anywhere inside the container.

A gas obeying the properties outlined in these assumptions is called an ideal gas or a perfect gas. Real gases show some deviations from these properties. Real gases available to us are good approximation of an ideal gas at low pressure and high temperature.

The kinetic theoretical definitions of mass and volume of a gas are obtained directly from the above assumptions:

1. The mass of a gas is defined as the sum of the masses of the constituent molecules.
2. The volume of a gas is defined as the inner volume of the gas container.

Mean free path: The straight line path described by a molecule between two collisions is called a free path.

• The gas molecules move randomly. As a result, the lengths of the free paths vary in an irregular manner. So, to get a concrete picture, the Idea of a mean-free path becomes essential.
• It is the average distance that a molecule can travel between two successive collisions.

Mean free path Definition: The mean value of the distance travelled by a molecule between two successive collisions Is called the mean free path (λ).

• In other words, if a molecule suffers N number of collisions with other molecules when it travels through a total distance d, then the mean free path is, \(\lambda=\frac{d}{N}\).
• The basic assumptions of kinetic theory describe every molecule as a point mass. But the value of mean free path becomes theoretically infinite if the molecules are treated as geometrical points.
• So some deviations from the assumptions, are necessary. Every molecule is assumed to be a very small hard sphere of non-zero radius. Then it is possible to get an effective theoretical value of the mean free path.

An expression for the free path: Let σ = diameter of each molecule of the gas n = number of molecules per unit volume = number density of the molecules.

Let us consider a particular molecule of the gas moving with a velocity v at an instant of time. It will collide with all molecules whose centres come at a distance of cr or less along its line of motion. Now we choose a cylinder of radius cr and of length v. Clearly,

1. The axis of this cylinder is the path described by the chosen molecule in unit time, and
2. All other molecules, whose centers come within this cylinder, collide with the chosen molecule in that unit time.

The volume of this cylinder = cross-sectional area x length = πσ²v, and the number of molecules having centres in this volume = πσ²vn. So, the number of collisions per unit of time, N = πσ²vm.

The mean free path of the molecules is, therefore,

= \(\frac{\text { path described by a molecule }(d)}{\text { number of collisions }(N)}=\frac{\nu}{\pi \sigma^2 v n}\)

i.e., \(\lambda=\frac{1}{\pi \sigma^2 n}\)….(1)

This equation (1) shows that the mean free path (λ) of gas molecules is

1. Inversely proportional to the number of gas molecules (n) in a unit volume of the gas and
2. Inversely proportional to the square of molecular diameter (σ).

So, \(\lambda \propto \frac{1}{n} and \lambda \propto \frac{1}{\sigma^2}\),

i.e., \(\lambda \propto \frac{1}{\sigma^2 n} or, \lambda=\frac{k}{\sigma^2 n}\)

The constant k in equation (1) is \(\frac{1}{\pi}\). However, it has been estimated by different scientists by different other methods. The estimates give different results, but the value of k is always slightly greater or slightly less than 1. Then, the approximate expression for the mean free path is \(\lambda=\frac{1}{\sigma^2 n}\).

Kinetic Theory Of Gases – Mean Speed And Root Mean Square Speed of Gas Molecule

Molecular velocity is a vector quantity. The number of gas molecules in any container is very large. So the velocity vectors are oriented randomly in all possible directions.

• As a result, the resultant velocity vector must be zero. Consequently, the mean velocity of the molecules is also zero. Clearly, this zero value is useless as it gives no information about the order of magnitude of the molecular velocities.
• Alternatively, we may take the magnitudes only of the molecular velocities to calculate the mean. Certainly, it is non-zero and a useful quantity. We also know the molecules move in straight lines between collisions.
• So the magnitude of molecular velocity is actually the molecular speed. The calculated mean velocity is essentially the mean speed of the molecules. However, mean speed is often loosely termed as mean velocity.

Let N be the number of molecules of a gas in a closed container and, at any instant, c₁, c₂, c₃…..c_N be the magnitudes of velocities of the N molecules, respectively.

So, mean velocity or mean speed of the molecules, \(\bar{c}=\frac{c_1+c_2+c_3+\cdots+c_N}{N}\)

Mean square velocity of the molecules is defined as the mean of the squares of velocities,

⇒ \(\overline{c^2}=\frac{c_1^2+c_2^2+c_3^2+\cdots+c_N^2}{N}\)

Root mean square speed or rms speed of the molecules, defined as the square root of mean square speed,

c = \(\sqrt{\overline{c^2}}=\sqrt{\frac{c_1^2+c_2^2+c_3^2+\cdots+c_N^2}{N}}\)

The mean velocity \(\bar{c}\) and the rms speed c of the gas molecules in a container are not equal. For example, let us take three molecules with velocities 40 m · s^-1, 80 m · s^-1, and 120 m · s^-1.

Then, \(\bar{c}=\frac{40+80+120}{3}=80 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

c = \(\sqrt{\frac{(40)^2+(80)^2+(120)^2}{3}}=86.4 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

In general, the rms speed is slightly greater than the mean velocity. In kinetic theory, the role of the rms speed is comparatively more important than that of the mean velocity.

 

 

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Deduction Of Difference Gas Laws From Kinetic Theory Of Gases

We have so far developed the kinetic theory of gases to a certain stage. Now it Is possible to see that the theoretical result of kinetic theory matches exactly with the thermodynamic gas laws obtained from experiments. This proves the success of the kinetic theory of gases.

1. Boyle’s law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V . \text { So, } p V \propto T\). If T = constant for a certain amount of gas, then pV = constant. This is Hoyle’s law.

2. Charles’ law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So } p V \propto T\). If p = constant for a certain amount of gas, then V ∝ T, This is Charles’ law.

3. Pressure law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So } p V \propto T\). If V = constant for a certain amount of gas, then p ∝ T. This is Charles’ law of pressure.

4. Joule’s law: According to the kinetic theory of gases, E = 3/2 RT. So E is a function of temperature only; it does not depend on the volume or pressure of the gas. This is Joule’s law or Mayer’s hypothesis, as discussed In the chapter First and Second Law of Thermodynamics,

5. Avogadro’s law: Let equal volumes of two gases be taken at the same pressure and temperature. The pressure, temperature, and volume are p, T, and V, respectively.

Now for the first gas,

N₁ = total number of molecules in the container,

m₁ = mass of each molecule,

c₁ = rms speed of the molecules.

Then, \(n_1=\frac{N_1}{V}\) = number of molecules per unit volume. According to kinetic theory, the pressure of the gas,

p = \(\frac{1}{3} m_1 n_1 c_1^2=\frac{1}{3} m_1 \frac{N_1}{V} c_1^2\)

Pressure being the same, so we get similarly for the second gas,

p = \(\frac{1}{3} m_2 \frac{N_2}{V} c_2^2 .\)

So, \(\frac{1}{3} m_1 \frac{N_1}{V} c_1^2=\frac{1}{3} m_2 \frac{N_2}{V} c_2^2\)

or, \(m_1 N_1 c_1^2=m_2 N_2 c_2^2\)…..(1)

Again, the temperature of the two gases is the same. So the average kinetic energy of a molecule is equal for the two gases. This means that

⇒ \(\frac{1}{2} m_1 c_1^2=\frac{1}{2} m_2 c_2^2 \text { or, } m_1 c_1^2=m_2 c_2^2\)….(2)

From relations (1) and (2), we get N₁ = N₂. So, equal So, pV = RT (1 mol of an ideal gas) volumes of different gases, at the same pressure, and Equation (4) is known as the ideal gas equation temperature, contains an equal number of molecules. This is Avognclro’s law.

6. Dalton’s law of partial pressure: The pressure of a gas mixture on the walls of its container is equal to the sum of the partial pressures exerted by constituent gases separately, at same temperature as that of the mixture, provided that the gases do not react chemically with each other—this is Dalton’s law.

Let several gases be mixed in a closed container. Their densities are, ρ₁, ρ₂, ρ₃…… and the molecular rms speeds are c₁, c₂, c₃ ….. respectively. Every gas molecule moves with its own kinetic energy, which does not depend on the motion of the other molecules. So, the net pressure on the container will be the sum of the pressures exerted by all individual molecules. Then, the pressure of the gas mixture is

p = \(\frac{1}{3} \rho_1 c_1^2+\frac{1}{3} \rho_2 c_2^2+\frac{1}{3} \rho_3 c_3^2+\cdots\)….(3)

= \(p_1+p_2+p_3+\cdots\)…..(3)

Here, p₁, p₂, p₃,……. are the partial pressures of the first second, third, … gases on the walls of the container. So equation (3) expresses Dalton’s law of partial pressure.

7. Graham’s law of diffusion: The rate of diffusion of a gas in a mixture is inversely proportional to the square root of the density of the gas—this is Graham’s law.

Let the densities of two gases be ρ₁ and ρ₂ and the rms velocities of the molecules be c₁ and c₂, respectively. The gases are allowed to diffuse with each other at the same temperature and same pressure. The diffusion is due to the motion of molecules so the rate of diffusion is clearly proportional to the rms speed of the molecules.

Now, p = \(\frac{1}{3} \rho_1 c_1^2=\frac{1}{3} \rho_2 c_2^2\)

or, \(\rho_1 c_1^2=\rho_2 c_2^2 or, \frac{c_1}{c_2}=\sqrt{\frac{\rho_2}{\rho_1}}\).

As, rate of diffusion r  ∝ c, we have \(\frac{r_1}{r_2}=\frac{c_1}{c_2}, i.e., \quad \frac{r_1}{r_2}\)= \(\sqrt{\frac{\rho_2}{\rho_1}}\)

or, \(r \propto \frac{1}{\sqrt{\rho}}\).

This is Graham’s law.

8. Ideal gas equation: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So, } p V \propto T \text {. }\). Then, pV = kT, where k is a constant. For 1 mol of an ideal gas, this constant k is called the universal gas constant, denoted by R.

So, pV= RT (1 mol of an ideal gas)…(4)

Equation (4) is known as the ideal gas equation.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Limitations Of Ideal Gas Laws

If a gas obeys Boyle’s law and Charles’ law accurately, then it is called an ideal gas. But in practice, real gases do not obey these ideal gas conditions in all circumstances.

Usually, a real gas behaves as an ideal gas at high temperatures and low pressures. But at low temperatures and high pressures, the behaviour deviates from that of an ideal gas.

The equation of state for 1 mol of an ideal gas is pV = RT. A real gas deviates from this equation chiefly due to the following two reasons:

1. The kinetic theory assumes that gas molecules are point masses and the volume of the molecules is negligible compared to the total volume of the gas. But every molecule, however small, has a finite volume. So their total volume is not always negligible.
   • In particular, at low temperatures and high pressures, the volume of a gas is comparatively small. In this case, the volume of the gas molecules becomes an important factor.
   • The effective volume for the motion of the molecules inside the gas container of volume V is reduced by an amount b (say). Then the equation of state for 1 mol of the gas becomes p(V-b) = RT.
2. The kinetic theory further assumes that the molecules do not attract one another. But in particular, at low temperatures and high pressures, the molecules come closer to one another.
   • As a result, the attractive forces are no longer negligible. Now, consider a molecule near the wall of the container. It experiences a resultant force towards the interior due to attractions by the other molecules. So it collides with the wall at a comparatively less velocity.

As a result, the pressure on the wall becomes less. Let p’ be the reduction of the pressure on the wall due to all these molecules, van der Waals established that \(p^{\prime}=\frac{a}{v^2}\) where a is a constant for a particular gas. Now, if p is the effective pressure on the wall, then the pressure would be \(p+p^{\prime} \text { or } p+\frac{a}{V^2}\) considering that the gas was ideal.

So the equation of state for 1 mol of a real gas becomes \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\)….(1)

Equation (1) is called the van der Waals equation for a real gas. The constants a and b are known as van der Waals constants; their values depend on the nature of the gas. For n mol of a real gas, let V be its volume. Then molar volume = V/n.

Then, we get \(\left\{p+\frac{a}{(V / n)^2}\right\}\left\{\frac{V}{n}-b\right\}=R T\)

or, \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T\)…(2)

This is the form of van der Waals equation for n mol of a real gas. It is to be noted that the van der Waals equation is only one of several equations of state of real gases, proposed by different scientists.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Degree Of Freedom (DOF)

Degree Of Freedom Definition: The minimum number of independent coordinates necessary to specify the instantaneous position of a moving body is called the degree of freedom of the body.

Degree Of Freedom Example:

1. Let us consider the motion of a particle falling freely under gravity. We take the initial point as the origin and the downward line as the z-axis. Then the position of the particle at any instant is specified by the z-coordinate only.

So, the number of degrees of freedom of the particle is 1. Essentially, the degree of freedom of every one-dimensional motion is, for example, the motion of a car along a road, an ant moving on a stationary rope, etc.

2. Motions of projectiles under gravity, orbital motion of planets around the sun, circular motion, motion of an ant on the floor of a room, etc., are examples of two-dimensional motions, in each case, any one point on the plane, Is chosen as the origin and two perpendicular axes x and y are considered. Then the position of the object at any Instant Is specified by the x- and y- coordinates. So, the number of degrees of freedom in two-dimensional motion Is 2.

3. According to the kinetic theory, all gas molecules of an Ideal gas are point masses and they are In a completely random to-and-fro motion. At least three coordinates (say x, y, z) are necessary to specify the position of a molecule at any Instant.

So, the number of degrees of freedom of an Ideal gas molecule is 3. Essentially, every three-dimensional motion has 3 degrees of freedom, for example, Brownian motion, motion of a fly in a room, etc.

• In the above examples, the particle, the object, and the gas molecule all are considered to be as a point mass, which cannot undergo rotation, if the body is rigid and has a finite size, it can undergo rotation also, about any axis, So, a rigid body will have degrees of freedom both due to its translatory motion and rotatory motion.
• Like translatory motion, the rotatory can also be resolved into three mutually perpendicular components. Thus a rigid body has six degrees of freedom, 3 for translatory motion and 3 for rotatory motion.
• Now, let us consider a system of two particles or two-point masses. Each particle has three degrees of freedom, so the system has six degrees of freedom. If the two particles remain at a fixed distance from each other, then there is one definite relationship between them.
• These definite relationships are known as constraints. As a result, the number of independent coordinates required to describe the configuration of the system is reduced by one. Hence, the system has (6-1) = 5 degrees of freedom.

These 5 degrees of freedom may be interpreted in another way: degrees of freedom for the translatory motion of the centre of mass = 3 and degrees of freedom for the rotatory motion of the two particles around the center of mass = 2.

In a system consisting of N particles, if the particles possess k independent relations i.e., constraints among them, then the number of degrees of freedom of the system is given by, f = 3N – k.

Degrees of freedom of different types of gases:

1. Monatomic gas: The molecule of a monatomic gas (for example, neon, helium, argon, etc.) consists of a single atom (a point mass). At least three coordinates (say x, y, z) are necessary to specify the position of the molecule at any instant in the three-dimensional space. So, the number of degrees of freedom of a monatomic gas molecule, f = 3×1-0 = 3.

2. Diatomic gas: The molecule of a diatomic gas like hydrogen, oxygen, nitrogen etc. has two atoms in it. Two point atoms have a fixed distance between them (neglecting the vibration of the atoms in the molecules)

i. e., the number of constraints is 1. Here, N = 2 and k = 1

∴ f = 3×2-1 = 5

3. Triatomic gas: Triatomic gas molecules are of two types:

(1) In a linear molecule such as CO₂, CS₂ > HCN, etc. the three atoms are arranged in a straight line. The number of independent relations between them is only two.

∴ f = 3×3-2 – 7 i.e., such a molecule has seven degrees of freedom.

2. In a non-linear molecule like H₂O, SO₂, etc. the three atoms are located at the three vertices of a triangle. Hence, there are three fixed distances among the three atoms.

∴ f = 3×3-3 = 6

Therefore, a non-linear triatomic molecule has six degrees of freedom.

Degrees of freedom in different cases:

 

 Kinetic Theory Of Gases Conclusion

The subject of study in which theoretical expressions on the bulk properties of a body are obtained from the application of Newton’s laws of motion on the internal molecular behavior is called the kinetic theory of matter. This approach has been highly successful for gases.

• The random and perpetual motion of very small particles present as impurities in a liquid or gas is known as Brownian motion. This happens due to random collisions of the Brownian particles with the molecules in matter. This motion furnishes evidence of the molecular model.
• The kinetic theory of gases relates the motion of the molecules of the gas to the volume, pressure, and temperature of the gas. Actually, this theory explains the macroscopic properties of the gases with the help of microscopic parameters.

Three basic assumptions of the kinetic theory of gases are—

1. Molecules are point masses; the intermolecular space is much larger than the space occupied by the molecules.
2. The attraction between molecules is negligible. As a result, the molecular potential energy is zero, i.e., the energy is purely kinetic.
3. Molecular motion is continuous and random. A molecule may have any velocity between zero and infinity in any direction. A molecule may have any velocity between zero and infinity in any direction.

The straight-line path described by a molecule between two successive collisions is called a firepath. The mean value of the lengths of different free paths of different molecules of a gas is called the mean free path.

The pressure of a gas in a container depends on

1. The mass of a molecule,
2. The number of molecules in unit volume and
3. The average velocity of the molecules.

Temperature is a property of a gas, which is proportional to the total kinetic energy of the gas molecules.

Absolute zero temperature is the temperature at which the internal energy of the gas becomes zero, i.e., the molecular motion stops entirely.

The velocity which is possessed by the highest number of gas molecules in a container is called the most probable velocity.

Gases obeying Charles’ law and Boyle’s law perfectly are called ideal or perfect gases.

In general, real gases deviate from this ideal behavior, especially at

1. High pressures and
2. Low temperatures.

The minimum number of independent coordinates necessary to specify the instantaneous position of a moving particle is called the degree of freedom of the tire particle.

Principle of equipartition of energy: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a molecule per degree of freedom is 1/2 kT (T = absolute temperature and k = Boltzmann constant (1.38 x 10^-23 J · K^-1).

Kinetic Theory Of Gases Useful Relations For Solving Numerical Examples

Mean free path of a gas molecule (λ):

1. \(\lambda=\frac{d}{N}\) where d = total distance traveled by a molecule, N = total number of collisions suffered by that molecule through the distance d.

2. \(\lambda=\frac{1}{\pi \sigma^2 n}\) where, σ = diameter of each molecule of the gas, n = the number of molecules per volume i.e., the number density of the molecule.

Let c₁, c₂,…..c_N be the magnitudes of instantaneous velocities of N molecules in a gas. Then,

1. Mean velocity, \(\bar{c}=\frac{c_1+c_2+\cdots+c_N}{N}\)

2. Root mean square speed or rms speed, \(c=\sqrt{\frac{c_1^2+c_2^2+\cdots+c_N^2}{N}}\)

At absolute temperature T,

\(\begin{array}{|c|c|c|}
\hline \begin{array}{c}
\text { Average speed of } \\
\text { a gas molecule } \\
(c)
\end{array} & \begin{array}{c}
\text { rms speed of a } \\
\text { gas molecule } \\
(c)
\end{array} & \begin{array}{c}
\text { Most probable } \\
\text { speed of a gas } \\
\text { molecule }\left(c_m\right)
\end{array} \\
\hline \sqrt{\frac{8 k T}{\pi m}} & \sqrt{\frac{3 k T}{m}} & \sqrt{\frac{2 k T}{m}} \\
\hline
\end{array}\)

 

∴ \(\bar{c}: c: c_m=\frac{2}{\sqrt{\pi}}: \sqrt{\frac{3}{2}}: 1\)

Pressure of an ideal gas, p = \(\frac{1}{3} \rho c^2=\frac{1}{3} m n c^2\), where m- mass of a molecule, n = number of molecules in unit volume and ρ = mn = density.

The internal energy of a gas in a container, which is equal to the sum of the kinetic energies of the molecules, is

U = E = \(\frac{3}{2} p V ; \quad \text { So, } p=\frac{2}{3} \frac{E}{V}=\frac{2}{3} u \text {, }\)

where u = energy per unit volume.

• If M is the molecular weight of a gas, the rms speed of the molecules is related to temperature T of the gas by the relation, \(c=\sqrt{\frac{3 R T}{M}} \quad \text { or, } \quad c \propto \sqrt{T}\)
• The equation of state for 1 mol of an ideal gas is p V = R T.
• For real gases, volume and pressure corrections lead to the van der Waals equation of state:

⇒ \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\)

• For n mol of real gas, van der Waals equation of state is \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T\) [where, a, b are small positive constants ‘a’ is related to the average force of attraction between the molecules and ‘b’ Is related to the total volume of the molecules.
• The average kinetic energy of a gas molecule, e = \(\frac{3}{2} k T\)
• If some amount of gas contains N molecules, the total internal energy i.e., total kinetic energy, E = \(\frac{3}{2}Nk T\)
• For 1 mol of a gas, N = N_A = Avogadro’s number.
• Then, \(E=\frac{3}{2} N_A k T=\frac{3}{2} R T \text {, where } R=N_A k \text {. }\)
• Molar specific heat at constant volume, \(C_v=\frac{d Q}{d T}=\frac{d E}{d T}\)

If n₁ mol of a gas is mixed with n₂ mol of another gas (they do not react with each other), then,

1. Molar specific heat of the mixture at constant volume, \(C_\nu=\frac{n_1 C_{\nu_1}+n_2 C_{\nu_2}}{n_1+n_2}\) where, \(C_{\nu_1}\) = molar specific heat at constant pressure of 1st and 2nd gases respectively.

2. Molar specific heat of the gas mixture at constant pressure, \(C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\) where, \(C_{p_1}\) and \(C_{p_2}\) are the molar specific heat at constant pressure of 1st and 2nd gases respectively.

3. The ratio of Cp and Cv for the mixture is,

⇒ \(\gamma=\frac{C_p}{C_v}=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{v_1}+n_2 C_{v_2}}\)

For 1 mol of a monatomic gas, \(C_\nu=\frac{d E}{d T}=\frac{3}{2} R ; C_p=C_\nu+R=\frac{5}{2} R ; \gamma=\frac{C_p}{C_\nu}=\frac{5}{3}\)

In general cases, if f is the number of degrees of freedom of an ideal gas molecule then \(C_\nu=\frac{f_2}{2} R ; C_p=\left(1+\frac{f}{2}\right) R ; \gamma=\frac{C_p}{C_\nu}=1+\frac{2}{f}\)

Kinetic Theory Of Gases Very Short Answer Type Questions

Question 1. What is the name of the smallest entity of matter that exhibits all the properties of that matter?
Answer: Molecule

Question 2. Which one of the following has the highest intermolecular force—solid, liquid, or gas?
Answer: Solid

Question 3. If a gas jar filled with a light gas like hydrogen is held side down on another gas jar filled with carbon dioxide, it is observed that the two gases produce a homogeneous mixture in the jars. What is the name of this process?
Answer: Diffusion

Question 4. How does the velocity of Brownian particles change due to the movement of the vessel?
Answer: No change occurs

Question 5. What is the direction of velocities of gas molecules according to the kinetic theory of gases?
Answer: All possible directions

Question 6. Gas molecules collide with each other and with the walls of the container. What is the type of these collisions? .
Answer: Perfectly elastic

Question 7. What do you call the straight line path described by a gas molecule between two successive collisions?
Answer: Free path

Question 8. Both vaporization and vapor pressure prove the ________ of molecules of a liquid.
Answer: Mobility

Question 9. Brownian motion supports the _______ of the matter.
Answer: Kinetic theory

Question 10. In Brownian motion in a medium, if the particles decrease in size, how does their velocity vary?
Answer: Increases

Question 11. Does the velocity of the particles increase or decrease, when the viscosity increases in Brownian motion in a medium?
Answer: Decreases

Question 12. Which gases obey the basic assumptions of the kinetic theory?
Answer: Ideal gases

Question 13. According to the kinetic theory of gases, every gas molecule behaves as a ________
Answer: Point mass

Question 14. According to the kinetic theory of gases, the velocity of gas molecules varies from ______ to ______
Answer: zero, infinity

Question 15. Which is greater rms speed or mean velocity?
Answer: rms speed

Question 16. Does the pressure of a gas increase or decrease when the velocity of the gas molecules increases?
Answer: Increases

Question 17. On which other factor does the pressure of a gas depend, besides the number of molecules per unit volume and the temperature of the gas?
Answer: Mass of the gas molecules

Question 18. Is the most probable velocity of gas molecules higher or lower than the mean velocity?
Answer: Lower

Question 19. Will the rms speed of oxygen and hydrogen gas molecules be the same at equal temperatures?
Answer: No, hydrogen gas will have a higher rms speed

Question 20. What is the ratio of the rms speeds of O₃ and O₂ at a certain temperature?
Answer: √2: √3

Question 21. By how many times will the pressure of a gas kept in a gas container of constant volume increase to double the rms speed of the gas molecules?
Answer: 4 times

Question 22. The velocities of the three gas molecules are 4cm · s^-1, 8 cm · s^-1, and 12 cm · s^-1, respectively. Calculate their rms speed.
Answer: 8.64 cm · s^-1

Question 23. What is the relation between rms speed and molecular mass of a gas?
Answer: rms speed is inversely proportional to the square root of the molecular mass of that gas

Question 24. Hydrogen and oxygen gases are kept in two vessels at the same temperature and pressure. What is the ratio of the rms speed of their molecules?
Answer: 4:1

Question 25. If a gas molecule of mass m and velocity u collides perpendicularly with a wall of a container, what will be the value of momentum of the molecule after the collision?
Answer: mu

Question 26. If n number of molecules, each having mass m and velocity u, perpendicularly hit the walls of a container in every second, what will be the value of the applied force?
Answer: 2mnu

Question 27. In the kinetic theory of gases, _____ is more important than mean velocity.
Answer: rms speed

Question 28. What is the name of the force that acts among the molecules of matter?
Answer: Intermolecular force

Question 29. Under which conditions do real gases behave as ideal gases?
Answer: At low pressure and high temperature

Question 30. Which property of a gas is proportional to the net internal energy of the gas molecules?
Answer: Temperature of the gas

Question 31. At which temperature does the kinetic energy of gas molecules become zero?
Answer: Absolute zero temperature

Question 32. To which gases is the van der Waals’ equation applicable?
Answer: Real gases

Question 33. If the temperature of a gas is increased at constant volume, how will the number of collisions of the molecules per unit time change?
Answer: Increase

Question 34. What is the dimension of an in van der Waals’ equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T?\)
Answer: ML^-3T^-2

Question 35. What is the dimension of b in van der Waals’ equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T?\)
Answer: L³

Question 36. What is the relation between the pressure p of a gas and its energy density?
Answer: \(\left[p=\frac{2}{3} u\right]\)

Question 37. According to the kinetic theory of gases, as there is no attractive force between the gas molecules, the entire energy of them is ______
Answer: Kinetic energy

Kinetic Theory Of Gases  Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 Is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The root mean square speeds of the molecules of different ideal gases at the same temperature are the same.

Statement 2: The average translational kinetic energy of molecules of a different ideal gas is same at the same temperature.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: The rms speed of oxygen molecules (O₂) at an absolute temperature T is c. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed remains unchanged.

Statement 2: The rms speed of the molecules of a gas is directly proportional to √T/M.

Answer: 1. Statement 1 is true, and statement 2 Is true statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume.

Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: The mean free path of gas molecules, varies inversely with the density of the gas.

Statement 2: The mean free path of gas molecules is defined as the average distance traveled by a molecule between two successive collisions.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The following show \(\frac{pV}{T}\) versus p graph for a certain mass of O₂ gas at two temperatures T₁ and T₂. It follows from the graph that T₁ > T₂.

Statement 2: At higher temperatures, real gas behaves more like an ideal gas.

Answer: 1. Statement 1 is true, and statement 2 Is true statement 2 is a correct explanation for statement 1.

Question 6.

Statement 1: For an ideal gas, at a constant temperature, the product of the pressure and the volume is constant.

Statement 2: The mean square velocity of die molecules is inversely proportional to mass.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 7.

Statement 1: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume.

Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Kinetic Theory Of Gases Match Column 1 with Column 2

Question 1.

Answer: 1. C, 2. B, D 3. A, 4. A

Question 2.

Answer: 1. C, 2. B, 3. A

Question 3.

R = universal gas constant, f = number of degrees of freedom, T = temperature.

Answer: 1. C, 2. B, 3. A

Question 4. Match the following columns according to the graph.

Answer: 1. C, 2. A, 3. B

Kinetic Theory Of Gases Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. The pressure exerted by an ideal gas is p = \(\frac{1}{3} \frac{M}{V} c^2\), where the symbols have their usual meanings. Using standard gas equation, pV = RT, we find that \(c^2=\frac{3 R T}{M} \quad \text { or } \quad c^2 \propto T\) Average kinetic energy of translation of 1 mol of gas = \(\frac{1}{2} M c^2=\frac{3 R T}{2}\)

1. Average thermal energy of a helium atom at room temperature (27°C) is (given, Boltzmann constant k= 1.38 x 10^-23 J · K^-1)

1. 2.16 x 10²¹ J
2. 6.21 x 10²¹ J
3. 6.21 x 10^-21 J
4. 6.21 x 10^-23 J

Answer: 3. 6.21 x 10²¹ J

2. Average thermal energy of 1 mol of helium at this temperature is (given, gas constant for 1 mol = 8.31 J · mol^-1 · K^-1)

1. 3.74 x 10³ J
2. 3.74 x 10^-3 J
3. 3.47 x 10⁶ J
4. 3.47X10^-6 J

Answer: 1. 3.74 x 10³ J

3. At what temperature, when pressure remains unchanged, will the rms speed of hydrogen double its value at STP?

1. 819 K
2. 819 °C
3. 1000K
4. 1000°C

Answer: 2. 819 °C

4. At what temperature, when pressure remains unchanged, will the rms speed of a gas be half its value at 0°C?

1. 204.75 K
2. 204.75 °C
3. -204.75 K
4. -204.75 °C

Answer: 4. -204.75 °C

Question 2. If c₁, c₂, c₃ …. are random speeds of gas molecules at a certain moment then average velocity c_av = \(\frac{c_1+c_2+c_3+\cdots+c_n}{n}\) and root mean square speed of gas molecules,

⇒ \(c_{\mathrm{rms}}=\sqrt{\frac{c_1^2+c_2^2+c_3^2+\cdots+c_n^2}{n}}=c.\)

Further, c² ∝ T or, c ∝ √T.

At 0 K, c =0, i.e., molecular motion stops,

1. If three molecules have velocities 0.5km · s^-1, 1 km · s^-1 and 2 km · s^-1, the ratio of rms speed and average velocity is

1. 0.134
2. 1.34
3. 1.134
4. 13.4

Answer: 3. 1.134

2. The temperature of a certain mass of a gas is doubled. The rms speed of its molecules becomes n times, where n is

1. \(\sqrt{2}\)
2. 2
3. \(\frac{1}{\sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 1. \(\sqrt{2}\)

3. KE per molecule of the gas in the above question becomes x times, where x is

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. 4
4. 2

Answer: 4. 2

4. KE per mole of hydrogen at 100°C (given R = 8.31 J · mol^-1 · K^-1) is

1. 4946 J
2. 4649 J
3. 4496 J
4. 4699J

Answer: 2. 4649 J

5. At what temperature, when pressure remains constant, will the rms speed of the gas molecules be increased by 10% of? rms speed at STP?

1. 57.3 K
2. 57.3 °C
3. 557.3 K
4. -57.3 °C

Answer: 2. 57.3 °C

Question 3. A cubical box of side lm contains helium gas (atomic weight = 4) at pressure 10 N • m^-2. During an observation time of 1s, an atom traveling with rms speed parallel to one of the edges of the cube was found to make 500 hits with a particular wall without any collision with other atoms.

1. Evaluate the temperature of the gas.

1. 125K
2. 160K
3. 181K
4. 185K

Answer: 2. 160K

2. Evaluate the average kinetic energy per atom.

1. 3.31 x 10^-21 J
2. 3.75 x 10⁶ J
3. 3.81 x 10^-15 J
4. 3.22 x 10³ J

Answer: 1. 3.31 x 10^-21 J

3. Evaluate the total mass of the helium gas in the box.

1. 9×10^-4 kg
2. 5x 10^-3 kg
3. 3 x 10^-4 kg
4. 7 x 10^-3 kg

Answer: 3. 7 x 10^-3 kg

Kinetic Theory Of Gases Integer Answer Type Questions

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. Two identical cylinders contain helium at 3.5 standard atmospheres and argon at 2.5 standard atmospheres respectively. If both these gases are filled in one of the cylinders, what would be the pressure of the mixture?
Answer: 6

Question 2. The rms speed of molecules of a gas at -73 °C and 1 standard atmosphere pressure is 100 m · s^-1. The temperature of the gas is increased to 527°C and pressure is doubled. The rms speed becomes k times. What is the value of k?
Answer: 2

Question 3. The density of a gas is 6 x 10^-2 kg· m^-3 and the root mean square speed of the gas molecules is 500 m · s^-1. The pressure exerted by the gas on the walls of the vessel is n x 10³ N · m^-2. Find the value of n.
Answer: 5

Question 4. A gas has molar heat capacity c = 37.55 J · mol^-1 · K^-1, in the process pT = constant. Find the number of degrees of freedom of the molecules of the gas.
Answer: 5

Question 5. A vessel has 6g of hydrogen at pressure p and temperature 500 K. A small hole is made in it so that hydrogen leaks out How much hydrogen (in g) leaks out if the final pressure is \(\frac{p}{2}\) and temperature falls to 300K?
Answer: 1

 

Energy

A body can do work by the transfer of some amount of energy.

Energy Definition: The total amount of work that a body can do is the same as its total amount of energy. Hence, energy and work are equivalent physical quantities, having the same units and dimensions. Both are scalar quantities.

Energy is manifested in nature in different forms:

1. Mechanical energy
2. Heat energy
3. Light energy
4. Sound energy
5. Magnetic energy
6. Electrical energy
7. Chemical energy
8. Atomic energy

In all-natural events, energy changes from one form to another. In fact, every natural phenomenon can be interpreted as a sequence of energy transformations.

Energy Examples:

1. In an electric bulb, electrical energy is converted into heat energy at first, and then into light energy.
2. In an electric fan, electrical energy changes into mechanical energy.
3. In a telephone, sound energy is converted into electrical energy (during speaking), and electrical energy into sound energy (during listening).
4. In hydroelectric plants, water stored in a high reservoir flows downwards and its potential energy changes into kinetic energy. This kinetic energy rotates the turbines which generate electrical energy.
5. Water from sea, river and other water bodies evaporate and forms clouds. This water eventually returns to the earth in the form of rain. The potential energy of the clouds converts into the kinetic energy of the raindrops.
6. On burning coal, chemical energy is converted into heat and light energy.

We observe such numerous incidents of energy transformations in daily life. But neither energy is created nor destroyed in such incidents. Energy can only get transferred from one body to another or can change its form. As soon as a body loses energy, another body gains it in equal amounts. Hence the total energy in the universe remains the same. This is the law of conservation of energy.

Law Of Conservation Of Energy: Energy cannot be created or destroyed. It can only be converted from one form to another.

• It means that the total energy in the universe remains constant.
• After the introduction of Einstein’s theory of relativity, the law of conservation of energy was corrected and the law of conservation of mass energy was established.
• This means that mass and energy in the universe are not conserved individually.
• Only mechanical energy is discussed in detail in the following sections.

Work And Energy – Conservation Force

Conservation Force Definition: A system in which the total mechanical energy i.e., the sum of its kinetic and potential energy [remains conserved, is called a conservative system. The forces acting in such a system are called conservative forces.

Conservation Force Example: Gravitational force, restoring force of spring or elastic force, electrostatic force, force between the magnetic poles, etc., are conservative forces.

Work is done to lift a body against gravity and is stored as potential energy. While returning to its initial position, the body uses this stored energy and does work that is exactly equal to the initial work done while lifting. Hence energy is conserved.

Work Done In A Dosed Path Under A Conservative Force: For a conservative system, the work done depends only on the initial and final positions. It does not depend on the path taken in reaching the final position from the initial position.

• This means that the sum of the kinetic and potential energies remains constant throughout.
• It also implies that if a body goes around a complete loop so that its final and initial positions are the same, then the total work done is zero. For any conservative force, the work done is reversible.
• Hence if work done against a force can be restored, the force is called conservative. Also if work done by a force in a closed path equals to zero, the force is called conservative.

Thus in a conservative system, it is possible to restore the initial work done.

Suppose a body gets displaced by dx along the x-axis under the action of a conservative force F (which may vary with position) also acting along the x-axis. Therefore the work done by the conservative force dW = F(x)dx.

Since change in potential energy is negative of work done, then change in potential energy for this displacement dx is given by dU = -F(x)

In a three-dimensional reference frame, this equation is dU = \(-\vec{F} \cdot d \vec{r}\)….(1)

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System

Suppose a conservative force \(\vec{F}\), acting on a particle of mass m moves, it from point A to point B.

If the velocity of the particle at any point of its motion in the force field is v, then \(\vec{F} \cdot d \vec{r}=m \frac{d \vec{v}}{d t} \cdot \vec{v} d t=m \vec{v} \cdot d \vec{v}=m d\left(\frac{1}{2} v^2\right)\)

∴ Work done, to take the particle from A to B, by force \(\vec{F}\),

W = \(\int_A^B \vec{F} \cdot d \vec{r}=\left[\frac{1}{2} m v^2\right]_A^B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(1)

where v_A and v_B are the velocities of the particle at A and B respectively.

Equation (1), we know \(\vec{F} \cdot d \vec{r}=-d U\)

∴ \(\int_A^B \vec{F} \cdot d \vec{r}=-\int_A^B d U=U_A-U_B\)

= difference in potential energy between the points A and B…(2)

Comparing (1) and (2), \(U_A-U_B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(2)

Equation (3) establishes that in a conservative field when the kinetic energy of a particle increases, its potential energy decreases.

∴ \(\frac{1}{2} m v_A^2+U_A=\frac{1}{2} m v_B^2+U_B\)…(4)

If K_A and K_B are the kinetic energies of the particle at A and B respectively, then

K_A+U_B = K_B+ U_B or, K + U = constant (E).

Hence in a conservative field i.e., in a conservative system (where no dissipation of energy occurs due to forces like friction, etc.), at each point sum of potential energy and kinetic energy remains constant.

In other words, total mechanical energy (E) of a particle in a conservative system remains constant. This is the statement of the principle of conservation of mechanical energy.

Total mechanical energy In a free fall under gravity remains constant: Let a body of mass m be at rest at point A. DE, the earth’s surface is taken as the reference plane and the height of point A from DE is h. As the body is at rest it has potential energy only and no kinetic energy.

Potential energy at A = mgh

Kinetic energy at A = 0

Total mechanical energy at A = mgh + 0 = mgh

Now the body is released and it starts falling freely. When the body reaches a point B, it acquires a velocity v.

At B, the body has both potential energy (due to its height) and kinetic energy (due to its motion).

Let AB = x

∴ Potential energy at B = mg(h- x)

Kinetic energy at B = \(\frac{1}{2} m v^2=\frac{1}{2} m \cdot 2 g x\)

= mgx (because \(v^2=u^2+2\) as and u=0)

∴ Total mechanical energy at B = mg(h-x) + mgx = mgh

= total energy at A.

Also at C, when the body is about to touch the reference plane, its potential energy becomes zero. If the velocity acquired at that point is V, then kinetic energy at C

= \(\frac{1}{2} m V^2=\frac{1}{2} m \cdot 2 g h\left[because V^2=2 g h\right]\)

= mgh

Total mechanical energy at C = mgh + 0 = mgh

∴ Total energy at A = total energy at B = total energy at C.

Therefore, the total mechanical energy (= potential energy + kinetic energy) of a freely falling body, under the action of i gravity, remains conserved at all positions.

As the body touches the ground, both its potential energy and kinetic energy become zero; the entire mechanical energy transforms into heat, sound, and other forms of energy.

Total Mechanical Energy Of A Body, Falling Under Gravity Along A Frictionless Inclined Surface, Remains Constant: Let a body of mass m be at rest at a point A on a frictionless plane of inclination θ. Earth’s surface CD is taken as the reference plane.

Let the height of point A from the reference plane, GA = h. Being at rest, the body has no kinetic energy at A.

The potential energy at A = mgh Kinetic energy at A = 0

∴ Total mechanical energy at A = mgh + 0 = mgh

On releasing the body, it falls along the incline and reaches a point B such that, AB = x. Let the velocity of the body at B be v.

Acceleration of the body along the incline, a = gsinθ.

∴ \(v^2=0+2 g \sin \theta \cdot x=2 g x \sin \theta \quad \text { [as } v^2=u^2+2 a s \text { ] }\)

∴ Kinetic energy at \(B=\frac{1}{2} m v^2=\frac{1}{2} \cdot m \cdot 2 g x \sin \theta\)

= mgx sinθ

The perpendicular height of B from CD plane = FB = GE = GA-EA = h-x sinθ

(because \(\sin \theta=\frac{E A}{A B} \quad \text { or, }E A=A B \sin \theta=x \sin \theta\))

∴ The potential energy at B = mg(h-x sinθ)

∴ Total mechanical energy at B

= mg(h- xsinθ) + mgx sinθ

= mgh = total energy at A

At point C, i.e., where the body just touches the plane CD, it has no potential energy. If the velocity of the body at that moment is V, then

V² = 0 + 2g sinθ · AC = 2gsinθ · AC

and hence, kinetic energy at C = \(\frac{1}{2}\) m · 2g sinθ · AC = mgh

[sinθ = \(\frac{G A }{A C}\) = \(\frac{h}{A C}\) or, AC sinθ = h]

∴ Total energy at C = 0 + mgh = mgh

Thus, total mechanical energy at A = total mechanical energy at B = total mechanical energy at C.

Hence, the total mechanical energy of a body, moving along a frictionless inclined plane under gravity, is conserved.

Total Mechanical Energy Of A Hydrogen Gas-Filled Balloon Rising Upwards Remains Constant: Let us assume that air and earth form a system. If the earth’s surface is taken as the plane of reference, when the balloon is at rest on the ground, both its potential and kinetic energy = 0, i.e., total mechanical energy = 0.

Now the balloon is released. Upthrust, T due to air on the balloon is more than the weight, mg of the balloon filled with gas and so the balloon rises up. Let the mass of the gas-filled balloon = m and the mean resultant upward force on it = F – T – mg.

∴ Acceleration of the balloon = \(\frac{F}{m}\). If the velocity of the balloon is v at a height h from the earth surface, then

v2² = 2\(\frac{F}{m}\)h [according to the formula v² = u² + 2as]

Hence, kinetic energy at h = \(\frac{1}{2}\)= \(\frac{1}{2}\)m 2 \(\frac{F}{m}\)h = Fh

We know that when a stone falls freely under gravity, the work done is negative. Similarly, when the balloon rises by itself (without the help of any external agent) under the effect of the mean upward resultant force F, the work done is also negative. Here, the upthrust is not applied by an external agent.

So, at height h the change in potential energy = -Fh. As the initial potential energy of the balloon on the ground was zero, the total potential energy at a height h = -Fh.

∴ The total mechanical energy of the balloon at height h = potential energy + kinetic energy = – Fh + Fh = 0 = total energy of the balloon on the ground before its release.

Thus, the total mechanical energy remains conserved for a hydrogen gas-filled balloon when it is rising up. It is to be noted that, with the increase in height h, the kinetic energy increases but the potential energy decreases equally.

Actually, the expression mgh for potential energy should be modified for a balloon as (mg- T)h, where T is the upward thrust exerted on the balloon by the air surrounding it. For a hydrogen-filled balloon, the upward thrust is greater than its weight, i.e., T> mg. So the potential energy (mg-T)h is negative; this negative value goes on increasing with an increase in height h.

Another similar event is observed when a piece of wood is held at the bottom of a bucket full of water. When it is released, it floats up by itself. It can be shown that the total mechanical energy remains conserved. As the piece of wood rises upwards, its kinetic energy gradually increases and the potential energy decreases equally.

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System Numerical Examples

Example 1. After falling from a height of 200 m, water flows horizontally with a certain velocity. Ignoring any energy dissipation, find the velocity of flow.
Solution:

Potential energy changes into kinetic energy during the free fall of water.

Let mass of water = m, height = h, final velocity = v

Here, at a height of 200 m

potential energy (P.E.)_i = mgh and kinetic energy (K.E.)_i = 0

when water falls and flows horizontally, potential energy, (P.E.)_f = 0 and kinetic energy (K.E.)_f = \(\frac{1}{2}\)mv²

∴ From the conservation of mechanical energy we,

∴ mgh = \(\frac{1}{2}\)mv

or, \(\nu=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 200}\)

= \(62.61 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 62.61 m · s^-1

Example 2. The mass of the bob of a simple pendulum is 10 g and the effective length is 13 cm. The bob is pulled 5 cm away from the vertical and then released. What will be the kinetic energy of the bob when it passes through the lowest point?
Solution:

We observe that, the energy of the bob at point B = potential energy = mgh = 10 x 980 x AC.

Here OA = OB = 13 cm and CB = 5 cm.

∴ OC = \(\sqrt{13^2-5^2}\)

= \(\sqrt{144}=12 \mathrm{~cm}\)

∴ AC = OA-OC = 13 – 12 = 1 cm.

∴ Potential energy of the bob at B = 10 x 980 x 1 = 9800 erg

∴ The kinetic energy of the bob at the lowest point A = potential energy of the bob at B = 9800 erg.

Example 3. After a collision with an ideal spring, a body of mass 8 g, moving with a constant velocity of 10 cm· s^-1 comes to rest Force constant of the spring is 200 dyn · cm^-1. If the total kinetic energy of the body is spent in compressing the spring, find the compression.
Solution:

Here, the kinetic energy of the body transforms into the potential energy of the spring.

∴ \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\), where x = compression of the spring

∴ x = \(\sqrt{\frac{m v^2}{k}}=\sqrt{\frac{8 \times 10^2}{200}}=2 \mathrm{~cm}\) .

Example 4. The effective length of a pendulum is 50 cm and the mass of the bob is 4 g. The bob is drawn to one side until the string is horizontal and is then released. When the string makes an angle 60° with the vertical, what is the velocity and the kinetic energy of the bob?
Solution:

OB is the horizontal position of the string.

At B, the total energy of the bob

= potential energy = mgh = (4 x 980 x 50) erg

At C, the total energy of the bob = kinetic energy + potential energy

= kinetic energy + 4 x 980 x AD

(cos 60° = \(\frac{OD}{OC}\) or, OD = \(\frac{50}{2}\) = 25 cm

∴ AD = OA – OD – 50-25 = 25 cm)

According to the law of conservation of energy, the energy of the bob at C = energy of the bob at B

or, K.E. of the bob at C + P.E. of the bob at C = K.E. of the bob at B + P.E. of the bob at B

or, \(\frac{1}{2} m v^2+4 \times 980 \times 25=0+4 \times 980 \times 50\) (v = velocity of bob at C)

or, \(\frac{1}{2} m v^2=4 \times 980 \times 25=98000\)

∴ \(v^2=\frac{98000 \times 2}{4} \quad \text { or, } v=\sqrt{49000}=222.36 \mathrm{~cm} \cdot \mathrm{s}^{-1}\).

Example 5. The mass of the bob of a pendulum is 100 g and the length of the string is 1 m. The bob is initially held in such a way that the string is horizontal. The bob is then released. Find the kinetic energy of the bob when the string makes an angle of

1. 0° and
2. 30° with the vertical.

Solution:

When the string is horizontal, the height of the bob above its lowest position = 1 m = 100 cm.

The energy of the bob at point P = potential energy of the bob

= mgh = 100 x 980 x 100 = 98 x 10⁵ erg.

1. At 0° angle with the vertical, the string holds the bob at its lowermost position. Hence, energy at B = kinetic energy of the bob = initial potential energy = 98 x 10 erg.

2. When the string makes a 30° angle with the vertical, the height of the bob from its lowermost position

= BD = BA – DA = 100- AC cos30°

= \(100\left(1-\frac{\sqrt{3}}{2}\right)=13.4 \mathrm{~cm}\)

Potential energy of the bob at C = 100 x 980 x 13.4 erg;

kinetic energy at this position = initial potential energy- potential energy at C

= 98 x 10⁵ -(100 x 980 x 13.4) = 8486800 erg.

Example 6. A body of mass 1 kg falls to the ground from the roof of a building 20 m high. Find its

1. Initial potential energy,
2. Velocity when it reaches the ground,
3. Maximum kinetic energy and
4. Kinetic and potential energies at a position 2 m above the earth’s surface.

Solution:

1. Initial potential energy of the body = mgh = 1 x 9.8 x 20 = 196 J.

2. Suppose the body touches the ground with velocity v. Potential energy at roof level = kinetic energy just before touching the ground.

∴ 196 = \(\frac{1}{2}\) x 1 x v²

∴ v² = 392 and v = 19.8 m · s^-1

3. Maximum kinetic energy = initial potential energy = 196 J

4. Potential energy at a height of 2 m = 1×9.8×2 =19.6J

∴  Kinetic energy at that height of 2 m = decrease in initial potential energy = 196 – 19.6 = 176.4 J.

Example 7. A pump lifts 200 L of water per minute through a height of 5 m and ejects it through an orifice 2 cm in diameter. Find the velocity of efflux of water and the power of the pump.
Solution:

Volume of water lifted by the pump in is = \(\frac{200}{60}=\frac{10}{3} \mathrm{~L}=\frac{10^4}{3} \mathrm{~cm}^3 \text {. }\)

Mass of this volume of water = \(\frac{10^4}{3} g=\frac{10}{3} \mathrm{~kg}\)

∴ Increase in potential energy in 1 second

= \(\frac{10}{3}\) x 9.8×5 J · s^-1 = 163.33 J

∴ Power of the pump to raise water up to the height of 5m = 163.33 W.

If the velocity of efflux is v, then \(\pi r^2 \times v=\frac{10^4}{3}\)

or, \(\nu=\frac{10^4 \times 7}{3 \times 22 \times(1)^2}=1061 \mathrm{~cm} \cdot \mathrm{s}^{-1}=10.61 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

The kinetic energy of the amount of water thrown out per second

= \(\frac{1}{2}\) x \(\frac{10}{3}\) x(10.61)² J =187.62 J

∴ Power of efflux = 187.62 W

∴ Total power of the pump = 163.33 + 187.62 = 350.95 W.

Example 8. A body of mass 10 kg is raised to a height of 10 m with an upward force of 196 N. Find the work done by the upward force and the work done against gravitation. Show that the total energy in this case is equal to the work done by the upward force. [g = 9.8 m · s^-2]
Solution:

Work done by the upward force, W = force x displacement

= 196 N x 10 m = 1960 N · m = 1960 J.

Upward acceleration of the body in the absence of gravity,

a’ = \(\frac{\text { upward force }}{\text { mass }}=\frac{196}{10}=19.6 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ Effective upward acceleration, a = a’ -g = 19.6-9.8 = 9.8 m · s^-2

If the body starts from rest, and attains a velocity v at the height of 10 m, then from v² = u² + 2 as, v² = 2 x 9.8 x 10 m² · s^-2

∴  Kinetic energy of the body at this height,

K = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) x 10 x 2 x 9.8 x 10 = 980 J

Work done against gravitational force,

W’ = force due to gravity x displacement = mass x acceleration due to gravity x displacement = 10 x 9.8 x 10 = 980 J

This work done against gravitational pull gets stored as potential energy V of the body. Hence, V = 980 J.

∴ Total mechanical energy of the body at a height of 10 m = V + K= 980 + 980 = 1960 J = W

So, a part of the work done by the upward force changes into kinetic energy of the body, and the other part transforms itself into the stored potential energy. Thus, the total energy and work done by the upward force are equal.

Example 9. A body of mass 10 kg moving with a speed of 2.0 m · s^-1 on a frictionless table strikes a mounted spring and comes to rest. If the force constant of the spring be 4 x 10⁵ N · m^-1, then what will be the compression on the spring?
Solution:

The kinetic energy of the body is E = \(\frac{1}{2}\)mv², where m and v be the mass and the speed of the body respectively.

On striking the spring, the kinetic energy of the spring due to compression is completely converted into the potential energy of the spring.

If the spring is compressed through a distance x then its potential energy is U = \(\frac{1}{2}\)kx²

∴ \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\)kx2

or, x = \(v \sqrt{\frac{m}{k}}=2 \times \sqrt{\frac{10}{4 \times 10^5}}=10^{-2} \mathrm{~m}=1 \mathrm{~cm}\)

Example 10. Shows two blocks of masses m₁ = 3 kg and m₂ = 5 kg, both moving towards the right on a frictionless surface with speeds u₁ = 10 m · s^-1 and u₁ = 4 m · s^-1 respectively. To the back side of m₂ an ideal spring of force constant 1000 N · m^-1 is attached. Calculate the maximum compression of the spring when the blocks collide.

Solution:

m₁ = 3kg, m₂ = 5kg, u₁ = 10 m · s^-1, u₂ = 4 m ·  s^-1

Force constant, k = 1000 N · m^-1

Let v be the speed of the combination.

Using the law of conservation of linear momentum, m₁u₁ + m₂u₂ =(m₁ + m₂)v

∴ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{3 \times 10+5 \times 4}{3+5}=6.25 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Also, let x be the maximum compression of the spring when the blocks collide.

∴ From conservation of mechanical energy, we get \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2}\left(m_1+m_2\right) v^2+\frac{1}{2} k x^2\)

or, \(500 x^2=380-312.5=67.5\)

∴ x = \(\sqrt{\frac{67.5}{500}} \approx 0.367 \mathrm{~m}\)

Work And Energy – Non-Conservative Force

Non-Conservative Force Definition: In the presence of resistive forces in a system, mechanical energy does not remain conserved and gets dissipated. Such a system is called a non-conservative system and the resistive force is called a non-conservative force or dissipative force.

Non-Conservative Force Example: Frictional force is a non-conservative force.

• Frictional force resists the motion of an object. Thus on slid¬ing a body over a rough surface, work done against friction does not get stored in the body as potential energy.
• Frictional force always acts opposite to the direction of motion. To move a body from its initial position to a final position, some amount of work is done to overcome friction.
• If the body is brought back to its initial position along the same path, again some work is done to overcome friction. Thus, each time the body moves, some energy is lost. It can therefore be stated that, it is not possible to restore the initial work done in a non-conservative system.

Work Done In A Closed Path Under A Non-Conservative Force: The total work done to move a body under a non-conservative force along a closed path once completely, is positive or negative but never zero. For example, to slide a body over a rough surface from one point to another, work has to be done against friction.

To return the body to its initial position by sliding it over the same surface following any path, work has to be done against friction again. So, the total amount of work done is not zero in a closed path. Hence, a force is called non-conservative when work done against it cannot be restored. Alternatively, when the work done by a force in a closed path is not zero, the force is called nonconservative.

Work And Energy – Mass Energy Equivalence

The principle of mass-energy equivalence can be obtained from Einstein’s famous theory of relativity.

Mass is a form of energy. Mass can be converted into energy and vice versa. If m amount of mass of a substance is completely converted to energy, then the amount of energy liberated is, E = mc² (c = velocity of light in vacuum = a constant)

According to this equation, the equivalent energy of mass m is E, and the equivalent mass of energy E is m

Mass Energy Equivalence Example:

1. In CGS system c = 3 x 10¹⁰ cm · s^-1

∴ Equivalent energy of 1 g mass

= 1 x (3 x 10¹⁰)²= 9 x 10²⁰ erg = 9 x 10¹³ J

Again in \(\mathrm{SI}\), \(c=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence equivalent energy of 1 kg mass = \(1 \times\left(3 \times 10^8\right)^2=9 \times 10^{16} \mathrm{~J} .\)

2. Mass of an electron = \(9.1 \times 10^{-28} \mathrm{~g}\)

Equivalent energy of the mass of 1 electron = \(9.1 \times 10^{-28} \times\left(3 \times 10^{10}\right)^2 \mathrm{erg}\)

= \(\frac{9.1 \times 10^{-28} \times 9 \times 10^{20}}{1.6 \times 10^{-12}} \mathrm{eV}\)

= \(0.511 \times 10^6 \mathrm{eV}=0.511 \mathrm{MeV} .\)

The increase or decrease in energy of an electron while crossing a potential difference of 1V is called 1 electronvolt (eV). 1 eV = 1.6 x 10^-12 erg.

Rest Mass: Einstein’s theory of relativity also informs us that the mass of a substance is not a constant quantity but depends on the velocity of the substance. Especially if the velocity of an object becomes comparable to that of light, then its mass increases significantly.

That is why to measure the true mass of an object as an intrinsic property, the object should be at rest with respect to the observer. The mass of the object thus measured is called its rest mass and the equivalent energy is called rest energy. For example, the rest mass of an electron =9.1 x 10^-28 g, and the rest energy is 0.511 MeV.

Unit Of Mass And Energy: As mass and energy are equivalent to each other, their units are also the same. Sometimes mass is given in units of energy, and energy is given in units of mass. For example, ‘energy of 1g’ denotes 9 x 10²⁰ erg amount of energy; or mass of ‘9 x 10¹⁶ J’ represents a mass of 1 kg. From this equivalence, it can be stated that the rest mass of an electron is 0.511 MeV.

Law Of Conservation Of Mass-Energy: During the mutual transformation of mass and energy, the law of conservation of mass or the law of conservation of energy cannot be applied separately. It becomes the law of conservation of mass energy.

The total amount of mass energy in nature Is constant, it can never be created or destroyed. It can only change from one form to another.

Mass can be transformed into energy only within an atom. The energy thus obtained from mass is the source of atomic energy. When gamma rays with an energy of a few MeV or more enter the electric field of a heavy nucleus, it can be transformed into an electron and a positron (an electron-like particle but of positive charge). This is an example of the transformation of energy to mass.

Two-Dimensional Collisions Numerical Examples

Example 1. Two particles of masses m₁ and m₂, moving with velocities u₁ and u₁, respectively, and making an angle θ between them, collide with each other. After the collision, the 1st particle travels in the initial direction of motion of the 2nd, and vice-versa. Find the velocities of the two particles after collision. Under what condition, would this collision be elastic?
Solution:

Suppose v₁, v₂ are the velocities of the two particles, respectively, after collision. The particles before and after collision move as shown. It also shows the chosen directions of the x and the y-axis.

For momentum conservation along the x-axis, we get, \(m_1 u_1 \cos \frac{\theta}{2}+m_2 u_2 \cos \frac{\theta}{2}=m_1 v_1 \cos \frac{\theta}{2}+m_2 v_2 \cos \frac{\theta}{2},\)

or, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)….(1)

Similarly, along the y-axis, we get, \(-m_1 u_1 \sin \frac{\theta}{2}+m_2 u_2 \sin \frac{\theta}{2}=m_1 v_1 \sin \frac{\theta}{2}-m_2 v_2 \sin \frac{\theta}{2}\)

or, \(-m_1 u_1+m_2 u_2=m_1 v_1-m_2 v_2\)……(2)

Adding equations (1) and (2), \(2 m_2 u_2=2 m_1 v_1\)

or, \(v_1=\frac{m_2}{m_1} u_2\)…(3)

Subtracting equation (2) from (1), \(2 m_1 u_1=2 m_2 v_2\)

or, \(v_2=\frac{m_1}{m_2} u_1\)…..(4)

The kinetic energy before collision is, \(K_1=\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\)

and that after collision is \(K_2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1\left(\frac{m_2}{m_1} u_2\right)^2+\frac{1}{2} m_2\left(\frac{m_1}{m_2} u_1\right)^2\)

= \(\frac{1}{2} \frac{m_2}{m_1} m_2 u_2^2+\frac{1}{2} \frac{m_1}{m_2} m_1 u_1^2\)

Here, K₁ ≠K₂; so the collision is inelastic, in general. As a m special case, it would be an elastic collision if K₁ = K₂. It is possible only when m₁ = m₂, i.e., the two particles are of equal masses.

Example 2. A bomb explodes and splits up Into three fragments. Two fragments, each of mass 200 g, move away from each other making an angle of 120°, at a speed of 100 m · s^-1. Find the direction and velocity of the third fragment whose mass is 500 g. Also, find out the energy released in an explosion.
Solution:

The velocity of fragments A and B along OA and OB. A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, in the direction opposite to the resultant of OA and OB.

If the velocity of the third piece is v, then taking the components along the line CD in the CGS system, 500 v = 200 x 10⁴ cos60° + 200 x 10⁴ cos60°

or, v = \(\frac{200 \times 10^4}{500}=4 \times 10^3 \mathrm{~cm} \cdot \mathrm{s}^{-1}=40 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence, the velocity of the third fragment is 40 m · s^-1. It moves so as to make an angle 120° with each of OA and OB. The energy released due to the explosion is the kinetic energy of the three fragments.

∴ Energy released = \(\frac{1}{2} \times 200 \times\left(10^4\right)^2\)

+ \(\frac{1}{2} \times 200 \times\left(10^4\right)^2+\frac{1}{2} \times 500 \times\left(4 \times 10^3\right)^2\)

= \(2400 \times 10^7 \mathrm{ergs}=2400 \mathrm{~J} .\)

Example 3. A spaceship while flying in space, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction; the other two fly off at 60° to the original direction, on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragments.
Solution:

Let the mass of the spaceship be 3m and the initial speed be u.

Hence, mass of each fragment = m.

Let their velocities be v₁, v_2, and v₃ after the explosion.

Applying the conservation of momentum law along the direction perpendicular to the original direction of motion,

0 = mv₂ sin60°- mv₃ sin60°

or, mv₂ sin60° – mv₃ sin60°

or, v₂ = v₃ = v (say)

Applying the law of conservation of linear momentum along the original direction of motion,

3 mu = mv₁ + 2mv cos60°

or, 3u = v₁ + v…(1)

The kinetic energy of the spaceship before explosion E= 1/2 x 3mu²

Energy released during the explosion, \(E_r=\frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2\)

According to the problem, \(E_r=2 E \quad \text { or, } \frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2=2 \times \frac{1}{2} \cdot 3 m u^2\)

or, \(v_1^2+2 v^2-3 u^2=6 u^2\)

or, \(v_1^2+2 v^2=9 u^2\)….(2)

Solving (1) and (2), v₁ =u and v = 2 u

Hence, the kinetic energy of the first fragment after the explosion

= \(\frac{1}{2} m v_1^2=\frac{1}{2} m u^2=\frac{1}{2} \cdot \frac{2}{3} \cdot E=\frac{1}{3} E\)

the kinetic energy of each of the other two fragments

= \(\frac{1}{2} m v^2=\frac{1}{2} m(2 u)^2=2 m u^2=2 \cdot \frac{2}{3} E=\frac{4}{3} E\)

 

Work And Energy Conclusion

If a body undergoes a displacement due to a force acting on it, work is said to be done, and the product of force and displacement is the measure of the work done.

Work Done Against A Force: Consider a particle on which some forces are acting. When an external agent causes a displacement opposite to the direction of the resultant of these forces, work is said to be done against the force.

Work Done By A Force: Consider a particle on which some forces are acting. The resultant of these forces can cause a displacement. This is said to be work done by a force.

• If a force acts at right angles to the direction of displacement of a body, no work is done by the force. This force is called a no-work force.
• The rate of doing work with respect to time is called power. That is work done in unit time is called power.
• Energy is the capacity to do work.
• The ability of a body to do work due to its speed, special position, or special configuration, or all of these, is called its mechanical energy.
• Mechanical energy is of two types— kinetic energy and potential energy.
• The ability of a body to do work due to its speed is called its kinetic energy.
• The ability of a body to do work due to its special position or configuration is called its potential energy.
• The ability of a body to do work, acquired due to its rise against gravity, is called its gravitational potential energy.
• Gravitational potential energy depends on the chosen plane of reference. This energy may also have a negative value.
• The ability of a body to do work, gained due to its special shape, is called elastic potential energy.

Law Of Conservation Of Energy: Energy can neither be created nor destroyed.

Mass-Energy Equivalence: In this universe, the total sum of mass and energy is a constant. This is the law of conservation of mass energy. Mass and energy are equivalent; one can be converted into the other.

• If the total momentum and the total kinetic energy of a system are conserved, the collision is termed as an elastic collision.
• If the total momentum is conserved, but the total kinetic energy is not, it is an inelastic collision.

Coefficient Of Restitution: The coefficient of restitution is defined as the ratio of the velocity with which the two bodies separate after a collision to their velocity of approach before the collision.

1. For elastic collision, e = 1.
2. For perfectly inelastic collision, e = 0.
3. For partially elastic collision, 0 < e < 1.

A system in which the total mechanical energy remains conserved is called a conservative system. Forces acting in such a system are called conservative forces.

In a system where resistive forces are present, the mechanical energy is not conserved. Such systems are called non-conservative systems. Forces of resistance are called dissipative forces.

Work And Energy Useful Relations For Solving Numerical Problems

When a force \(\vec{F}\) acting on a body is associated with a displacement \(\vec{s}\), work done, W = \(\vec{F}\) · \(\vec{s}\) = Fs cosθ, where θ is the angle between \(\vec{F}\) and \(\vec{s}\).

For a variable force \(\vec{F}\), if the displacement of a particle is from A to B, the total work done is

W = \(\int_A^B \vec{F} \cdot d \vec{s}\)

= \(\int_A^B F \cos \theta d s\)

Power (P) = \(\frac{\text { work }(W)}{\text { time }(t)}\) = force (F) x velocity of the body(v)

Kinetic energy = \(\frac{1}{2}\) mv²

For a moving object of mass m, and kinetic energy E, momentum p = √2mE

Gravitational potential energy = mgh

Within the elastic limit, the elastic potential energy of a spring (stretched or compressed by x) = \(\frac{1}{2}\)kx², where k is the spring constant.

Efficiency of a machine

= \(\frac{\text { work output from the machine }}{\text { energy input to the machine }} \times 100 \%\)

In a conservative field, total energy (E) = kinetic energy (K) + potential energy (V) = constant.

The energy equivalent of a mass m is, E = mc², where c is the velocity of light in a vacuum.

Law of conservation of momentum during a linear collision between two bodies: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In case of elastic collision

1. \(u_1-u_2=v_2-v_1\) and
2. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

Coefficient of restitution, e = \(\frac{v_2-v_1}{u_1-u_2}\)

In case of collision of a falling body with a fixed horizontal plane, total distance traveled before coming to rest, d = \(h \frac{1+e^2}{1-e^2}\)

Work And Energy Very Short Answer Type Questions

Question 1. A force is acting on a body in motion, but is not doing any work. Give an example of such a force.
Answer: Centripetal force on a body, in a uniform circular motion

Question 2. Is work a vector or a scalar quantity?
Answer: Scalar quantity

Question 3. What is the amount of work done by a force when the body moves in a circular path?
Answer: Zero

Question 4.In a tug-of-war game, which of the teams does effective work?
Answer: The stronger team

Question 5. A person is carrying a bucket of water and is in a lift moving up with uniform velocity. Is the person doing any work on the bucket of water? Will the energy of the bucket and water remain constant?
Answer: No

Question 6. A motor drives a belt at a constant velocity of v m · s^-1. If m kg of sand falls on the belt per second, what is the rate of work done by the force exerted by the belt on the sand?
Answer: 1/2mv³W

Question 7. A boy tried to lift a bucketful of water but failed. What is the work done by him?
Answer: Zero

Question 8. What is the work done by the tension in the string during the oscillation of a simple pendulum?
Answer: Zero

Question 9. A box was lifted vertically through a height of 6 m in 3 s. If the box had been lifted in a zig-zag way in 5 s, the work done would have been the same. Is the statement true or false?
Answer: True

Question 10. 1 kg · m = ________ J.
Answer: 9.8

Question 11. A force \(\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) \mathrm{N}\) acts on a particle, and the particle moves from the origin to a point \(\vec{r}=(2 \hat{i}-\hat{j}) \mathrm{m}\). What will be the work done on the particle?
Answer: 7

Question 12. How many joules are in 1 MeV?
Answer: 1.6 x 10^-13J

Question 13. What is the unit of energy?
Answer: Joule

Question 14. Does the kinetic energy of a ball, thrown inside a moving train, depend on the speed of the train?
Answer: No

Question 15. Which type of energy is lost in doing work against friction?
Answer: Mechanical energy

Question 16. A small car and a lorry are moving with the same kinetic energy. Brakes are applied to produce the same force against the motion. Which one will cover a greater distance before stopping?
Answer: Both cover the same distance

Question 17. When a body falls on the ground from a height, it becomes slightly warm—why?
Answer: Kinetic energy changes to heat energy

Question 18. Is the resistance due to air a conservative force?
Answer: No

Question 19. What happens to internal energy, when the temperature of the body increases?
Answer: Increases

Question 20. What type of energy is stored in the spring of a watch?
Answer: Potential energy

Question 21. The kinetic energies of a heavy and a light object are the same. Momentum of which object will be higher?
Answer: Heavy

Question 22. Momenta of a light and a heavy body are the same. Which body has greater kinetic energy.?
Answer: Lighter

Question 23. If E is the kinetic energy of a body of mass m, what will be its momentum?
Answer: \(\sqrt{2 m E}\)

Question 24. An object breaks up into two masses m₁ and m₂ due to explosion. The two fragments move in opposite directions. What will be the relation between the kinetic energy and the masses?
Answer: Inversely

Question 25. What is the loss of KE of a freely falling body of mass m, during the t th second?
Answer: [1/2 mg²(2t – 1)]

Question 26. The increase in momentum of a body is 100%. What will be the increase in its kinetic energy?
Answer: 300

Question 27. The increase in kinetic energy of a body is 69%. What will be the increase in its momentum?
Answer: 30

Question 28. Which physical quantity in conserved during both the elastic and inelastic collisions?
Answer: Momentum

Question 29. Two objects coalesce after a collision with each other. What is the coefficient of restitution?
Answer: 0

Question 30. The coefficient of restitution between a ball and a horizontal floor is e = 1/2. If the ball falls from a height of 10 m, after its impact with the floor, the ball bounces up to a height of ______.
Answer: 2.5m

Work And Energy Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The absolute PE of a system as measured by two different persons at the same time can be different.

Statement 2: The value of the absolute PE of a system depends upon the reference value chosen.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: Work done by normal contact force can be non-zero.

Statement 2: Normal contact force is always perpendicular to the displacement of the object. (Here displacement is measured with respect to a frame of reference attached to the two surfaces in contact.)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: In a circular motion, work done by the centripetal force is not always zero.

Statement 2: If the speed of the particle increases or decreases in a circular motion, net force, acting on the particle is not directed toward the center.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: When a body moves uniformly in a circle its momentum goes on changing but its kinetic energy remains constant.

Statement 2: \(\vec{p}=m \vec{v}, \mathrm{KE}=\frac{1}{2} m v^2\). In circular motion \(\vec{v}\) changes, v² does not change.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the lowest point is \(\sqrt{5 g R}\).

Statement 2: Mechanical energy is conserved and for the minimum velocity at the lowest point, the velocity at the highest point will be zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 6.

Statement 1: Work done by constant force is equal to the magnitude of force multiplied by displacement.

Statement 2: Work done is a scalar quantity. It may be positive, negative, or zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 7.

Statement 1: If work done by a conservative force is negative then potential energy associated with that force should increase.

Statement 2: This is from the reaction Δu = -W. Here Au is change in potential energy and W is work done by conservative force.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Work And Energy Match Column A With Column B.

Question 1. The displacement time graph of a body is shown.

Answer: 1. A, 2. B, C, D, 3. B, C, D, 4. C

Question 2. A force F = kx (where k is a positive constant) is acting on a particle. In Column A displacements (x) are given and in Column B work done by the force is given.

Answer: 1. B, 2. A, 3. C

Question 3.

Answer: 1. B, 2. D, 3. C

Question 4. The system is released from rest. Friction is absent and string is massless. In time t = 0.3 s (take g = 10 m · s^-2)

Answer: 1. C, 2. A, 3. D, 4. B

Question 5.

Answer: 1. A, B, C, 2. B, C, 3. A, C, 4. A, B, C

Question 6. A particle is suspended from a string of length R. It is given a velocity u = \(3 \sqrt{g R}\) at the bottom.

Answer: 1. C, 2. B, 3. A, 4. E

Work And Energy Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A body of mass 2 kg starts from rest and moves with uniform acceleration. It acquires a velocity 20 m · s^-1 in 4 s.

1. Power exerted on the body at 2 s is

1. 50 W
2. 100 W
3. 150 W
4. 200 W

Answer: 2. 100 W

2. Average power transferred to the body in the first 2 s is

1. 50 W
2. 100W
3. 150 W
4. 200 W

Answer: 1. 50 W

Question 2. A ball of mass m is dropped from a height H above a level floor as shown. After striking the ground it bounces back and reaches up to the height h.

1. During the collision, the part of the KE which appears in other forms (other than KE or PE) is (this part of the energy is termed as lost energy as it cannot be utilized properly)

1. mgH
2. mgh
3. mgH – mgh
4. Zero

Answer: 3. mgH – mgh

2. The speed of the ball just after the collision is

1. \(\sqrt{2 g H}\)
2. \(\sqrt{2} g h\)
3. \(\sqrt{2 g(H-h)}\)
4. None of these

Answer: 2. \(\sqrt{2} g h\)

3. If the lost energy in the collision is half of the value computed in Question (1), and H = \(\frac{3 h}{2}\), then the height attained by the ball after the collision is

1. \(\frac{7 h}{4}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{3 h}{2}\)
4. \(\frac{9 h}{5}\)

Answer: 1. \(\frac{7 h}{4}\)

Question 3. A block of 2.5 kg is pulled 2.20 m along a frictionless horizontal table by a constant force of 16 N directed at 45° above the horizontal.

1. Work done by the applied force is

1. 25 J
2. 27 J
3. 24.9 J
4. 22.5 J

Answer: 3. 24.9 J

2. Work done by the normal force exerted by the table is

1. 24.9 J
2. Zero
3. 27 J
4. 27.5 J

Answer: 2. Zero

3. Work done by the force of gravity is

1. 24.9 J
2. 27 J
3. Zero
4. 27.5 J

Answer: 3. Zero

 

Oscillation And Waves

Simple Harmonic Motion Preliminary Topics

Periodic Motion Definition: Any motion, that repeats itself at regular intervals of time, is known as periodic motion.

The motion of planets around the sun, the hands of a clock, and the blades of a revolving electric fan are some examples of periodic motion. A characteristic of this motion is that each moves along a definite circular or elliptical path repeatedly in a regular time interval.

• The motion along an elliptical path is called elliptical or orbital periodic motion, while the other motions are called rotational periodic motion.
• On the other hand, if the bob of a simple pendulum is pulled aside slightly and then released, it swings to the other side passing through its equilibrium position.
• On its way back, the bob again passes through its equilibrium position and returns to its point of release.
• Then it goes on repeating this to and fro motion. Thus the bob covers a definite path repeatedly. If the angular displacement of oscillation of the pendulum is less than 4°, the amplitude or range of oscillation is small compared to the radius of curvature of the oscillatory path and so the path of the bob may be taken as a straight line.

Hence, this motion is called linear periodic motion. The motion of an elastic spring and that of a piston in the cylinder of an automobile engine are examples of linear periodic motion.

Oscillation Or Vibration: If a body undergoing periodic motion has an equilibrium position somewhere inside its path, it experiences no net external force at that point. Hence, if it is left there at rest, it remains there forever.

Now, if the body is given a small displacement from the equilibrium position, a restoring force comes into play which tries to bring the body back to its equilibrium point, giving rise to oscillations or vibrations.

Oscillation Or Vibration Definition: If a particle that executes periodic motion moves to and fro along the same path, the motion is called oscillation or vibration.

Simple Pendulum: A pendulum is suspended from a rigid support. Let P be the point of suspension and O be the mean or equilibrium position of the body. If the bob is pulled to position B and then released, it oscillates along the path BOC.

If the angular displacement, i.e., ∠OPB or∠OPC is less than 4°, then the path BOC or COB may be taken as a straight line. The pendulum is then called a simple pendulum and its oscillation is a linear periodic motion.

Elastic Spring: The upper end of an elastic spring is attached with a rigid support, Let P be the point of suspension. If a heavy body is suspended from its lower end, the spring stretches and the body hangs at rest, at some position of equilibrium O.

If the body is slightly pulled to B and then released, the spring executes an up-and-down oscillatory motion along the straight path BOC and COB. The oscillation of the elastic spring is therefore a linear periodic motion.

Stretched String: A string XY is attached to two rigid supports at points X and Y, It remains at a position of equilibrium along the straight line XY, If the string is pulled slightly upwards or downwards and then released, it vibrates about its position of equilibrium.

• If we consider a point O on the string, it is seen that the point vibrates along the straight path BOC, This vibration is also a linear periodic motion. Although any kind of oscillation or vibration is a periodic motion, the converse is not true. All periodic motions are not oscillations or vibrations.
• For example, the earth completes one revolution around the sun in 1 year, but it is not a to-and-fro motion about any mean position. Hence the motion is periodic but not oscillatory.

Some Quantities Related To Oscillation: It is evident from the different examples of oscillation above that the motion of the particle is restricted to a line segment, say BC.

Complete Oscillation: If a vibrating particle starts its motion from any point on its path towards a certain direction, returns to the same point, and then follows the same path in the same direction then it is said to have executed one complete oscillation or one complete vibration.

• If the particle starts its motion from B and after tracing the paths BOC and COB returns to B, then the particle executes one complete oscillation. It is seen that for a complete oscillation, the particle moves along the entire straight path twice.
• So, if the particle starts its motion from D and after tracing the paths DOC, COB, and BD, finally returns to D, then also it can be said that the particle has executed one complete oscillation. A complete oscillation is also known as a period.

Time Period: Time period of oscillation of a vibrating particle is defined as the time taken by it to execute one complete oscillation. Its dimension is T and its unit in all systems is second (s).

Frequency: The frequency of oscillation of a vibrating particle is defined as the number of complete oscillations executed by it in 1 second.

In time T the particle executes one complete oscillation, Thus the number of complete oscillations executed by the particle In ls is \(\frac{1}{T}\). Hence, the frequency n of the particle is n = \(\frac{1}{T}\).

The dimension of frequency is T^-1 and its unit in all systems is second^-1 or per second. This unit of frequency Is called hertz (Hz). So, 1 Hz = ls^-1.

Amplitude: The amplitude of oscillation of a vibrating particle is defined as Its maximum displacement from Its equilibrium position.

• Amplitude, A = OB or OC. The dimension of amplitude is 1, and Its units In the CGS system and SI are centimeters (cm) and meters (m) respectively. It is to be noted that in the above discussions, sometimes we have used the term ‘oscillation1 and sometimes ‘vibration’. In fact, oscillation and vibration are synonymous.
• Usually, when the time period of the particle is large, i.e., frequency is low, the motion of the particle is called oscillation.

Amplitude Example: Oscillation of a simple pendulum or an elastic spring. On the other hand, when the time period of the particle is small, i.e., frequency is high, the motion of the particle is called vibration. Examples of vibration are the vibration of a stretched string or that of a tuning fork.

Periodic Functions: A function f(t) is periodic if the function repeats itself after a regular interval of the independent variable t.

The simplest examples of a periodic motion can be represented by any of the following functions.

f(t) = \(A \cos \frac{2 \pi}{T} t\)….(1)

and \(g(t)=A \sin \frac{2 \pi}{T} t\)….(2)

Here T is the time period of the periodic motion.

To check the periodicity of these functions t is to be replaced by (t+ T), In the above equations simultaneously.

Hence equation (1) gives us,

f(t+T) = \(A \cos \left[\frac{2 \pi}{T}(t+T)\right]=A \cos \left[\frac{2 \pi}{T} t+2 \pi\right]\)

= \(A \cos \frac{2 \pi}{T} t[because \cos (\theta+2 \pi)=\cos \theta]\)

= f(t)

And equation (2) gives us,

g(t+T) = \(A \sin \left[\frac{2 \pi}{T}(t+T)\right]=A \sin \left(\frac{2 \pi}{T} t+2 \pi\right)\)

= \(A \sin \frac{2 \pi}{T} t[because \sin (\theta+2 \pi)=\sin \theta]\)

= g(t)

So, for a periodic function with period T, f(t+T) = f(t) or, g(t+T) = g(t)

The result will be the same if we consider a linear combination of sine and cosine functions of period T,

i.e., f(t) = \(A \sin \frac{2 \pi}{T} t+B \cos \frac{2 \pi}{T} t\)

Another example of a periodic function is, f(t)= sinωt + cos2ωt + cos4ωt But, f(t) = e^-ωt is not a periodic function, because it decreases monotonically with the increase in time and tends to zero as t → ∞

Displacement As A Function Of Time: Displacement can be represented by a mathematical function of time. In the case of periodic motion, this function is periodic in nature. One of the simple periodic functions is, f(t) = Acoscot

When the argument cot, is increased by an integral multiple of 2π radians, the value of the function remains the same.

Properties Of Simple Harmonic Motion Or SHM: Simple harmonic motion is the simplest form of oscillation. From the properties of simple harmonic motion, we can analyze any complex oscillation or vibration.

Any type of oscillatory motion can be considered to be the result of two or more simple harmonic motions acting on a particle. Thus, it is of great importance to discuss SHM in detail.

Restoring Force: When a vibrating particle is at its position of equilibrium, the resultant force acting on it is zero. For example, when a simple pendulum is at its equilibrium position the downward force due to the weight of the bob is balanced by the upward tension of the string.

• So, the resultant force acting on the bob is zero. If the bob is displaced slightly from its equilibrium position and released, then a resultant force acts on the bob which tries to bring it to its equilibrium position.
• This force is called the restoring force. Since force is a vector quantity, this restoring force has a magnitude and a direction.
• If the magnitude and the direction of the restoring force satisfy the following two conditions, the motion of the particle is termed simple harmonic motion (SHM),

1. The restoring force Is always directed towards the position of equilibrium of the particle.
2. The magnitude of the restoring force is proportional to the displacement of the particle from its position of equilibrium.

Equation Of Simple Harmonic Motion: Suppose, a particle is executing linear periodic motion along x-axis, and the point O, which is the origin (x = 0), is the position of equilibrium of the particle. Let D be any point on the path of the particle with position coordinate x.

According to condition (2), if F is the restoring force acting on the particle at D, then F ∝ x

Again from condition (1), as the restoring force F is directed towards the equilibrium position, it is taken as negative since the displacement OD = x is taken as positive.

So, F = -kx …..(1)

k is called the force constant and it is positive. Therefore, the magnitude of the restoring force acting on the particle when it is at a position of unit displacement is called the force constant. The units of k are dyn · cm^-1 (CGS) and N · m^-1 (SI).

If m is the mass of the particle and a is its acceleration then F = ma. So, from equation (1) we get, ma = -kx

or, a = \(-\frac{k}{m} x=-\omega^2 x\)…(2)

Here, \(\omega=+\sqrt{\frac{k}{m}}=\text { constant }\)…… (3)

Any one of the equations (1) or (2) is called the equation of simple harmonic motion. As the forms of these equations are identical, it can be said that the properties of acceleration of the particle and those of the restoring force are identical. Simple harmonic motion can be defined with reference to the properties of acceleration.

Simple Harmonic Motion Definition: The motion of a particle is said to be simple harmonic if its acceleration

1. Is proportional to its displacement from the position of equilibrium and
2. Is always directed towards that position.

It is to be noted that the acceleration of a particle executing simple harmonic motion is expressed as a = -ω²x. Conversely, if the acceleration of a particle obeys the equation a = -ω²x, then we can say that the motion of the particle is simple harmonic.

 

Oscillation And Waves

Simple Harmonic Motion Energy Of Simple Harmonic Motion

Let m be the mass of a particle executing simple harmonic motion, A be the amplitude and T be the time period of the motion. The particle possesses kinetic energy due to its velocity all along its path of motion except at the extremities.

Again restoring force acts on the particle all along its path of motion except at the position of equilibrium. So, the work that is to be done to move the particle against the restoring force remains stored in the particle as potential energy.

Kinetic Energy: At a displacement x from the position of equilibrium, the velocity of the particle, v = \(\omega \sqrt{A^2-x^2}\)

So, the kinetic energy of the particle of mass m at that instant, K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)….(1)

1. When the particle is just at the position of equilibrium, the kinetic energy at that instant, K = \(\frac{1}{2}\)mω²A² this is the maximum value of the kinetic energy.
2. As the particle reaches any end of its path, X = ± A, the kinetic energy at that instant, K = 0 this is the minimum value of the kinetic energy.

So, the kinetic energy of a particle executing SHM is maximum at the position of equilibrium and zero at the two ends, i.e., at the two extremities of its path of motion.

Potential Energy: When the particle is at a distance x from its position of equilibrium, the restoring force acting on it is, F = mω²x.

Again, when the particle is just at the position of equilibrium, x = 0, the restoring force F = 0, i.e., no force acts on the particle.

So, within the displacement from 0 to x, the average force acting on the particle =  \(\frac{0+m \omega^2 x}{2}=\frac{1}{2} m \omega^2 x\)

Now potential energy of the particle, U = work done to move the particle through the distance x against this average force = average force x displacement acting on the particle = \(\frac{1}{2} m \omega^2 x \cdot x=\frac{1}{2} m \omega^2 x^2\)…(2)

1. When the particle is just at the position of equilibrium, x = 0, the potential energy at that instant, U = 0  this is the minimum value of the potential energy.
2. As the particle reaches any end of its path, x = ± A, the potential energy at that instant, U = \(\frac{1}{2}\) mω²A² this is the maximum value of the potential energy.

Calculation Of Potential Energy With The Help Of Calculus: When the displacement of the particle is x from the position of equilibrium, the restoring force acting on it is, mω²x. Let dx be a further infinitesimal displacement of the particle such that the above force acting on the particle remains constant throughout this displacement.

So, work done in moving the dx = mω²x · dx

Hence, the total work done in moving it from 0 to x, i.e., the potential energy of the particle is,

U  = \(\int_0^x m \omega^2 x d x=\frac{1}{2} m \omega^2 x^2\)

Total Mechanical Energy: Total energy of a particle executing simple harmonic motion,
E = K+ U = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2 .\)

• Since m and co are constants, if the amplitude A remains unchanged, then E = constant, i.e., the total energy of the particle does not depend on its displacement. As the particle moves away from the position of equilibrium, kinetic energy gradually gets transformed into potential energy.
• While returning from the extreme position towards the position of equilibrium, its potential energy gradually gets converted into kinetic energy. This conversion is in accordance with the law of conservation of energy.
• Due to any external cause (for example, air resistance), if the amplitude of the motion decreases, the total energy will also decrease. In that case, the energy of the particle is transferred to the surroundings (for example, different air particles).
• The conversion of kinetic energy to potential energy and vice versa with the change in displacement of the particle executing SHM are shown in the following table. Using this table, the energy-displacement curve is plotted.

From the table, we see that when the displacement of the particle is ± \(\frac{A}{\sqrt{2}}\)

then kinetic energy = potential energy = \(\frac{1}{4} m \omega^2 A^2=\frac{E}{2}\)

Oscillation And Waves

Simple Harmonic Motion Energy Of Simple Harmonic Motion Numerical Examples

Example 1. When a particle executing SHM is at a distance of 0.02 m from its mean position, then its kinetic energy is thrice its potential energy. Calculate the amplitude of motion of the particle.
Solution:

According to the question, K = 3U

i.e., \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=3 \times \frac{1}{2} m \omega^2 x^2\) or, \(A^2-x^2=3 x^2 or, A^2=4 x^2\) or, \(A=| \pm 2 x|=0.04\)

[since x = 0.02m]

∴Amplitude = 0.04 m.

Example 2. When a particle executing SHM is at a distance of 0.02 m from its position of equilibrium, then its kinetic energy is twice its potential energy. Calculate the distance from the position of equilibrium where its potential energy is twice its kinetic energy.
Solution:

In the first case, when x = 0.02 m, K = 2 U

i.e., \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=2 \times \frac{1}{2} m \omega^2 x^2\)

or, \(A^2-x^2=2 x^2 or, A^2=3 x^2\)

or, \(A^2=3 \times(0.02)^2=0.0012\)

In the second case, 2 K=U

i.e., \(2 \times \frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2 x^2\)

or, \(2\left(A^2-x^2\right)=x^2 or, 3 x^2=2 A^2\)

or, \(x^2=\frac{2}{3} A^2=\frac{2}{3} \times 0.0012=.0008\)

or, x = \(\sqrt{0.0008}=0.0282 \mathrm{~m} \approx 0.03 \mathrm{~m}\)

Example 3. A particle of mass 0.2 kg is executing SHM along the x-axis with a frequency of \(\frac{25}{\pi}\)Hz. If its kinetic energy is 0.5 J at x = 0.04 m, then find its amplitude of vibration.
Solution:

Here, n = \(\frac{25}{\pi} \mathrm{Hz}\); so, \(\omega=2 \pi \cdot \frac{25}{\pi}=50 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

We know, K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right) or, \frac{2 K}{m \omega^2}=A^2-x^2\)

or, \(A^2=\frac{2 K}{m \omega^2}+x^2=\frac{2 \times 0.5}{0.2 \times(50)^2}+(0.04)^2=0.0036 or, A=0.06 \mathrm{~m}\).

Example 4. The total energy of a particle executing SHM is 3 J. A maximum force of 1.5 N acts on it the Time period and epoch of the SHM are 2 s and 30° respectively. Establish the equation of this SHM and also find the mass of the particle.
Solution:

Total energy = \(\frac{1}{2} m \omega^2 A^2=3\)…(1)

Maximum force = mass x maximum acceleration

= m x ω²A = 1.5

Dividing (1) by (2) we get, \(\frac{1}{2}\) A = 2 or, A = 4 m

Time period, T = 2s

∴ \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Epoch, \(\alpha=30^{\circ}=\frac{\pi}{6}\)

So, the equation of SHM is x = Asin(ωt+α)

or, x = \(4 \sin \left(\pi t+\frac{\pi}{6}\right) \mathrm{m}\)

Again from equation (2), m = \(\frac{1.5}{\omega^2 A}=\frac{1.5}{\pi^2 \cdot 4}=0.038 \mathrm{~kg}\)

Example 5. The equation of motion of a particle executing SHM is x = Asin(ωt+ θ). Calculate the velocity and acceleration of the particle. If m is the mass of the particle, then what is the maximum value of its kinetic energy?
Solution:

Given equation ofSHM, x = Asin(ωt+ θ)….(1)

So, velocity v = \(\frac{d x}{d t}=\omega A \cos (\omega t+\theta)\)

From equation (1) we get, \(\sin (\omega t+\theta)=\frac{x}{A}\)

∴ \(\cos (\omega t+\theta)=\sqrt{1-\sin ^2(\omega t+\theta)}=\sqrt{1-\frac{x^2}{A^2}}=\sqrt{\frac{A^2-x^2}{A^2}}\)

∴ v = \(\omega A \sqrt{\frac{A^2-x^2}{A^2}}=\omega \sqrt{A^2-x^2}\)

Again, the acceleration of the particle,

a = \(\frac{d \nu}{d t} =\frac{d}{d t}\{\omega A \cos (\omega t+\theta)\}=-\omega^2 A \sin (\omega t+\theta)\)

= \(-\omega^2 x\)  (x = \(A \sin (\omega t+\theta)\))

As the maximum value of \(\cos (\omega t+\theta)\) is 1 ,

the maximum value of velocity, \(v_{\max }=\omega \mathrm{A}\)

∴ Maximum kinetic energy of the particle

= \(\frac{1}{2} m v_{\max }^2=\frac{1}{2} m \omega^2 A^2 .\)

Example 6. The amplitude of a particle of mass 0.1 kg executing SHM is 0.1m. At the mean position, its kinetic energy is 8×10^-3 J. Find the time period of its vibration.
Solution:

At the mean position, the potential energy of the particle =0

Hence, kinetic energy of the particle at the mean position = total energy = \(\frac{1}{2}\)A

∴ \(\frac{1}{2} m \omega^2 A^2=E \text { or, } \omega^2=\frac{2 E}{m A^2} \text { or, } \omega=\sqrt{\frac{2 E}{m A^2}}\)

i.e., time period,

T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m A^2}{2 E}}\)

= \(2 \times 3.14 \times \sqrt{\frac{0.1 \times(0.1)^2}{2 \times 8 \times 10^{-3}}}\)

= \(2 \times 3.14 \times \sqrt{\frac{1}{16}}=\frac{2 \times 3.14}{4}=1.57 \mathrm{~s} .\)

Example 7. An object of mass 10 kg executes SHM. Its time period and amplitude are 2 s and 10 m respectively. Find Its kinetic energy when it is at a distance of

1. 2 m and
2. 5 m respectively from its position of equilibrium.

Justify the two different results for (1) and (2).

Solution:

T = \(2 \mathrm{~s}\), i.e., \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Kinetic energy, K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)

1. When x = 2m, K = \(\frac{1}{2} \times 10 \times(\pi)^2 \times\left\{(10)^2-(2)^2\right\}\) = \(5 \times(3.14)^2 \times 96=4732.6 \mathrm{~J} .\)
2. When x = 5m, K = \(\frac{1}{2} \times 10 \times(\pi)^2 \times\left\{(10)^2-(5)^2\right\}\) = \(5 \times(3.14)^2 \times 75=3697.35 \mathrm{~J}\)

The velocity of a particle executing SHM becomes maximum at its mean position. As the displacement increases, i.e., as it moves away from the equilibrium position, its velocity decreases and so the kinetic energy also decreases.

Example 8. A particle of mass m executes SHM with amplitude and frequency n. What is the average kinetic energy of the particle during its motion from the position of equilibrium to the end?
Solution:

KE in the mean position = \(\frac{1}{2} m a^2 w^2\)

= \(\frac{1}{2} m a^2(2 \pi n)^2\)

= \(2 \pi^2 m a^2 n^2\)

KE at the end = 0

∴ Average kinetic energy = \(\frac{0+2 \pi^2 m a^2 n^2}{2}=\pi^2 m a^2 n^2\)

Oscillation And Waves

Simple Harmonic Motion Simple Pendulum

A simple pendulum is nothing but a small, heavy body suspended from a rigid support with the help of a long string. The heavy body remains in its lowest position A when the string is vertical.

• This position OA is called the position of equilibrium of the pendulum. The heavy body is called the pendulum bob. The support (O) from which the bob is suspended is called the point of suspension.
• The center of gravity of the suspended bob Is the point of oscillation, When the bob Is displaced from the equilibrium position by a little distance and then released, the pendulum oscillates about its equilibrium position on either side.

For convenience in the mathematical treatment of the properties of a pendulum, an Ideal simple pendulum Is considered. A simple pendulum will be ideal If

1. The string is weightless,
2. The string is inextensible,
3. No frictional resistance acts on the bob during its oscillation and
4. The bob is a point mass. Conforming to all the conditions stated above is not practically possible; we never get an ideal simple pendulum. Hence, for laboratory use, a small, heavy metal ball is tied to one end of a long, light string and the system is suspended from a rigid support.

Simple Pendulum Definition: If a small, heavy body, suspended from a rigid support by a long, weightless, and inextensible string, can be set into oscillation, then the arrangement is called a simple pendulum.

1. When the pendulum bob is slightly displaced from position A to B and then released, it starts oscillating along arc BAC and CAB. This means that the position of the pendulum periodically changes from OB to OC.
2. This to-and-fro oscillatory motion is periodic. But practical experience shows that due to air resistance and friction at the suspension point, this oscillation slowly subsides and ultimately the pendulum comes to rest along OA, its equilibrium position.

A Few Definitions Related To Simple Pendulum

Plane Of Oscillation: In the given diagram, the straight lines OA, OB, and OC lie on the same vertical plane. The pendulum does not leave that plane during oscillation. This plane is called the plane of oscillation.

Effective Length: The distance of the center of gravity of the bob from the point of suspension is called the effective length of the pendulum. In the case of a spherical bob of radius r, the center of gravity lies at the center of the sphere, and if the length of the string is l, then the effective length L = l+r.

Amplitude And Angular Amplitude: The maximum displacement of The pendulum hob on either side of its equilibrium position is called amplitude. In the given diagram, AB or AC is the amplitude of the pendulum.

The angle subtended at the point of suspension by the equilibrium position and the maximum displaced position of the pendulum bob is the angular amplitude of the pendulum. The angular amplitude ∠AOB = ∠AOC = θ.

It is to be noted that, the angular amplitude should be less than 4° so that the arc CAB is almost a straight line.

Complete Oscillation And Period Of Oscillation Or Time Period: Starting from an endpoint, when the pendulum bob again and returns to the same point, the pendulum completes one complete oscillation.

• The bob starting from point B reaches C and returns to B. This completes one complete oscillation. During one complete oscillation, the pendulum bob covers twice the total path of its movement. Hence, let us assume that the bob starts from point A towards point B.
• After reaching point B, the bob starts moving in the opposite direction, crosses A, and reaches point C, then from C the bob reaches A. A complete oscillation is executed in this manner also. After executing one complete oscillation the pendulum returns to its initial phase.
• The time taken by a pendulum to complete one oscillation is called the period of oscillation or the time period. In other words, the time period is the minimum time taken by the pendulum to return to its starting phase.

The movement from B to C is a half oscillation of the pendulum, and the time required for it is called the half-time period or half the period of oscillation.

Simple Pendulum Frequency: The number of complete oscillations executed in one second by a pendulum is its frequency. If the time period of a pendulum is T, then as per definition, the number of complete oscillations in time T = 1. Hence, in unit time, the number of complete oscillations = \(\frac{1}{t}\).

Now, from definition, the frequency n = \(\frac{1}{t}\).

The unit of frequency is s^-1 or hertz or Hz; the dimension is T^-1.

Motion Of A Simple Pendulum: A simple pendulum of effective length*¹ L is oscillating with m angular amplitude not exceeding 4°. The bob of the pendulum oscillates from B to C on either side of its position of equilibrium, O.

Let at any instant of motion, the bob of mass m be at P and its displacement from the position of equilibrium OP = x. If the angular displacement is θ, then θ = \(\frac{x}{L}\) rad, provided θ is small and sinθ ≈ θ ≈ tanθ.

At P, the weight mg of the bob acts vertically downwards. The component mg sinθ tries to bring the bob to the position of equilibrium. As this force acts in a direction opposite to that of displacement, it is the restoring force, expressed as

F = -mg sinθ

= -mgθ[since # is less than 4°]

= -mg\(\frac{X}{L}\)

Now, the acceleration of the bob,

a = \(\frac{F}{m}=\frac{-g}{L} x=-\omega^2 x \quad\left[\text { where, } \omega=\sqrt{\frac{g}{L}}\right]\)

As the motion of the bob obeys the equation, a = -ω²x, it can be said that the motion of a simple pendulum with an angular amplitude less than 4° is simple harmonic.

Time Period: Time period of the pendulum,

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{g}{L}}}\)

= \(2 \pi \sqrt{\frac{L}{g}}\)…(1)

Mechanical Energy Of The Pendulum: The kinetic energy of the pendulum,

K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \frac{g}{L}\left(A^2-x^2\right)\)

The potential energy of the pendulum,

U = \(\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \frac{g}{L} x^2\)

∴ The total mechanical energy of a simple pendulum,

E = K+ U = \(\frac{1}{2} m \omega^2 A^2=\frac{1}{2} m \frac{g A^2}{L}=\frac{m g A^2}{2 L}\)

Thus, if the angular amplitude is less than 4°, the total mechanical energy of a simple pendulum is

1. Directly proportional to the mass of the bob,
2. Inversely proportional to the effective length of the pendulum and
3. Directly proportional to the square of the amplitude of the SHM executed by the pendulum.

Tension On The String: When the pendulum oscillates about its point of suspension, a centripetal force is required for the circular motion of the bob. At the instant when the bob passes through the position of equilibrium O, its velocity becomes maximum. So, the upward centripetal force, Fc, becomes maximum at O. The resultant of downward weight mg and tension F’ on the string becomes equal to this force F.

Therefore, F_c = F’ – mg or, F’ = F_c + mg

The kinetic energy of the pendulum at P,

K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \frac{g}{L}\left(A^2-x^2\right)\)

So, \(m v^2=\frac{m g\left(A^2-x^2\right)}{L}\)

Therefore, the centripetal force at P,

∴ \(F_c=\frac{m v^2}{L}=\frac{m g\left(A^2-x^2\right)}{L^2}\)

∴ \(F^{\prime}=m g+F_c=m g+m g \frac{\left(A^2-x^2\right)}{L^2}\)

= \(m g\left(1+\frac{A^2-x^2}{L^2}\right)\)

When the bob passes through O (where x = 0 ), the tension F1 on the string becomes the maximum

∴ \(F_{\max }^{\prime}=m g\left(1+\frac{A^2}{L^2}\right)\)

Seconds Pendulum Pendulum Clock Definition: A simple pendulum of a time period of 2 seconds, or a half-time period of 1 second, is called a seconds pendulum.

A seconds pendulum has a time period, T = 2s

It is known, T = \(2 \pi \sqrt{\frac{L}{g}} or, 2=2 \pi \sqrt{\frac{L}{g}}\)

or, \(1=\pi \sqrt{\frac{L}{g}} or, L=\frac{g}{\pi^2}=\frac{981}{\pi^2}=99.40 \mathrm{~cm}=0.9940 \mathrm{~m}\).

Pendulum Dock Running Fast Or Slow: A pendulum clock marks time by means of its time period. From the equation T = \(2 \pi \sqrt{\frac{L}{g}},\), it can be shown that if

1. The effective length l of a pendulum changes, or
2. The value of the acceleration due to gravity changes, the time period of a second pendulum does not remain 2 s, but increases or decreases accordingly.

An increase in the time period means that the pendulum oscillates slowly, i.e., the clock goes slow. On the other hand, a decrease in the time period makes the clock run fast.

• For example, when a pendulum clock is taken from the poles to the equator, or from the sea level to the top of a mountain, or from the earth’s surface to deep inside a mine, the acceleration due to gravity g decreases; hence, the time period increases. Thus, in each case, the clock goes slow.
• A pendulum clock may run slow or fast depending on the temperature. At a higher temperature, due to the expansion of the metallic suspender and the bob, the effective length increases, and the time period also increases.

On the other hand, at low temperatures, the time period decreases. Therefore, a pendulum clock runs slow in summer and fast in winter. To get the correct time from the same pendulum at different temperatures, compensated pendulums are used.

Some Uses Of A Simple Pendulum

Finding The Value Of g: From equation (1) we get, g = \(4 \pi^2 \frac{L}{T^2}\)…(1)

Using this equation, the value of acceleration due to gravity (g) of a place can be determined. For different lengths of strings, the time period (T) and the effective length (L) of a pendulum can be evaluated by experiment.

Then \(\frac{L}{T^2}\) value in each case is determined and the average value of \(\frac{L}{T^2}\) is calculated. Substituting this average value in equation (1), the value of g can be obtained.

Determining The Height Of A Place: Let R = radius of the earth, h = height of a place (say, top of a hill), g = value of acceleration due to gravity on earth’s surface, g’ = value of acceleration due to gravity at height h, T =time period of a simple pendulum al a fixed place on the earth surface and T’ = time period of the same pendulum at height h.

From equation (1) Newtonian Gravitation and Planetary Motion,

⇒ \(\sqrt{\frac{g}{g^{\prime}}}=\frac{R+h}{R}=1+\frac{h}{R}\)…(2)

Also \(T=2 \pi \sqrt{\frac{L}{g}} and T^{\prime}=2 \pi \sqrt{\frac{L}{g^{\prime}}}\)

∴ \(\frac{T}{T^{\prime}}=\sqrt{\frac{g}{g}}\)

or,  \(\sqrt{\frac{g}{g^{\prime}}}=\frac{T^{\prime}}{T}\)

From (2) and (3),

1 + \(\frac{h}{R}=\frac{T^{\prime}}{T} \text { or, } \frac{h}{R}=\frac{T^{\prime}}{T}-1\)

or, h = \(\frac{R}{T}\left(T^{\prime}-T\right)\)

Knowing the value of R and determining the values of T and T’ using a stopwatch, the value of h can be determined from equation (4).

Finding The Depth Of A Mine: Let h = depth of the mine, g = value of acceleration due to gravity on earth’s surface, g’ = value of acceleration due to gravity at the depth h below earth’s surface, T = time period of a simple pendulum on the earth surface, T’ = time period of the same pendulum at depth h.

In this case, \(T=2 \pi \sqrt{\frac{L}{g}}\) and \(T^{\prime}=2 \pi \sqrt{\frac{L}{g^{\prime}}}\)

or, \(\frac{T}{T^{\prime}}=\sqrt{\frac{g^{\prime}}{g}} or, g^{\prime}=g \frac{T^2}{T^{\prime 2}}\)…(5)

From equation (4) Newtonian Gravitation and Planetary Motion, we get

⇒ \(g^{\prime}=g\left(1-\frac{h}{R}\right)\)…(6)

From equations (5) and (6), \(\frac{T^2}{T^{\prime 2}}=1-\frac{h}{R}\)

or, \(h=R\left(1-\frac{T^2}{T^{\prime 2}}\right)\)

Knowing the value of R and determining the values of T and T’ using a stopwatch, the value of h can be determined from equation (7).

Oscillation And Waves

Simple Harmonic Motion Simple Pendulum Numerical Examples

Example 1. A simple pendulum executes 40 complete oscillations in a minute. What is the effective length of the pendulum? g = 980 cm · s^-2.
Solution:

Time period (T) = \(\frac{60}{40}=\frac{3}{2} \mathrm{~s}\)

Now, T = \(2 \pi \sqrt{\frac{L}{g}} or, T^2=4 \pi^2 \frac{L}{g}\)

or, \(L=\frac{g T^2}{4 \pi^2}=\frac{980 \times\left(\frac{3}{2}\right)^2}{4 \times \pi^2}=55.9 \mathrm{~cm}\).

Example 2. What will be the percentage increase in the time period of a simple pendulum when its length is increased by 21%?
Solution:

Given, the increase in length of the pendulum = 0.21 L, where L is the initial length

Hence, increased length L’ = L + 0.21L = 1.211

Let, the time period change to T’ from T due to the change in length.

As T ∝ √L,

⇒ \(\frac{T}{\sqrt{L}}=\text { constant }\)

Thus, \(\frac{T}{\sqrt{L}}=\frac{T^{\prime}}{\sqrt{L}}\)

or, \(T^{\prime}=\sqrt{\frac{L^{\prime}}{L}} T=\sqrt{\frac{1.21 L}{L}} T=1.1 T\)

∴ Increase in time period = \(T^{\prime}-T=1.1 T-T\)

= 0.1 T = 10% of T

Hence, the time period increases by 10%.

Example 3. Two simple pendulums of lengths 100 cm and 101 cm are set Into oscillation at the same time. After what time does one pendulum gain one complete oscillation over the other?
Solution:

The length of the First pendulum is comparatively less, and hence, its time period is also less; thus the first pendulum oscillates faster.

By the time the second pendulum executes n oscillations, suppose the first one completes (n + 1) oscillations.

Hence, if T₁ and T₂ are time periods of the first and the second pendulums, \((n+1) T_1=n T_2 \text { or, } \frac{T_2}{T_1}=\frac{n+1}{n}=1+\frac{1}{n}\)…(1)

Again \(\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{101}{100}}=\left(1+\frac{1}{100}\right)^{1 / 2}\)

= \(1+\frac{1}{2} \times \frac{1}{100}\)=\(1+\frac{1}{200}\)…(2)

From equations (1) and (2) \(\frac{1}{n}=\frac{1}{200} \text { or, } n=200 \text {, i.e., } n+1=201\)

Thus, the required time = time of 201 complete oscillations of the first pendulum = \(201 \times T_1\)

= \(201 \times 2 \pi \sqrt{\frac{L_1}{g}}\)

= \(201 \times 2 \times \pi \sqrt{\frac{100}{980}} \approx 403 \mathrm{~s}=6 \mathrm{~min} 43 \mathrm{~s} .\)

Example 4. Find the length of a simple pendulum on the surface of the moon that has a time period same as that of a simple pendulum on the earth’s surface. The mass of Earth is 80 times that of the moon and the radius of Earth is 4 times that of moon.
Solution:

If masses of the earth and the moon are and respectively, \(\frac{M_1}{M_2}=80\)

Again, their radii are R₁ and R₂ (say).

So, \(\frac{R_1}{R_2}=4\)

Acceleration due to gravity on earth, \(g_1=\frac{G M_1}{R_1^2}\); acceleration due to gravity on moon, \(g_2=\frac{G M_2}{R_2^2}\)

∴ \(\frac{g_1}{g_2}=\frac{M_1}{M_2} \times \frac{R_2^2}{R_1^2}=\frac{M_1}{M_2} \times\left(\frac{R_2}{R_1}\right)^2=80 \times\left(\frac{1}{4}\right)^2=5\)

Also, in the case of a simple pendulum, time period on earth’s surface T = \(2 \pi \sqrt{\frac{L_1}{g_1}}\)

and time period on the moon’s surface T = \(2 \pi \sqrt{\frac{L_2}{g_2}}\)

∴ \(\sqrt{\frac{L_1}{g_1}}=\sqrt{\frac{L_2}{g_2}} \text { or, } \frac{L_1}{L_2}=\frac{g_1}{g_2}=5\)

∴ \(L_2=\frac{L_1}{5}\)

Hence, the length of the pendulum on the moon’s surface should be 1/5 th of its length on the earth’s surface.

Example 5. Two pendulums of time periods 1.8 s and 2 s are set Into oscillation at the same time. After how many seconds will the faster-moving pendulum execute one complete oscillation more than the other? How many oscillations will the faster-moving pendulum execute during this time?
Solution:

Suppose the faster pendulum executes one more oscillation than the other after t s.

Number of complete oscillations of the first pendulum in t s = \(\frac{t}{1.8}\)

Number of oscillations of the second pendulum in t s = \(\frac{t}{2}\)

According to the given condition, \(\frac{t}{1.8}\) – \(\frac{t}{2}\) = 1

or, 0.2t= 2×1.8 or, t = 18 s

Number of complete oscillations executed by the faster pen¬dulum (time period 1.8 s) = \(\frac{1.8}{1.8}\) = 10.

Example 6. A pendulum clock runs 20s slow per day. What should be the change in length of the clock so that it records the correct time? Take the pendulum as a simple pendulum.
Solution:

Half time period of a simple pendulum t = \(\frac{T}{2}=\pi \sqrt{\frac{L}{g}}\)

For a perfect seconds pendulum t = 1s

∴ 1 = \(\pi \sqrt{\frac{L}{g}} \text { or, } L=\frac{g}{\pi^2}\)….(1)

Also, 1d = 24 x 60 x 60 = 86400 s.

Number of half oscillations executed in a day by a pendulum that runs 20 s slow per day = 86400 – 20 = 86380.

Hence, the time period of that pendulum \(t_1=\frac{86400}{86380} \mathrm{~s}\)

If the length of this simple pendulum is L₁, then \(t_1=\pi \sqrt{\frac{L_1}{g}} \text { or, } L_1=\frac{g t_1^2}{\pi^2}\)…(2)

Subtracting equation (1) from equation (2), \(L_1-L=\frac{g}{\pi^2}\left(t_1^2-1\right)\) = \(\frac{980}{\pi^2}\left[\left(\frac{86400}{86380}\right)^2-1\right]\)

Hence, to get the correct time, the length of the pendulum is to be decreased by 0.46 mm.

Alternative Method: Half-time period of a perfect seconds pendulum = 1 s.

Number of Half oscillations of a clock that runs slow by t₀ s in a day = 86400 – t₀.

Hence, half time period = \(\frac{86400}{86400-t_0} \mathrm{~s} \)

∴ Increase in the value of half time period, dt = \(\frac{86400}{86400-t_0}-1=\frac{t_0}{86400-t_0} \mathrm{~s}\)

If the value of t₀ is negligibly smaller than 86400, dt \(=\frac{t_0}{86400} \mathrm{~s}\)

Now, t = \(pi \sqrt{\frac{L}{g}} or, \log t=\log \pi+\frac{1}{2} \log L-\frac{1}{2} \log g\)

Differentiating, \(\frac{d t}{t}=\frac{1}{2} \frac{d L}{L}-\frac{1}{2} \frac{d g}{g}\)…(1)

Here, dL = increase in length and dg = increase in acceleration due to gravity.

Putting values of t and dt in (1), \(\frac{t_0}{86400}=\frac{1}{2}\left(\frac{d L}{L}-\frac{d g}{g}\right)\)

or, \(t_0=43200\left(\frac{d L}{L}-\frac{d g}{g}\right)\)…(2)

This equation can be used as a rule for a seconds pendulum. For a second pendulum L = \(\frac{g}{\pi^2}\) and on the surface of the earth g = 980 cm · s^-2. If the clock runs fast, the value of t₀ is negative.

In the given problem, t₀ = 20 s;

If there is no change in the value of g, dg = 0

∴ \(t_0=43200 \times \frac{a L}{L}\)

or, \(d L=\frac{20 \times L}{43200}=\frac{20}{43200} \times \frac{980}{\pi^2}=0.046 \mathrm{~cm}=0.46 \mathrm{~mm}\)

Hence, the length of the defective clock has increased by 0.46 mm. Thus to get the correct time, its length needs to be decreased by 0.46 mm.

Example 7. A pendulum of length 60 cm is suspended inside an airplane. The aeroplane is flying up with an acceleration of 4 m · s^-2 making an angle of 30° with the horizontal. Find the time period of oscillation of the pendulum.
Solution:

Horizontal component of acceleration a of the plane = acos30° and vertical component = asin30°

Hence, the downward acceleration experienced by the pendulum bob = g- (-a sin30°) = g+ a sin30°

So, the acceleration of the pendulum bob

g’ = \(\sqrt{\left(g+a \sin 30^{\circ}\right)^2+\left(a \cos 30^{\circ}\right)^2} \)

= \(\sqrt{\left(9.8+4 \times \frac{1}{2}\right)^2+\left(4 \times \frac{\sqrt{3}}{2}\right)^2}=\sqrt{(11.8)^2+12}\)

= \(\sqrt{151.24}=12.3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ The time period of the pendulum

T = \(2 \pi \sqrt{\frac{L}{g^{\prime}}}=2 \times \pi \sqrt{\frac{0.60}{12.3}}=1.38 \mathrm{~s} .\)

Example 8. The effective length of a simple pendulum is 1 m and the mass of its bob is 5 g. If the amplitude of motion of the pendulum is 4 cm, what is the maximum tension on the string to which the bob is attached?
Solution:

The velocity of the bob is maximum at its position of equilibrium.

The maximum velocity \(v_{\max }=\omega A=\frac{2 \pi A}{T} .\)

At this position, centripetal force is also maximum, whose value is \(F_c=\frac{m \nu_{\max }^2}{L}=\frac{m \omega^2 A^2}{L}\)

[L = effective length of the pendulum]

The resultant of the weight mg of the bob and tension F on the string becomes equal to this force F_c .

∴ F = \(m g+F_c=m g+\frac{m \omega^2 A^2}{L}=m\left(g+\frac{\omega^2 A^2}{L}\right)\)

Again, \(T=2 \pi \sqrt{\frac{L}{g}} ; so, \omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{L}} or, \omega^2=\frac{g}{L}\)

∴ F = \(m\left(g+\frac{g A^2}{L^2}\right)=m g\left(1+\frac{A^2}{L^2}\right)\)

= \(5 \times 980\left(1+\frac{4^2}{100^2}\right)[because L=1 \mathrm{~m}=100 \mathrm{~cm}]\)

= \(4908 \mathrm{dyn}=0.04908 \mathrm{~N} \approx 0.05 \mathrm{~N}\)

Example 9. Prove that the change in the time period t of a simple pendulum due to a change AT of temperature is, \(\Delta t=\frac{1}{2} \alpha t \Delta T\), where α = coefficient of linear expansion.
Solution:

If L is the effective length of a simple pendulum, then its time period is, t = \(2 \pi \sqrt{\frac{L}{g}}\)

For a change ΔT of temperature, the length becomes, L’ = L(1 +αΔT)

Therefore the time period,

t’ = \(2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{L}{g}(1+\alpha \Delta T)}\)

= \(2 \pi \sqrt{\frac{L}{g}}(1+\alpha \Delta T)^{1 / 2}=t\left(1+\frac{1}{2} \alpha \Delta T\right)=t+\frac{1}{2} \alpha t \Delta T\)

[neglecting the terms containing a², a³, etc. since a is very small]

The change in time period, Δt = t’ -t = \(\frac{1}{2}\)αtΔT (Proved).

Alternative Method: t = \(2 \pi \sqrt{\frac{L}{g}}\)

log t = \(\log 2 \pi+\frac{1}{2} \log L-\frac{1}{2} \log g\)

Differentiating with respect to L, \(\frac{1}{t} d t=\frac{1}{2 L} d L\)

∴ \(\frac{d t}{t}=\frac{1}{2} \frac{d L}{L}\)…(1)

Now \(L_t=L(1+\alpha \Delta T)\)

or, \(L_t-L=L \alpha \Delta T or, d L=L \alpha \Delta T\)

or, \(\frac{d L}{L}=\alpha \Delta T\)…(2)

From equations (1) and (2) we get, \(\frac{d t}{t}=\frac{1}{2} \alpha \Delta T \text { or, } d t=\frac{1}{2} \alpha \Delta T \cdot t \text { (Proved). }\)

Example 10. The bob of a simple pendulum is made of brass and its time period is T. It is completely immersed in a liquid and is allowed to oscillate. If the density of the liquid is 1/8 th of the density of brass, what will be the time period of oscillation of the pendulum now?
Solution:

Initial time period of the simple pendulum, T = \(2 \pi \sqrt{\frac{L}{g}}\)

[L = effective length of the pendulum]

If m is the mass, V is the volume of the bob and d is the density of brass, then the apparent weight of the hob inside the liquid,

W₁ = W- buoyancy (weight of the displaced liquid)

= \(V d g-V \frac{d}{8} g=\frac{7}{8} V d g\)

If g₁ is the effective acceleration due to gravity in the immersed condition, then

⇒ \(W_1=m g_1=\frac{7}{8} V d g \text { or, } g_1=\frac{7}{8} \frac{m g}{m}=\frac{7}{8} g\)

∴ Final time period, \(T_1=2 \pi \sqrt{\frac{L}{g_1}}=2 \pi \sqrt{\frac{8 L}{7 g}}=\sqrt{\frac{8}{7}} T\).

Example 11. A brass sphere is hung from one end of a massless and inextensible thread. When the sphere is set into oscillation, it oscillates with a time period of T. If now the sphere is dipped completely into a non-viscous liquid, then what will be the time period of its oscillation? (The density of the liquid is 1/10th of that of brass)
Solution:

Let the volume of the sphere be V, the density of brass be ρ, the density of the liquid be ρ’.

∴ Apparent weight of the sphere when immersed in the liquid = real weight – weight of displaced liquid = Vρg-Vρ’g =Vg(ρ-ρ’)

∴ Apparent acceleration due to gravity of the sphere immersed in the liquid,

g’ = \(\frac{\text { apparent weight }}{\text { mass }}=\frac{V g\left(\rho-\rho^{\prime}\right)}{V \rho}=g\left(1-\frac{\rho^{\prime}}{\rho}\right)\)

According to the problem, \(\frac{\rho^{\prime}}{\rho}=\frac{1}{10}\);

hence \(g^{\prime}=g\left(1-\frac{1}{10}\right)=\frac{9}{10} g \text {. }\)

In the case of a simple pendulum, \(T\propto \frac{1}{\sqrt{g}}\); so, if the rime period of oscillation of the sphere, when unmersed in die liquid, is T’, then \(\frac{T^{\prime}}{T}=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{\frac{10}{9}} \text { or, } T^{\prime}=\frac{\sqrt{10}}{3} T \text {. }\)

Oscillation And Waves

Simple Harmonic Motion A Few Examples Of SHM

Oscillation Of A Mass Attached To A Vertical Elastic Spring: Let a body of mass m be attached to the bottom of a vertical elastic spring of negligible mass suspended from a rigid support, As a result of this, let the increase in length of the spring be l.

So the force constant of the spring.

k = force required for a unit increase in length = \(\frac{mg}{l}\)

Now the mass is pulled downwards through a distance x from its position of equilibrium O. If the extension of the spring does not exceed its elastic limit, then a reaction force, – kx, equal and opposite to the applied force is developed in the spring.

This force acts as the restoring force. If a is the acceleration of the suspended body, then restoring force -kx = ma

or, a = \(\frac{-k}{m} x=-\omega^2 x \quad\left[\text { where } \omega=\sqrt{\frac{k}{m}}\right]\)

As the motion of the body of mass m obeys the equation a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.

In this case, the time period of oscillation,

T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}} .\)

Now, k = \(\frac{m g}{l}\); therefore T = \(2 \pi \sqrt{\frac{m}{\frac{m g}{l}}}=2 \pi \sqrt{\frac{l}{g}}\)

Here, the initial increase in length of the spring due to suspension of the body of mass m is l. So by measuring this increase in length with a meter scale and the time period T with a stopwatch, acceleration due to gravity g can be calculated from the above relation.

Oscillation Of A Mass Attached To A Horizontal Elastic Spring: Let one end of an elastic spring of negligible mass be attached to a vertical support and its other end to a body of mass m. The body lies on a smooth horizontal plane. At this moment, no force acts on the body due to the spring as it is not stretched. So the body is at rest.

If the body is now moved towards the right, the spring will be elongated and a restoring force F will act on the body towards the left, trying to bring the mass to its equilibrium position.

If the force constant of the spring is k and the body is moved through a distance x towards the right, then F = – kx.

∴ Acceleration of the body, a = \(\frac{F}{m}=\frac{-k x}{m}=-\omega^2 x\left[\text { where } \omega=\sqrt{\frac{k}{m}}\right]\)

As the motion of the body obeys the equation, a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.

Time period of oscillation, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}}\)

It is to be noted that the time periods of vertical oscillation and horizontal oscillation of a spring are equal.

Oscillation Of A Liquid In A U-Tube: Consider that a U-tube of uniform cross-section a contains a liquid of density ρ. Let the length of the liquid column in each limb at equilibrium be L. Therefore the total length of the liquid column is 2L, if the horizontal separation between the two limbs is negligibly small.

Then the mass of the liquid column, m = 2Lαρ

If the liquid in one limb is depressed by x then the liquid level in the other limb will be raised by x. Hence the difference in the height of the liquid levels in the two limbs will be 2x.

Weight of this liquid head = 2xαρg.

This weight provides a restoring force trying to bring the liquid to its initial equilibrium. This force acts opposite to the direction of displacement x in the two limbs.

Thus, restoring force = -2xαρg.

Due to this force, if a is the acceleration of the liquid level, then ma = -2xαρg

or, 2Lαρa = -2xαρg

or, a = \(\frac{-g \cdot x}{L}=-\omega^2 x\left[\text { where } \omega=\sqrt{\frac{g}{L}}\right]\)

As the motion of the liquid level obeys the equation a = -ω²x, it follows a simple harmonic motion. So, if the liquid in one limb of a U-tube is depressed and then released, the up and down motion of the liquid column would be simple harmonic.

Time period of this motion, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{L}{g}}\)

Oscillation Of A Piston In A Gas Cylinder: Suppose, some amount of gas is enclosed in a cylinder fitted with a frictionless piston. Let the piston be initially at C, the position of equilibrium, the pressure of the gas enclosed be P and the length of the gas column be L.

The piston is now pushed down slightly to B very slowly and then released. The compressed gas will then expand and cause the piston to oscillate up and down.

When the piston is moved through a distance x from C to B, suppose the pressure of the enclosed gas increases from P to P + p and the volume decreases to (V- v) from V. If this change takes place isothermally, then according to Boyle’s law, PV = (P + p) (V- v)

or, PV = PV-Pv+ pV-pv

or, Pv = pV [neglecting pν as it is very small]

or, p = \(\frac{P v}{V}=\frac{P \alpha x}{\alpha L}=\frac{P x}{L}\)

[α = cross-sectional area of the piston]

An additional force acts on the piston for this excess pressure p and tries to bring the piston to its initial position of equilibrium.

So, the restoring force = \(-p \alpha=\frac{-P \alpha x}{L}\)

∴ Acceleration of the piston,

a = \(\frac{\text { restoring force }}{\text { mass of the piston }}=-\frac{P \alpha x}{L M}\)

[M = mass of the piston]

or, a = \(-\omega^2 x, \text { where } \omega=\sqrt{\frac{P \alpha}{L M}}\)

The motion of the piston obeys the equation a = -ω2x.

So this motion is simple harmonic. –

Time period of this motion, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{L M}{P \alpha}}=2 \pi \sqrt{\frac{L \alpha M}{P \alpha^2}}=2 \pi \sqrt{\frac{V M}{P \alpha^2}} .\)

 

Simple Harmonic Motion Synopsis

Periodic Motion: Any motion, which repeats itself at regular intervals of time is called periodic motion.

Oscillation Or Vibration: If a particle executing periodic motion moves to and fro along the same path, the motion is called oscillation or vibration.

Complete Oscillation: If an oscillating particle starting from any point on its path towards a certain direction returns to the same point and then moves in the same direction, it is said to have executed a complete oscillation.

Time Period: The time period of oscillation of a vibrating particle (T) is defined as the time taken by it to execute one complete oscillation.

Frequency: The frequency of oscillation of a vibrating particle (n) is defined as the number of complete oscillations executed by it in 1 second.

Amplitude: The magnitude of the maximum displacement of a vibrating particle on either side of its position of equilibrium is called the amplitude (A) of vibration.

Phase: The state of the motion at any instant of a particle executing simple harmonic motion is called its phase.

Initial Phase Or Epoch: Epoch or phase constant is the phase of the particle executing simple harmonic motion at the initial instant, i.e., at t = 0.

Simple Harmonic Motion: If the acceleration of a vibrating particle is

1. Directly proportional to the displacement of the particle from the position of equilibrium and
2. Is always directed towards the equilibrium position, then the motion is called a simple harmonic motion.

• All simple harmonic motions are periodic motions but all periodic motions are not simple harmonic.
• If it is possible to oscillate a small but heavy body suspended from a rigid support by means of a long, weightless, and inextensible string, then that system is called a simple pendulum.
• The motion of a simple pendulum is simple harmonic in nature if the angular amplitude of oscillation of the pendulum is less than 4°.
• A simple pendulum that has a time period of 2s or a half time period of Is is called a seconds pendulum.

Simple Harmonic Motion Useful Relations For Solving Numerical Problems

For a SHM, F ∝-x or, F = -kx; [where, F = restoring force, k = force constant, x = displacement of the particle from its equilibrium position,]

a = -ω²x and ω = 2πn [where, a = acceleration of the particle, n = frequency, ω = angular frequency]

Differential Equation Of Simple Harmonic Motion: \(\frac{d^2 x}{d t^2}=-\omega^2\)

The General Equation For Displacement In A Simple Harmonic Motion: x = A sin(ωt+ α)

(where, A = amplitude, α = initial phase]

The velocity of a particle executing simple harmonic motion, v = \(\pm \omega \sqrt{A^2-x^2}\) maximum velocity v_max = ± Aω [where x = A]; minimum velocity v_min = 0 (where x = ±A)

Acceleration, a = -ω²x, maximum acceleration, a_max = ω²A [where x = ±A]; minimum acceleration, a_min = 0 (where x = 0)

Kinetic energy, K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)

Potential energy, U = \(\frac{1}{2} m \omega^2 x^2\)

Total energy, E = \(K+U=\frac{1}{2} m \omega^2 A^2=\text { constant }\)

Time period, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x}{a}}=2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}\)

Frequency, n = \(\frac{1}{T}\)

Time period of a simple pendulum oscillating at an angular amplitude less than 4° is T = \(2 \pi \sqrt{\frac{L}{g}}\)

[where, L = effective length of the pendulum = length from the point of suspension of the pendulum to the center of gravity of the bob.]

In case of oscillation of a mass attached to a vertical elastic spring, time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{l}{g}}\)

[where, k = spring constant,  l = initial elongation of the spring due to the attachment of the mass m]

In case of oscillation of a mass attached to a horizontal elastic spring, time period, T = \(2 \pi \sqrt{\frac{m}{k}}\)

In case of oscillation of a liquid in a U-tube, time period, T = \(2 \pi \sqrt{\frac{L}{g}}\)

[Where L = length of the liquid column in each limb at equilibrium].

Simple Harmonic Motion Very Short Answer Type Questions

Question 1. If the time period of an SHM is 2 s, then what will be its frequency?
Answer: \(\frac{1}{2}\)Hz

Question 2. If the frequency of an SHM is 200 Hz, then what will be its time period?
Answer: \(\frac{1}{200}s\)

Question 3. What is the unit of force constant of SHM in SI?
Answer: N • m^-1

Question 4. The motion of the earth around the sun is a ______ motion.
Answer: Periodic

Question 5. What is the maximum displacement of a vibrating particle from the equilibrium position called?
Answer: Amplitude

Question 6. What is the phase difference between the displacement and the velocity of a particle executing SHM?
Answer: 90°

Question 7. What is the phase difference between the displacement and the acceleration of a particle executing SHM?
Answer: 180°

Question 8. If the time period is T then what will be the time taken by a particle executing SHM to traverse from the position of equilibrium to an extremity?
Answer: \(\frac{T}{4}\)

Question 9. What will be the change in the time period of a simple pendulum if the metallic bob of the pendulum is replaced by a wooden bob of the same radius? provided both bobs are of uniform density
Answer: Time Period Remains The same

Question 10. In which direction is the acceleration of a particle executing SHM directed?
Answer: Equilibrium

Question 11. At which points of its path, the velocity of a particle executing SHM becomes zero?
Answer: Extreme

Question 12. At which position, the velocity of a particle executing SHM becomes maximum?
Answer: Equilibrium

Question 13. At which position, the acceleration of a particle executing SHM become zero?
Answer: Equilibrium

Question 14. At which points of its path, the acceleration of a particle executing SHM becomes maximum?
Answer: Extreme

Question 15. A simple pendulum is oscillating in a vertical plane with a small amplitude. State whether the total energy at any point in its motion will be equal to that at an extreme point.
Answer: Yes

Question 16. At which position, the kinetic energy of a particle executing SHM becomes maximum?
Answer: Equilibrium

Question 17. At which points of its path, the potential energy of a particle executing SHM becomes maximum?
Answer: Extreme

Question 18. Total mechanical energy of a simple pendulum is directly proportional to the mass of the pendulum. Is the statement true or false?
Answer: True

Question 19. How is the total mechanical energy of a simple pendulum related to the length of the pendulum?
Answer: Inversely proportional

Question 20. Total mechanical energy of a simple pendulum is directly proportional to the amplitude of the pendulum. Is the statement true or false?
Answer: False

Question 21. The bob of a simple pendulum is made of iron. A powerful magnetic pole is placed below the bob in its equilibrium position. How will the time period of the pendulum change?
Answer: Decrease

Question 22. A body attached to a spring is executing SHM. If the force constant of the spring is increased then what will be the change in the frequency of oscillation?
Answer: Frequency will increase

Question 23. If a straight tunnel is bored from the north pole to the south pole of the earth and if a body is dropped into that tunnel then what time will the body take to move from one end to the other end of the tunnel?
Answer: 42 min

Question 24. What is the type of motion of a body along the tunnel passing through the centre of the earth?
Answer: Simple harmonic

Simple Harmonic Motion Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The total energy of a particle performing simple harmonic motion could be negative.

Statement 2: The potential energy of a system could be negative.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: The spring constant of a spring is k. When it is divided into n equal parts, then the spring constant of each piece is k/n.

Statement 2: The spring constant is independent of the material used for the spring.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: A particle performs a simple harmonic motion with amplitude A and angular frequency ω. To change the angular frequency of the simple harmonic motion to 3ω and amplitude to A/2, we have to supply an extra energy of (5/4) mω²A², where m is the mass of the particle executing simple harmonic motion.

Statement 2: The angular frequency of simple harmonic motion is independent of the amplitude of oscillation.

Answer: 4. Statement 1 is false, statement 2 is true.

Simple Harmonic Motion Match Column 1 With Column 2

Question 1. A particle of mass 2 kg is moving on a straight line under the action of force F = (8 – 2x)N. The particle is released from rest at x = 6m. For the subsequent motion match the following (all the values in Column 2 are in their SI units.)

Answer: 1. C, 2. D, 3. D, 4. B

Question 2. Two particles ‘ P ’ and ‘ Q ’ start SHM at t = 0. Their positions as a function of time are given by \(x_p=A \sin \omega t; \quad x_Q=A \sin \left(\omega t+\frac{\pi}{3}\right)\)

Answer: 1. B, 2. A, 3. D, 4. C

Simple Harmonic Motion Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A block of mass m is connected to a spring of spring constant k as shown. The block is found at its equilibrium position at t = 14 and it has a velocity of 0.25 m · s^-1 at t = 2s. The time period of oscillation is 6s.

1. The amplitude of oscillation is

1. \(\frac{3}{2 \pi} \mathrm{m}\)
2. \(3 \mathrm{~m}\)
3. \(\frac{1}{\pi} \mathrm{m}\)
4. \(1.5 \mathrm{~m}\)

Answer: 1. \(\frac{3}{2 \pi} \mathrm{m}\)

2. Determine the velocity of a particle at t = 5s.

1. -0.4 m · s^-1
2. 0.5 m · s^-1
3. -0.25 m · s^-1
4. None of these.

Answer: 3. -0.25 m · s^-1

Question 2. Two identical blocks A and B, each of mass m = 3 kg, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown. Another identical block C moving with velocity v₀ = 0.6 m · s^-1 collides with A and sticks to it. As a result, the motion of the system takes place in some way.

1. After the collision of C and A, the combined body and block B would

1. Oscillate about the center of mass of the system and the center of mass is at rest
2. Oscillate about the center of mass of the system and the center of mass is moving
3. Oscillate but about different locations other than the center of mass
4. Not oscillate

Answer: 2. Oscillate about the center of mass of the system and the center of mass is moving

2. Oscillation energy of the system, i.e., part of the energy which is oscillating (changing) between potential and kinetic forms, is

1. 0.27 J
2. 0.09 J
3. 0.18 J
4. 0.45 J

Answer: 2. 0.09 J

3. The maximum compression of the spring is

1. 3√30 mm
2. 3√20 mm
3. 3√10 mm
4. 3√50 mm

Answer: 3. 3√10 mm

Question 3. A particle suspended from a vertical spring oscillates 10 times per second. At the highest point of oscillation, the spring becomes unstretched. Take g = π² m • s^-2

1. The maximum speed of the particle is

1. 5π cm · s^-1
2. 4π cm · s^-1
3. 3π cm · s^-1
4. 2π cm · s^-1

Answer: 1. 5π cm · s^-1

2. The speed of the particle when the spring is stretched by 0.2 cm is

1. 15.4 cm · s^-1
2. 12.8 cm · s^-1
3. 10.8 cm · s^-1
4. 11.4 cm · s^-1

Answer: 1. 15.4 cm · s^-1

Question 4. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown. The pipe is in the horizontal plane. The centers of the balls can move in a circle of radius 0.06m. Each spring has a natural length of 0.06πm and a spring constant of 0.1 N · m^-1. Initially, both the balls are displaced by an angle θ = \(\frac{\pi}{6}\) radian with respect to the diameter PQ of the circle.

1. The frequency of oscillation of ball B is

1. π HZ
2. π^-1 HZ
3. π² HZ
4. π^-2 Hz

Answer: 1. π HZ

2. Speed of ball A when A and B are at the two ends of the diameter PQ

1. 0.05 m · s^-1
2. 0.071 m · s^-1
3. 0.0628 m · s^-1
4. 0.083 m · s^-1

Answer: 3. 0.0628 m · s^-1

3. The total energy of the system is

1. 4 x 10^-4 J
2. 5 x 10^-3 J
3. 4 x 10^-3J
4. 5 x 10^-4 J

Answer: 1. 4 x 10^-4 J

Question 5. A man has an antique pendulum clock of 1832 which bears the signature of the purchaser. He does not want to replace it in the fond memory of his great-grandparents. It ticks off one second in each side-to-side swing. It keeps the correct time at 20 °C. The pendulum shaft is made of steel and its mass can be ignored as compared to the mass of the bob. The linear expansion coefficient of steel is 1.2x 10^{-5 °}C^-1

1. What is the fractional change in length if the shaft is cooled to 10 °C?

1. 0.01 %
2. 1.2 x 10^-1 %
3. 1.2 x 10^-3 %
4. 1.2 x 10^-4 %

Answer: 1. 1.2 x 10^-2 %

2. How many seconds will the clock gain or lose in a day at 10 ºC?

1. Gains 5.2 s
2. Loses 5.2 s
3. Gains 10.4 s
4. Loses 10.4 s

Answer: 1. Gains 5.2 s

3. How closely must the temperature be controlled so that it does not gain or lose more than a second in a day?

1. ± 0.2 °C
2. ± 0.1 °C
3. ± 1 °C
4. ± 2 °C

Answer: 4. ± 2 °C

4. The pendulum mentioned in the paragraph is called _____ and its time period is_______

1. Seconds pendulum, 1s
2. Seconds pendulum, 2s
3. 2 Second pendulum, 2s
4. None

Answer: 2. Seconds pendulum, 2s

Simple Harmonic Motion Integer Type Question And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression 4v² = 25-x², what should be the value of (T/π)? [T is the time period (in second) of the SHM.]
Answer: 4

Question 2. Starting from the origin, a body oscillates simply harmonically with a period of 2s. After a certain time (t) its kinetic energy will be 75% of the total energy. What should be the value of 1/t (in s^-1)?
Answer: 6

Question 3. A particle executing SHM can be expressed by the equation x = 3cosωt+ 4sinωt. Find the amplitude of the resultant SHM.
Answer: 5

Question 4. Two pendulums of lengths 100 cm and 225 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
Answer: 2

Question 5. At a certain temperature, the pendulum of a clock keeps the correct time. The coefficient of linear expansion for the pendulum material = 1.85 x 10^-5 K^-1. How much will the clock gain or lose in 24 h if the ambient temperature is 10 °C higher?
Answer: 8

 

 Viscosity And Surface Tension

The flow of fluids is of two types

1. Laminar or streamline flow and
2. Turbulent flow.

Laminar or Streamline Flow Definition: A smooth, uninterrupted flow in ordered layers, without any energy transfer between the layers, is called a laminar streamline or steady flow.

The velocity of a fluid along its flow, in general, depends on its position and on time. This means that the fluid velocity may be different at different points, and at any particular point, it may also change with time.

Mathematically speaking, for a one-dimensional fluid flow along the x- direction, the fluid velocity \(\vec{v}\) is a function of position x and time t, i.e., \(\vec{v}\) = f(x, t).

For a laminar or streamline motion, the fluid satisfies the condition that \(\vec{v}\) is a function of x only, and not of r. It means that, at any particular point along the fluid flow, the magnitude and the direction of the fluid velocity do not change with time, although the velocity may be different at different points. In short, \(\vec{v}\) = f(x), but \(\vec{v}\) ≠ f(t)

In Fig a laminar flow for a liquid is shown. At points A, B, C, and D, let the flow velocities at any instant be ν_A, ν_B, ν_c, and ν_D respectively. Also at any subsequent time, a liquid particle that reaches the point A will have the velocity ν_A similarly, at point B, the velocity will be ν_B at point C the velocity will be ν_C and at point D. the velocity will be ν_D. It means that each particle of the liquid follows the velocity of its preceding particle and moves along the same path.

Streamline: In the case of streamline motion, the paths along which the particles of the fluid move are called streamlines. A tangent drawn at any point on this path indicates the direction of motion of the fluid at that point.

Properties of streamlining:

1. Two streamlines never intersect each other. Otherwise, at the point of intersection of two streamlines, two tangents can be drawn and hence two directions of motion of the particle are possible. Hut, in streamlined motion, any particle can move in one direction only and hence two streamlines can never intersect.

2. In the flow tube, where the streamlines are crowded together, the velocity of flow is higher. Where they are spaced, the velocity of flow is lower.

A special case of streamlined flow is a steady flow, for which the fluid velocity is a constant at all points along this flow at all times. So this velocity in neither a function of time, not of position. Example: a sufficiently slow liquid flow along a narrow uniform horizontal tube.

Turbulent Flow: In general, the motion of a fluid is streamlined, if its velocity does not exceed a definite limiting value. The limiting value of velocity is called the critical velocity.

If the velocity of a fluid exceeds the critical velocity, then the flow becomes turbulent, and in some regions, eddies and vortices are formed. This kind of flow is called a turbulent flow.

Turbulent Flow Definition: If the velocity of a fluid along its flow continuously and randomly changes in magnitude and direction, then it is called a turbulent or disorderly flow.

The path of fluid particles in turbulent flow is shown in Fig At every point along the flow, both the magnitude and the direction of fluid velocity change with time.

Turbulent Flow Experiment: Reynolds demonstrated the difference between streamline and turbulent flow by a simple experiment. a discharge pipe Q is attached horizontally to a vertical water-filled cylinder P. The flow rate may be varied with a valve at the end of the pipe.

• To make the flow behaviour visible, we use KMnO₄ solution, which is injected centrally into the horizontal pipe through a very narrow tube (diameter < 1 mm), positioned to avoid additional turbulence.
• If the velocity of the transparent liquid is low, the coloured liquid is observed to travel continuously in the form of a thread, indicating a streamline flow.
• As the flow of the transparent liquid is increased gradually, the coloured thread gets disrupted and later on the coloured liquid begins to move randomly, or forms eddies and vortices and mixes with the transparent liquid.
• The velocity of the transparent liquid at which this disturbance starts is called the critical velocity. It depends on the nature of the liquid, the cross-section of the tube, etc.

Viscosity And Surface Tension Viscosity: When a liquid flows slowly over a fixed horizontal surface, i.e., when the flow is laminar, the layer of the liquid in contact with the fixed surface remains at rest due to adhesion.

• The layer just above it moves slowly over the lower one, the third layer moves faster over the second one, and so on. The velocities of the layers of liquid increase with the increase in distance from the horizontal rigid surface.
• For two consecutive horizontal layers inside the liquid, the upper layer moves with a velocity greater than that of the lower one.
• The upper layer tends to accelerate the lower layer, while the lower layer tends to retard the upper one. In this way, the two adjacent layers tend to decrease their relative velocity—as if a tangential force acts on the upper layer and tries to oppose its motion.
• This tangential force is called viscous force. Therefore, to maintain a constant relative motion between the layers, an external force must act. If no external force acts, then the relative motion between the layers will cease and the flow of the liquid will stop.

Viscosity Definition: The property by virtue of which a liquid opposes the relative motion between its adjacent layers is called viscosity of the liquid.

Comparison of viscosity with friction: Viscosity is a general property of a fluid. The frictional force acting between two solid surfaces resembles in many ways the viscosity of a liquid.

• Hence, viscosity is called internal friction of a liquid. Like friction, the viscous force is absent if a liquid is at rest.
• The difference between the frictional force in solids and viscosity in liquids is that the viscous force depends on the area of liquid surface while the frictional force does not.

Viscosity and mobility of different liquids: Viscosities of different liquids are different. If alcohol and oil are poured separately into two identical vessels and stirred, then oil will come to rest earlier. This shows that the viscosity of oil is greater.

The greater the viscosity of a liquid, the lesser is its mobility. For example, the viscosity of honey is more than that of water and hence honey flows much slower than water. Coal tar has the least mobility.

Velocity profile: The surface formed by joining the end points of the velocity vectors of different layers of any section of a flowing liquid is called its velocity profile. Velocity profile for flow above a horizontal surface is shown in Fig.

Velocity profile of a non-viscous liquid: An ideal liquid is non-viscous. For such a liquid, there is no resistance due to viscosity. The velocities of the different layers are the same.

Every particle in a given cross-section of the liquid moves forward with the same velocity. On joining the ends of these velocity vectors, we get a plane surface. Therefore, we can say that the velocity profile of a non-viscous liquid is linear (on 2D graph).

Velocity profile of a viscous liquid: When a viscous liquid flows through a horizontal tube, the layer of liquid in contact with the wall of the tube remains stationary due to adhesion. So the velocity of that layer is zero.

• The layer of the liquid which flows along the axis of the tube has the maximum velocity. As we progress from the centre towards the walls, the velocity decreases.
• Therefore, on joining the ends of the velocity vectors, we get a parabolic surface. The velocity profile of a viscous liquid is a parabola (on 2D graph).

Coefficient of Viscosity: Let PQ be a solid horizontal surface. A liquid is in streamline motion over the surface PQ. Two liquid surfaces CD and MN are at distances x and (x+dx) respectively from the fixed solid surface. The velocity of layer CD is ν and that of layer MN is ν+ dν.

Due to the viscosity of the liquid, an opposing force acts between these two layers and tries to slow down the relative motion of the layers. If this opposing viscous force is F, then for streamline motion of the liquid, Newton proved that

1. F ∝ A; A = area of cross-section of the liquid surface, and
2. \(F \propto \frac{d v}{d x} ; \frac{d v}{d x}\) = velocity gradient = rate of change of velocity with distance perpendicular to the direction of flow.

∴ \(F \propto A \frac{d v}{d x} \text { or, } F=-\eta A \frac{d v}{d x}\) ….(1)

Here, η is a constant known as the coefficient of viscosity. Its value depends on the nature of the liquid.

Equation (1) is known as Newton’s formula for the streamline flow of a viscous liquid. Liquids that obey this law are called Newtonian liquids and liquids that do not obey this law are called non-Newtonian liquids.

From equation (1), we get, \(\eta=\frac{F}{A \frac{d v}{d x}}\)

If A = 1 and \(\frac{d v}{d x}=1\), then η = F; from this, we can define the coefficient of viscosity.

Coefficient of Viscosity Definition: The coefficient of viscosity of a liquid is defined as the required tangential force acting per unit area to maintain unit relative velocity between two liquid layers unit distance apart.

Units of coeffcient of viscocity: \(\eta=\frac{F}{A \frac{d v}{d x}}=\frac{F d x}{A d \nu}\)

So, unit of \(\eta=\frac{\mathrm{N} \cdot \mathrm{m}}{\mathrm{m}^2 \cdot\left(\mathrm{m} \cdot \mathrm{s}^{-1}\right)}=\mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}=\mathrm{Pa} \cdot \mathrm{s}\)

Unit:

• dyn · s · cm^-2 CGS System or g · cm^-1 · s^-1
• N · s · m^-2 or Pa · s or kg · m^-1 · s^-1 SI

Relation: \(1 \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}=\frac{1 \mathrm{~kg}}{1 \mathrm{~m} \times 1 \mathrm{~s}}=\frac{1000 \mathrm{~g}}{100 \mathrm{~cm} \times 1 \mathrm{~s}}\)

= 10 g · cm^-1 · s^-1

Poise and decompose: The coefficient of viscosity of a liquid is 1 poise, when a tangential force of 1 dyn is required to maintain a relative velocity of 1 cm · s^-1 between two parallel layers of the liquid 1 cm apart where each layer has an area of 1 cm².

So, 1 poise is the CGS unit of the coefficient of viscosity η.

1 poise = 1 dyn • s • cm^-2 = 1 g • cm^-1 • s^-1.

As, 1 kg · m^-1 · s^-1 = 10g · cm^-1 • s^-1 = 10 poise,

the SI unit of η is called 1 decapoise = 10 poise.

The coefficient of viscosity of a liquid is 1 decompose, when a tangential force of 1 newton is required to maintain a relative velocity of 1 m · s^-1 between two parallel layers separated by  distance of 1 m, where each layer has an area of 1 m².

Dimension of coefficient of viscosity: \([\eta]=\frac{[\mathrm{F}]}{[\mathrm{A}]\left[\frac{d \nu}{d x}\right]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \cdot \frac{\mathrm{LT}^{-1}}{\mathrm{~L}}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Effect of pressure and temperature on the coefficient of viscosity:

Effect of pressure: Usually, viscosity increases with pressure. In less viscous liquids, the viscosity increases at a low rate with pressure.

• But for highly viscous liquids, an increase in pressure results in a rapid rise in its viscosity. However, water behaves differently and, with an increase in pressure, its viscosity decreases.
• From the kinetic theory of gases, it is known that a change in pressure does not affect the viscosity of a gas. But for a large increase (or decrease) in pressure, viscosity is affected.

Effect of temperature: Usually, the coefficient of viscosity of liquids decreases with a rise in temperature. The relation between temperature and coefficient of viscosity is rather complicated. One commonly used equation relating these two is

⇒ \(\eta_t=\frac{A}{(1+B t)^C}\)

where, η_t = coefficient of viscosity of a liquid at t°C and A, B, and C are constants for a particular fluid.

For gases, the coefficient of viscosity increases with an increase in temperature.

Critical Velocity and Reynolds Number

Critical velocity: The maximum velocity of a fluid, up to which the flow of the fluid is streamlined and beyond which the flow becomes turbulent, is regarded as the critical velocity for that fluid.

On gradually increasing the velocity of a fluid, the streamline flow does not become turbulent abruptly. Rather this change occurs gradually.

With the help of experimental demonstration and also by dimensional analysis, it can be proved that the critical velocity (ν_c) of a fluid is

1. Inversely proportional to the density (ρ) of the fluid,
2. Directly proportional to the coefficient of viscosity (η) of the fluid, and
3. Inversely proportional to the characteristic length (l) of the channel. So,

⇒ \(v_c \propto \frac{\eta}{\rho l} \text { or, } v_c=N_c \cdot \frac{\eta}{\rho l}\) …..(1)

In the case of a tube, the characteristic length is the diameter of the tube while, for a canal, the characteristic length is its breadth.

If, for a liquid, ρ and η are known and its critical velocity ν_c can be determined experimentally during its flow through a tube of diameter l, then from equation (1), the value of the constant N_c for that liquid can be determined. This value is nearly 2300.

For any velocity ν of the fluid flow, equation (1) can also be written in an equivalent form as

⇒ \(v=N \cdot \frac{\eta}{\rho l} \text { or, } N=\frac{\rho l v}{\eta}\)…….(1)

N is called the Reynolds number.

Special cases:

1. If ν<ν_c, i.e., the velocity of fluid flow is less than the critical velocity, then comparing equations (1) and (2), we can say that N<N_c. It means that the value of Reynolds number is less than 2300. So, if the value of Reynolds number is less than 2300, then the flow will be streamlined.

2. On the other hand, if ν>ν_c, i.e., the velocity of the fluid is greater than the critical velocity, then N >N_c, and hence the value of Reynolds number will be greater than 2300. If Reynolds number is greater than 2300, then the flow will be turbulent.

Dimension of Reynolds number: From equation (2) we get, dimension of N

= \(\frac{\text { dimension of } \rho \times \text { dimension of } l \times \text { dimension of } \nu}{\text { dimension of } \eta}\)

= \(\frac{M L^{-3} \cdot L \cdot L T^{-1}}{M L^{-1} T^{-1}}=1\)

So, N is a dimensionless quantity; it is a pure number.

Reynolds number: A dimensionless number N= \(\frac{\rho l v}{\eta}\) can be formed by combining the characteristic length (l) of a fluid channel and the velocity (v), density (ρ) and coefficient of viscosity (η) of the fluid the magnitude of N determines whether the fluid flow is streamlined or turbulent. This number N is called the Reynolds number.

• In the above discussion, 2300 is an approximate value of Nc. Usually, for N < 2000, the fluid flow is streamlined, and for N> 3000 the fluid flow is turbulent. If N lies between 2000 and 3000, the streamline flow of a fluid gradually changes into turbulent flow.
• In the above discussion, 2300 is an approximate value of N_c. Usually, for N < 2000, the fluid flow is streamlined, and for N > 3000 the fluid flow is turbulent. If N lies between 2000 and 3000, the streamline flow of fluid gradually changes into turbulent flow.
• As N is a pure number, its value does not depend on the system of units chosen. For a particular flow, the value of N remains the same.

If the radius of a tube of flow is considered, instead of its diameter, then the effective value is, N_c ≈ 1150.

Viscosity And Surface Tension Numerical Example

Example: A plate of area 100 cm2 is floating on an oil of depth 2 mm. The coefficient of viscosity of oil is 15.5 poise. What horizontal force is required to move the plate horizontally with a velocity of 3 cm · s^-1?
Solution:

The viscous force, F = \(\eta A \frac{d v}{d x}\)

Here, A = 100 cm², η = 15.5 poise,

dν = 3 cm · s^-1 and dx = 2 mm = 0.2 cm.

∴ F = 15.5 x 100 x 3/0.2 = 23250 dyn

So the required horizontal force is 23250 dyn.

Terminal Velocity of a Body in a Viscous Medium and Stokes’ Law: When a body falls through a viscous medium (liquid or gas), it drags a layer of the fluid adjacent to it due to adhesion. But fluid layers at a large distance from the body are at rest.

• As a result, there is relative motion between different layers of the fluid at different distances from the body. But the viscosity of the fluid opposes this relative motion.
• The opposing force due to viscosity increases with increase in the velocity of the body due to the gravitational acceleration g. If the body is small in size, then after an interval of time the opposing upward force (i.e., viscous force and buoyant force) becomes equal to the downward force (weight of the body).
• Then the effective force acting on the body becomes zero and the body begins to fall through the medium with a uniform velocity, called the terminal velocity. A graph representing the change in velocity of a falling object with time is shown in Fig.

Stokes’ law: Stokes proved that, if a small sphere of radius r is falling with a terminal velocity ν through a medium of coefficient of viscosity η, then the opposing force acting on the sphere due to viscosity is

F = 6 πηrν ………..(1)

Equation (1) expresses Stokes.

1. To establish Stokes’ law, the following assumptions are
   made.
2. The fluid medium must be infinite and homogeneous. E3D The sphere must be rigid with a smooth surface.
3. The sphere must not slip when falling through the medium.
4. The fluid motion adjacent to the falling sphere must be streamlined.
5. The sphere must be small in size, but it must be greater than the intermolecular distance of the medium.

Equation for terminal velocity: if the density of the material of the sphere is ρ, then the weight of the sphere = \(\frac{4}{3} \pi r^3 \rho g .\)

If the density of the fluid medium is σ, then the upward buoyant force acting on the sphere = \(\frac{4}{3} \pi r^3 \sigma g\)

∴ The resultant downward force acting on the sphere =

= \(\frac{4}{3} \pi r^3 \rho g-\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3(\rho-\sigma) g\)….(2)

If the sphere attains terminal velocity, then

⇒ \(6 \pi \eta r \nu=\frac{4}{3} \pi r^3(\rho-\sigma) g \text { or, } \nu=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\) ….(3)

So, from equation (3), we see that the terminal velocity obeys the following rules.

1. Terminal velocity is directly proportional to the square of the radius of the sphere.
2. It is directly proportional to the difference of densities of the material of the sphere and that of the medium.
3. It is inversely proportional to the coefficient of viscosity of the medium.

If the density of the body is less than the density of the medium, i.e., ρ < σ, then it is clear that the terminal velocity becomes negative. Hence, the velocity of the body will be in the upward direction. For this reason, air or other gas bubbles move upwards through water.

Applications of Stokes’ law:

1. Falling of rain drops through air: Water vapour condenses on the particles suspended in air far above the ground to form tiny water droplets. The average radius of these tiny water droplets is 0.001 cm (approx.)

• Assuming the coefficient of viscosity of air as 1.8 x 10^-4  poise (approx.) the terminal velocity of these droplets is calculated as 1.2 cm · s^-1 (approx.) which is negligible. So, these water droplets float in the sky. Collectively these droplets form clouds.
• But as they coalesce to form larger drops, their terminal velocities increase. For example, the terminal velocity of a water droplet of radius 0.01cm becomes 120 cm · s^-1 (approx.). As a result, they cannot float any longer and so they come down as rain.

2. Coming down with the help of a parachute: When a soldier jumps from a flying airplane, he falls with acceleration due to gravity but due to viscous drag in air, the acceleration goes on decreasing till he acquires terminal velocity.

The soldier then descends with constant velocity and opens his parachute close to the ground at a pre-calculated moment, so that he may land safely near his destination.

Terminal Velocity Numerical Examples

Example 1. An oil drop of density 950 kg · m^-3 and radius 10^-6 m is falling through air. The density of air is 1.3 kg · m^-3 and its coefficient of viscosity is 181 x 10^-7 SI unit. Determine the terminal velocity of the oil drop, [g = 9.8 m · s^-2]
Solution:

Terminal velocity, \(\nu=\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{\eta}\)

[Here, ρ = 950 kg · m^-3 ; r = 10^-6 m; σ = 1.3 kg · m^-3; η = 181 x 10^-7 SI]

= \(\frac{2}{9} \cdot \frac{\left(10^{-6}\right)^2(950-1.3) \times 9.8}{181 \times 10^{-7}}\)

= \(1.14 \times 10^{-4} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 2. An air bubble of radius 1 cm is rising from the bottom of a long liquid column. If its terminal velocity is 0.21 cm · s^-1, calculate the coefficient of viscosity of the liquid. Given that the density of the liquid is 1.47 g · cm^-3. Ignore the density of air.
Solution:

Coefficient of viscosity of the liquid, \(\eta=\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{v}\)

[Here, r = 1cm; v = -0.21 cm · s^-1; ρ = 0; cσ = 1.41 g · cm^-3]

= \(\frac{2}{9} \times \frac{(1)^2(0-1.47) \times 980}{-0.21}=1524.4 \text { poise. }\)

 

Viscosity And Surface Tension Surface Tension Of Liquids

All liquids possess a special property—a tendency to minimise its surface area. This tendency of a liquid surface to contract its area is called surface tension. From our practical experience, we know that water droplets, or a small amount of mercury always takes the shape of a sphere.

• In the absence of external forces, all liquids always take a spherical shape. For a given volume, the surface area of a sphere is the least and hence a liquid drop has a natural tendency to take the shape of a sphere.
• If a clean dry needle is placed horizontally on the surface of water, then it is observed that the needle floats on water. The water surface under the needle is slightly depressed.
• Insects like spiders and mosquitoes are able to walk on the surface of water. Where their legs touch, the water surface becomes slightly depressed.
• From such observations, it seems that the surface of water behaves like a stretched rubber membrane.

Experimental demonstration: A wire loop is dipped into a soap solution. A thin soap film will be formed in the loop when it is taken out of the solution. This film acts as the free surface of the liquid.

A wet cotton thread loop, after being dipped in soap solution, is put on the film. No change is seen in the shape of the loop. The soap film inside the loop is now punctured with a fine needle and it is observed that the cotton thread pulls itself into a circle. What is the reason behind this observation?

• Initially, there was soap film inside and outside the cotton loop. Every point on the loop experienced equal and opposite forces tangential to the surface of the film. These two forces balanced each other. As a result, no resultant force acted on the loop.
• After the film inside the loop was removed, the inward force vanished and only the film outside the loop exerted a force on the thread. We know that among all plane surfaces having the same boundary length, the area of a circle is the greatest.
• Hence, a circle formed by the loop occupies the maximum area. So, the area of the film in between the loop and the thread reduces to a minimum. From this, it can be inferred clearly that the film has a tendency to minimize its area.
• It can be concluded that a tension always acts on the free surface of a liquid and that the free surface behaves as a stretched-thin membrane. Due to this tension, the free surface of any liquid has a tendency to contract so as to occupy the minimum area. This tension is known as surface tension.

Surface tension: Surface tension is the property of the free surface of a liquid due to which the liquid behaves as a stretched thin membrane and has a tendency to contract so as to minimise the surface area.

• As a reason behind the origin of surface tension, it can be said that the molecules of a liquid attract each other by cohesive force [the force of attraction which acts between the molecules of same material is called cohesive force).
• Equal cohesive force acts on the molecule inside the liquid from all directions.
• Consequently, the resultant cohesive force on the molecule is zero. But, no cohesive forces act on the molecules of the free surface of the liquid to the outward direction. So, the cohesive force inside the liquid is not balanced.
• As a result, a resultant cohesive force acts on each molecule of the free surface in the downward direction. Thus, the free surface of a liquid tends to have the least surface area.
• Let us imagine a straight line on the free surface of a liquid. Due to the tendency of the liquid surface to contract, the molecules on the opposite sides of the line try to move away from each other. This can be seen in the following experiment.

If a matchstick is placed on a water Surface, it remains at rest. But when a drop of alcohol is put on water on one side of the stick, the stick moves in the opposite direction.

1. The water surface exerts an equal pressure on both sides of the stick. This force is normal to the stick.

2. When a drop of alcohol is put on water, the force on that side is weaker and the stick is pulled away by the stronger tangential force towards the opposite side.

Hence, surface tension can also be defined as follows:

Surface tension Definition: The tangential force per unit length on a liquid surface, that acts along the normal on either side of an imaginary line on that surface, is called the surface tension of the liquid.

Units:

• dyn · cm^-1 CGS System
• N · m SI

Relation: \(1 \mathrm{~N} \cdot \mathrm{m}^{-1}=\frac{1 \mathrm{~N}}{1 \mathrm{~m}}=\frac{10^5 \mathrm{dyn}}{10^2 \mathrm{~cm}}=10^3 \mathrm{dyn} \cdot \mathrm{cm}^{-1}\)

Dimensional: \([\text { Surface tension }]=\frac{[\text { force }]}{\text { [length }]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{MT}^{-2}\)

Surface Energy: We know that on the free surface of a liquid, the surface tension always tries to minimize the surface area. So, to increase the area of the surface of the liquid, an external force is needed.

• The external force does work to increase the area of the surface of the liquid, and the work done remains stored inside the surface of the liquid as potential energy.
• The surface energy of a liquid is measured by the work done to increase the area of the surface of a liquid by unity.

Units:

• erg · cm^-2 CGS System
• J · m^-2 SI

Relation: \(1 \mathrm{~J} \cdot \mathrm{m}^{-2}=\frac{1 \mathrm{~J}}{1 \mathrm{~m}^2}=\frac{10^7 \mathrm{erg}}{10^4 \mathrm{~cm}^2}=10^3 \mathrm{erg} \cdot \mathrm{cm}^{-2}\)

Dimension: \([\text { Surface energy }]=\frac{[\text { work }]}{[\text { area }]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2}=\mathrm{MT}^{-2}\)

Relation between surface tension and surface energy: A rectangular wire frame PQRS is taken. A wire AB can move along PQ and SR.

When the frame is dipped into a soap solution and taken out, a thin film is formed within the frame.

As a result, the surface tension acts normally on the wire AB and tangentially to the surface of the film. This force tries to contract the film surface and pulls the wire AB towards QR. To keep the wire AB in its position, an equal but opposite force needs to be applied on it.

Let the length of the wire AB be l; the surface tension of the liquid be T.

∴ The net force acting on the wire AB in the direction QR = 2lT. [The film has two surfaces and surface tension acts on each surface, therefore the net force has the factor of 2]

∴ To keep the wire AB still, the force required to be applied in the opposite direction, F = 2lT

Now, the amount of work done in displacing the wire AB through a short distance δx against the surface tension (assuming the force F to be a constant throughout the displacement) so that it comes to the new position A’B’ is

Fδx = 2lTδx

Due to this, the total increase in the area of the film surface = 2 lδx.

This work remains stored as potential energy on the film surface.

∴ Work done for unit increase in area against the surface tension = \(\frac{2 l T \delta x}{2 l \delta x}=\) = T

So, the potential energy stored per unit area or the surface energy is numerically equal to the surface tension of the liquid. Note that the temperature is assumed to be constant.

Alternative definition of surface tension: Keeping the temperature constant, the amount of work done in increasing the area of a liquid surface by unity is called the surface tension of that liquid at that temperature.

Units:

• erg · cm^-2 CGS System
• J · m^-2 SI Units

Dimension: According to the alternative definition,

⇒ \([\text { surface tension }]=\frac{[\text { work }]}{[\text { area }]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2}=\mathrm{MT}^{-2}\)

Total surface energy: in the discussion, it was assumed that, during increase in the area of a liquid surface under the influence of an applied force, the temperature remains constant. But, in practice, some molecules from inside the liquid rise to its surface during the expansion of the surface.

• A resultant attraction exerted by the other molecules inside the liquid acts on these moving molecules. Therefore, these molecules on reaching the surface lose their linear kinetic energy and the average linear kinetic energy of the total liquid decreases.
• Since the temperature is directly proportional to the average kinetic energy, the temperature of the surface of the liquid decreases with an increase in its area. To keep the temperature constant, the liquid surface absorbs heat from its surroundings.

To increase the area of a liquid surface keeping the temperature constant, energy may be supplied in two ways

1. Mechanical energy to increase the surface area and
2. Heat energy to keep the temperature constant. The total of these two energies should actually be the surface energy.

So, the increase in potential energy per unit surface area or stored surface energy (E) = mechanical energy or work done (T) + heat (h) required for unit area

i.e., E = T+h……(1)

From thermodynamics, it can be proved that, h = \(-\theta \frac{d T}{d \theta}\)

[θ = temperature in absolute scale and dT/dθ = rate of increase in surface tension due to increase in temperature]

∴ E = \(T-\theta \frac{d T}{d \theta}\)…..(2)

Now, with the increase in temperature, the surface tension decreases and hence \(-\theta \frac{d T}{d \theta}\) is a negative quantity.

So, h is a positive quantity.

∴ E = \(T+\theta \frac{d T}{d \theta}\)…..(3)

when only the magnitudes are considered.

Again, at the absolute zero temperature, i.e., when θ = 0, E = T.

So, at any temperature except absolute zero, the total surface energy of a liquid is always greater than the surface tension of that liquid.

Factors Affecting Surface Tension of a Liquid: Surface tension of a liquid depends on the following factors.

1. Temperature of the liquid: With an increase in the temperature, the surface tension of almost all liquids decreases. For a small change in temperature, the relation between surface tension and temperature is

T’ = \(T\left[1-\alpha\left(t^{\prime}-t\right)\right]\)

Here, T and T’ are the surface tensions at temperatures t and t’ respectively]

For a given liquid, α is a constant quantity. It is called the temperature coefficient of surface tension.

It is experimentally verified that at a specific temperature of every liquid, the surface tension of the liquid disappears. This temperature is called the critical temperature of that liquid.

2. Pollution: If impurities are present on a liquid surface, then the surface tension of that liquid usually decreases. For example, when an oil or a fat-like substance is poured over water, it forms a thin film over the surface of water. This decreases the original surface tension of water.

3. Presence of dissolved substances: if a liquid contains dissolved inorganic substances, then the surface tension of that liquid increases. Again, if the liquid contains dissolved organic matter, then its surface tension decreases. For example, the surface tension of pure water is 0.072 N · m^-1.

If common salt (inorganic substance) is dissolved in water, then its surface tension becomes 0.083 N · m^-1 (approx.), but the surface tension of soap-water (organic substance) is approximately 0.030 N · m^-1.

4. Medium above the liquid surface: The surface tension of a liquid depends on the nature of the medium above the free surface of that liquid. For example, the surface tension of water is about 72 dyn · cm^-1 in the presence of dry air above the surface of water, but is about 70 dyn · cm^-1 when there is moist air above the surface of water at the same temperature.

5. Presence of electric charge: The surface tension of a liquid decreases due to the presence of electric charge on the surface of the liquid.

Some Phenomena in Connection with Surface Tension

1. Camphor darts to and fro when put on the Surface Of water: Camphor is soluble in water. When put on water, the portion that comes into contact with the water begins to dissolve. The part which gets dissolved in water contaminates the water and the surface tension of that part decreases. Due to this difference in surface tension, a net unbalanced force acts on the piece of camphor, and consequently, the piece of camphor darts to and fro on the surface of water.

2. Hair of a paint brush cling together when the brush is brought out of water: The hair of a brush lie apart while immersed in water because, inside water, the surface tension is absent. But when the brush is brought out of the water, a thin film of water clings to the hair, and the surface tension tries to contract the area of the film hence the hair clings together.

3. Turbulent sea calms down If oil is poured on the water: The surface tension of pure water is more than that of oily water. When oil is poured over sea water, the oil spreads in the direction of motion of the waves leaving uncovered sea water at the rear. Hence, the surface tension of the water ahead of the waves is lower than that of the water behind the waves. The water at the rear pulls the water at the front and, as a result, high waves become lower.

4. When oil is poured on water, it spreads readily Over the entire surface: Since the surface tension of pure water is greater than the surface tension of oil, a tensile force acts on the surface of oil. Due to this tensile force, oil spreads readily over the entire surface of water.

5. When chalk dust is sprinkled on water and a few drops of alcohol is added, then the dust particles rapidly spread on the water surface: Alcohol decreases the surface tension. Due to unequal surface tension on different parts of the water surface, the chalk particles spread rapidly on the surface of water.

6. Water cannot seep in through the cloth of raincoats, umbrellas and tents: The minute pores in the cloth of raincoats etc. trap air molecules. However, these pores are too small to let rain droplets enter, because the droplets retain their spherical shape due to surface tension, and the diameters of the spheres are greater than that of the pores. So, the rainwater falling on the cloth simply flows off.

7. A needle Coin float on water surface: A needle floats due to the surface tension of water. The surface of water where the needle is placed experiences a slight depression due to the surface tension of water. So, the water exerts an upward force on the needle which balances its weight (acting downwards). Therefore, a needle can float if it is placed carefully on a calm water surface.

Viscosity And Surface Tension Surface Tension Of Liquids Numerical Examples

Example 1. The surface tension of water at 20 °C is 72 dyn · cm^-1 and for water dT/dθ = -0.146 dyn · cm^-1 · K^-1. What is the total surface energy of water?
Solution:

We know that the total surface energy of water,

E = T – θ dT/dθ = 72 – 293 x (-0.146) [20°C = 293 K]

= 72 + 42.778 = 114.778 erg · cm^-2.

Example 2. A drop of water of radius 1 mm is to be divided into 10⁶ point drops of equal size. How much mechanical work should be done? The surface tension of water = 72 dyn · cm^-1.
Solution:

Let the radius of each point drop be r

∴ \(\frac{4}{3} \pi r^3 \times 10^6=\frac{4}{3} \pi\left(\frac{1}{10}\right)^3 \quad \text { or, } r=0.001 \mathrm{~cm}\)

The surface area of the original drop = \(4 \pi\left(\frac{1}{10}\right)^2 \mathrm{~cm}^2\)

and the total surface area of 10⁶ point drops = 10⁶ x 4π(0.001)² = 4π cm².

∴ Increase in surface area = \(4 \pi-4 \pi\left(\frac{1}{10}\right)^2=4 \pi \times 0.99 \mathrm{~cm}^2\)

= \(4 \pi-4 \pi\left(\frac{1}{10}\right)^2=4 \pi \times 0.99 \mathrm{~cm}^2\)

∴ Mechanical work done = increase in surface area x surface tension

= 72 x 4π x 0.99 = 895.73 erg

Example 3. 1000 water droplets having a radius of 0.01 cm each coalesce to form a single big drop. What will be the decrease in energy? The surface tension of water = 72 dyn · cm^-1
Solution:

Let the radius of the single big drop be R.

∴ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.01)^3 \times 1000 \quad \text { or, } R=0.1 \mathrm{~cm}\)

Surface area of the big drop = 4π (0.1)² cm²

Total surface area of 1000 droplets

= 4π (0.01)² x 1000 cm²

∴ Decrease in area

= 4π (0.01)² x 1000-4π(0.1)²

= 4π(0.1 -0.01) = 47 x 0.09 cm³

∴ Decrease in energy = 4π x 0.09 x 72 = 81.43 erg.

Example 4. A rectangular glass slab measures 0.1 m x 0.0154 m x 0.002 m and its weight in air is 80.36 x 10^-3 N. The slab is immersed half in water keeping its length and thickness horizontal. What will be the apparent weight of the slab? The surface tension of water is 72 x 10^-3 N · m^-1.
Solution:

While it is immersed, the following forces act on the glass slab

1. Weight of the slab acting downwards,
2. Upward buoyant force due to the weight of displaced water and
3. Downward force due to surface tension.

Now, weight of the slab = 80.36 x 10^-3 N

Buoyant force = weight of displaced water

= \(0.1 \times \frac{0.0154}{2} \times 0.002 \times 1000 \times 9.8\)

= 15.092 x 10^-3 N

Force due to surface tension

= 2 x (0.1 + 0.002) x 72 x 10^-3 N

= 14.688 x 10^-3 N

∴ The apparent weight of the slab

= 80.36 x 10^-3 – 15.092 x 10^-3 + 14.688 x 10^-3

= 79.956 x 10^-3 N.

Example 5. The radius of a soap bubble is increased from 1 cm to 3 cm. What amount of work is done for this? The surface tension of soap-water is 26 dyn • cm^-1•
Solution:

Work done = increase in area x surface tension =

47{(3)²-(1)²} x 26 x 2 [the soap-bubble has two surfaces]

= 5227.6 erg = 5.2276 x 10^-4 J.

Example 6. Determine the surface energy of a liquid film formed on a ring of area 0.15 m². The surface tension of the liquid = 5 N · m^-1.
Solution:

Surface energy, E = 2 x surface tension x area =2 x 5 x 0.15= 1.5 J

Example 7. Determine the surface energy of a soap-water film formed on a frame of area 10^-3 m². Surface tension of soap-water = 70 x 10^-3 N · m^-1.
Solution:

Surface energy = 2 x surface tension x area

= 2 x 70 x 10^-3 x 10^-3 = 14 x 10^-5 J.

Example 8. What will be the work done to form a soap bubble of radius 5 cm? The surface tension of soap-water = 70 dyn · cm^-1
Solution:

Work done =2 x 4πr² x T =8πr² T

[r = radius of the soap-bubble and T = surface tension of soap-water]

= 8 x 22/7 x (5)² x 70 = 44000 erg

= 0.0044 J.

Example 9.  If a large number of water droplets of diameter 2rcm each coalesce to form a large water drop of diameter 2jRcm, then prove that the rise in tem-perature of water is \(\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)\). Here, T is the surface tension of water and J is the mechanical equivalent of heat.
Solution:

If the number of small water droplets is n, then the dissipation of surface energy, W = (4πr²n-4πR²)T.

We know that W = JH and H = heat absorbed = msθ

[where m = mass of the larger water drop, s = specific heat of water, 6θ = temperature increase]

H = \(\frac{W}{J}=\frac{T}{J} \times 4 \pi\left(n r^2-R^2\right)\)

or, \(\frac{4}{3} \pi R^3 \cdot 1 \cdot \theta=\frac{T}{J} \times 4 \pi\left(n r^2-R^2\right)\)

⇒ \({\left[because m=\frac{4}{3} \pi R^3 \cdot 1=\frac{4}{3} \pi R^3 \text { and } s=1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}-1\right]}\)

or, \(\theta= \frac{3 T}{J} \frac{\left(n r^2-R^2\right)}{R^3}\)

⇒ \({\left[\text { here, } n \cdot \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \text { or, } R^3=n r^3 \text { or, } n=\frac{R^3}{r^3}\right]}\)

= \(\frac{3 T}{J}\left[\frac{R^3}{r^3} \cdot \frac{r^2}{R^3}-\frac{1}{R}\right]\)

= \(\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)\)

Example 10. Water is filled upto a height h in a beaker of radius R as shown in the Fig. 3.24. The density of water is p, the surface tension of water is T and the atmospheric pressure is P₀. Consider a vertical section ABCD of the water column through a diameter of the beaker. What is the force on water on one side of this section by water on the other side of this section?

Solution:

The force acted on the liquid of left side by the liquid of right side is equal to the resultant of the following two forces.

1. Force due to surface tension =2RT (right side).

2. Impulse due to the exerted pressure by the liquid of height h

= \(\left(P_0+\frac{\rho g h}{2}\right) \times 2 R h\) [where \(\frac{\rho g h}{2}\)= average pressure on the plane A B C D]

= \(2 P_0 R h+\dot{R} \rho g h^2\) (along left side)

∴ The resultant force on the plane ABCD = \(2 P_0 R h+R \rho g h^2-2 R T\)

Example 11. When water in a beaker is gradually heated, a bubble formed at the lower surface of the beaker starts to rise up from the bottom of the beaker. The radius of the spherical bubble is R and the radius of the circular region of the bubble touched with the lower surface of the container is r(r<<R). Show that, the value of r will be \(R^2 \sqrt{\frac{2 \rho_u g}{3 T}}\) just before the detached from the lower surface of the container [where, ρ_w = density of water, T = surface tension of water]. Consider, though the density and surface tension are unchanged with the increase in temperature, but the density of air changes significantly.

Solution:

At that instant when the bubble is just detached from the lower surface of the beaker, the buoyancy force = the force due to surface tension.

Now, the net force due to surface tension = (2πr)T sinθ (directed downward)

and the buoyancy force = \(\frac{4}{3} \pi R^3 \rho_w g\) (directed upward)

In this case, \(\frac{4}{3} \pi R^3 \rho_w g=(T)(2 \pi r) \sin \theta\)

[as θ is very small, sin/θ ≈ tanθ = r/R]

or, \(\frac{4}{3} \pi R^3 \rho_w g=(T)(2 \pi r)\left(\frac{r}{R}\right)\)

or, \(\frac{4}{3} \pi R^3 \rho_w g=T \cdot 2 \pi \frac{r^2}{R}\)

or, \(r^2=\frac{2\left(R^4 \rho_w g\right)}{3 T}\)

∴ r = \(R^2 \sqrt{\frac{2 \rho_w g}{3 T}}\)

Pressure Difference Between The Two Sides Of A Curved Liquid Surface

1. Suppose the free surface of a liquid is plane. A molecule lying on its surface is attracted by other surface molecules equally in all directions. So the resultant tangential force on the molecule is zero.

2. If the free surface of the liquid is concave, then every molecule on the surface experiences an upward resultant force due to attraction by other surface molecules.

3. If the liquid surface is convex, then the resultant force on a molecule on the surface due to attraction by other surface molecules will be directed downwards.

• Obviously, there must be a difference of pressure between the two sides of a curved surface for equilibrium of it. This difference of pressure i.e., the excess pressure force will balance the resultant force due to surface tension. The pressure on the concave side must be greater than the pressure on the convex side.
• We shall now calculate the excess pressure on the concave side of spherical surfaces in case of a liquid drop, an air bubble in a liquid, and a soap bubble.

Excess pressure inside a liquid drop: Let us consider a liquid drop of radius R of a liquid of surface tension T. Every molecule on its surface experiences a resultant pull normally inwards due to surface tension.

So the internal pressure of the drop becomes greater than the pressure outside it. The internal excess pressure of the drop produces a force acting outwards which balances the force due to surface tension and maintains the equilibrium of the drop.

Suppose, external pressure on the drop =P, internal pressure of the drop = (P+ p).

So, the excess pressure inside the drop = p.

Suppose this internal excess pressure acting normally outwards increases the radius of the drop from R to R + ΔR i.e., it increases the surface area of the drop. Here ΔR is taken to be so small that the pressure inside the drop may be taken as unchanged.

Work done by the excess pressure,

W = Excess pressure x area x displacement

= p · 4πR² · ΔR ………(1)

Increase of surface area of the liquid drop,

⇒ \(\Delta A =4 \pi(R+\Delta R)^2-4 \pi R^2\)

= \(4 \pi\left\{R^2+2 R \cdot \Delta R+(\Delta R)^2-R^2\right\}\)

= \(8 \pi R \cdot \Delta R\) ; [neglecting the term \((\Delta R)^2\) which is very small]

∴ Increase in surface energy,

E = increase in surface area x surface tension

= 8πR · ΔR · T …….(2)

This increase in surface energy of the liquid drop takes place at the cost of work done by the excess pressure i.e., E = W.

So, from equation (1) and (2) we have, p · 4πR² · ΔR = 8π R ⋅ ΔR · T

or, p = 2T/R …..(3)

Excess pressure inside an air bubble in a liquid: Let us consider an air bubble of radius R formed in a liquid of surface tension T. Like a liquid drop the air bubble has also one surface in contact with the liquid.

So proceeding similarly as in the case of a liquid drop we can prove that the excess pressure inside the air bubble in a liquid is given by p = 2T/R.

Excess pressure inside a soap bubble: Let us consider a thin soap bubble of radius R formed from a soap solution of surface tension T.

Suppose, the pressure outside the bubble = P, internal pressure =(P+p)

So, the excess pressure inside the bubble = p. Suppose, the radius of the bubble increases from R to R + ΔR due to this internal excess pressure acting normally outwards, i.e., the surface area of the bubble increases.

Here ΔR is taken to be so small that the pressure inside the bubble may be taken as unchanged.

Work done by the excess pressure,

W = excess pressure x area x displacement = p · 4πR² · ΔR …..(1)

The soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble.

So, increase of surface area of the soap bubble,

⇒ \(\Delta A =2\left[4 \pi(R+\Delta R)^2-4 \pi R^2\right]\)

= \(8 \pi\left[R^2+2 R \cdot \Delta R+(\Delta R)^2-R^2\right]\)

= \(16 \pi R \cdot \Delta R\) ; [neglecting the term] \((\Delta R)^2\) which is very small]

∴ Increase in surface energy,

E = increase in surface area x surface tension

= 16 πR · ΔR · T……….(2)

This increase in surface energy of the soap bubble takes place at the cost of work done by the excess pressure i.e., E = W

So, from equations (1) and (2) we have,

⇒ \(p \cdot 4 \pi R^2 \cdot \Delta R=16 \pi R \cdot \Delta R \cdot T\)

p = \(\frac{4 T}{R}\)

Viscosity And Surface Tension Bubble  Numerical Examples

Example 1. Find the excess pressure inside a rainwater drop of diameter 0.02cm. The surface tension of water = 0.072 N · m^-1.
Solution:

Water drop has only one curved surface.

So, excess pressure of a water drop, p =2T/r where, T = surface tension of water

r = radius of a water drop = 0.002/2 = 0.01 cm = 0.01 x 10m

∴ p = \(=\frac{2 \times 0.072}{0.01 \times 10^{-2}}=1440 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 2. Surface tension of soap solution = 27 dyn • cm^-1. Calculate the excess pressure (in N • m^-2) inside a soap bubble of radius 3 cm.
Solution:

The excess pressure inside a soap bubble,

p = \(\frac{4 T}{r}=\frac{4 \times 27}{3}=36 \mathrm{dyn} \cdot \mathrm{cm}^{-2}=3.6 \mathrm{~N} \cdot \mathrm{m}^{-2} \text {. }\)

Example 3. Find the pressure inside an air bubble of radius 0.1mm just inside the surface of water. Surface tension of water = 72 dyn · cm^-1
Solution:

Excess pressure inside an air bubble

= \(\frac{2 T}{r}=\frac{2 \times 72}{0.01}\)

[T = 72 dyn · cm^-1 and r = 0.1 mm = 0.01 cm] = 14400 dyn · cm^-2.

Atmospheric pressure = 76 x 13.6 x 980 dyn · cm^-2

∴ Total pressure inside an air bubble = (76 x 13.6 x 980 + 14400)

= 1.0274 x 106 dyn · cm^-2.

Example 4. The excess pressure inside a soap bubble of radius 8mm raises the height of an oil column by 2mm. Find the surface tension of the soap solution. Density of the oil = 0.8 g · cm^-3.
Solution:

Excess pressure in a soap bubble (p) = 4πr.

Again, p = hρg, h = height of the oil column.

Now, \(\frac{4 T}{r}=h \rho g \text { or, } T=\frac{h \rho g \times r}{4}=\frac{0.2 \times 0.8 \times 980 \times 0.8}{4}\)

[h = 2mm = 0.2cm; ρ = 0.8 g • cm^-3, g = 980cm · s^-2 and r = 8mm = 0.8cm],

∴ T = 31.36 dyn · cm^-1

Example 5. In an isothermal process, two soap bubbles of radii a and b combine and form a bubble of radius c. If the external pressure is p, then prove that the surface tension of the soap solution is \(T=\frac{p\left(c^3-a^3-b^3\right)}{4\left(a^2+b^2-c^2\right)}\)
Solution:

We know, the excess pressure inside the soap bub¬ble = internal pressure – external pressure.

∴ For the bubble of radius a, excess pressure, \(\frac{4 T}{a}=p_a-p\)

∴ \(p_a=\left(p+\frac{4 T}{a}\right)\)

Similarly, for the bubble of radius b, \(p_b=\left(p+\frac{4 T}{b}\right)\)

For the bubble of radius c, \(p_c=\left(p+\frac{4 T}{c}\right)\)

Boyle’s law is applicable in isothermal process.

According to this law, \(p_a V_a+p_b V_b=p_c V_c\)

or, \(\left(p+\frac{4 T}{a}\right) \times \frac{4}{3} \pi a^3+\left(p+\frac{4 T}{b}\right) \times \frac{4}{3} \pi b^3\)

= \(\left(p+\frac{4 T}{c}\right) \times \frac{4}{3} \pi c^3\)

or, \(\left(p+\frac{4 T}{a}\right) a^3+\left(p+\frac{4 T}{b}\right) b^3=\left(p+\frac{4 T}{c}\right) c^3\)

or, \(4 T\left(a^2+b^2-c^2\right)=p\left(c^3-a^3-b^3\right)\)

∴ T = \(\frac{p\left(c^3-a^3-b^3\right)}{4\left(a^2+b^2-c^2\right)} .\)

Example 6. Two soap bubbles of radii 0.04 m and 0.03 m are combined in such a way that a common surface is formed between the two bubbles. What is the radius of curvature of the common surface?
Solution:

Let, the radii of the two soap bubbles are r₁ and r₂, and the internal pressures are p₁ and p₂ respectively.

The radius of the common surface = r, atmospheric pressure = p₀

For the first bubble, \(p_1-p_0=\frac{4 T}{r_1}\)

and for the second bubble, \(p_2-p_0=\frac{4 T}{r_2}\)

where, T = surface tension of soap solution.

Subtracting (1) from (2) we get,

⇒ \(p_2-p_1 = 4 T\left(\frac{1}{r_2}-\frac{1}{r_1}\right)=4 T\left(\frac{1}{0.03}-\frac{1}{0.04}\right)\)

= \(\frac{4 \times 100}{12} \times T\) ……(3)

But, for the common surface, \(p_2-p_1=\frac{4 T}{r}\)……..(4)

Comparing (3) and (4) we get, \(\frac{4 T}{r}=\frac{4 \times 100}{12} \times T\)

∴ r = 0.12 m.

 

Viscosity And Surface Tension Conclusion

Streamline flow: A smooth, uninterrupted flow of fluid in ordered layers, without any energy transfer between the layers, is called a laminar or streamline flow.

Streamline: In a smooth flow, the path along which any fluid particle moves is called a streamline.

Turbulent flow: If the velocity of a fluid along its flow continuously and randomly changes in magnitude and direction then it is called a turbulent or disorderly flow.

Viscosity: The property by virtue of which a liquid resists the relative motion between its adjacent layers is called viscosity of that liquid.

Velocity profile: The surface obtained by joining the terminal points of the velocity vectors of different layers of a flowing liquid at any section of it is called its velocity profile.

• In case of flow of a non-viscous liquid along a tube, the velocity profile becomes flat
• In case of flow of a viscous liquid along a tube, the velocity profile becomes parabolic.

Velocity gradient: In a horizontal streamline flow, the rate of change of velocity with distance \(\left(\frac{d u}{d x}\right)\) in a direction perpendicular to the flow of the liquid is called the velocity gradient. Dimension of velocity gradient \(\left(\frac{d u}{d x}\right)=\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}=\mathrm{T}^{-1} .\)

Coefficient of viscosity: The tangential viscous force acting per unit area between two parallel liquid layers having unit velocity gradient between them is called the coefficient of viscosity of that liquid.

Units of coefficient of viscosity:

CGS system: dyn • s • cm^-2 or g • cm^-1 • s^-1 or poise

SI: N • s • m^-2 or Pa • s or kg • m^-1 • s^-1 or decapoise

Dimension of coefficient of viscosity: ML^-1T^-1.

• Usually the viscosity of a liquid increases with the increase in pressure and decreases with the increase in temperature.
• Pressure has almost no effect on the viscosity of a gas. With the increase in temperature, the viscosity of a gas increases.

Critical velocity: When the velocity of a fluid does not exceed a certain limiting value, the flow of the fluid remains streamline, but when the velocity exceeds that particular limiting value, the flow becomes turbulent. The limiting value of that velocity is called critical velocity.

Reynolds number N is a dimensionless quantity. For N< 2000, the fluid motion is streamlined. If N> 3000, then the fluid motion becomes turbulent. As the value of N gradually changes from 2000 to 3000, the pure streamline flow gradually changes into a fully turbulent flow.

Terminal velocity: When a body falls through a viscous medium under the influence of gravity, the viscosity of the medium offers resistance against its motion.

If the body is small, then after a certain time the magnitude of this upward viscous force become equal to the net force creating the motion. Then the body attains a uniform velocity through the medium. This uniform velocity of the body is called the terminal velocity.

Equation of continuity: In the case of streamline flow of a fluid (liquid or gas), the mass of fluid flowing per second through any cross-section of the tube of flow always remains constant. This is called the equation of continuity.

Bernoulli’s theorem: For the streamline flow of an ideal liquid, the sum of the potential energy, the kinetic energy, and the energy due to pressure per unit volume of the liquid remains constant at every point on the streamline. It leads to the relation

⇒ \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}=\text { constant }\)

or, velocity head + elevation head + pressure head = constant.

Torricelli’s theorem: The velocity of efflux of a liquid, confined in a container through an orifice at some depth of the container is equal to the velocity acquired by a body falling freely from rest under gravity from the free surface of the liquid to the orifice.

Surface tension: Surface tension is a property of the free surface of a liquid due to which it behaves as a stretched thin membrane and has a tendency to con¬tract so as to minimise the surface area.

Units of surface tension:

• CGS system: dyn · cm^-1
• SI: N • m^-1

Dimension of surface tension: MT^-1.

Surface energy: The potential energy per unit area of the surface film is called surface energy.

or, surface energy = \(\frac{\text { work done in increasing the surface area }}{\text { increase in surface area }} \text {. }\)

• The surface energy per unit area is numerically equal to the surface tension of a liquid (if temperature remains constant)
• At any temperature except absolute zero, the total surface energy of a liquid is always greater than the surface tension.
• The surface tension of all liquids decreases with the rise in temperature. At the critical temperature of a liquid, the surface tension vanishes.
• If a liquid surface is contaminated with impurities, then the surface tension of that liquid usually decreases.
• If an inorganic substance is dissolved in a liquid, then the surface tension increases, but if an organic substance is dissolved, then the surface tension decreases.
• The presence of electric charges on the surface of a liquid causes a decrease in the surface tension.

Due to surface tension, capillary action is observed in liquids.

• When a liquid is in contact with a solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact for that specific pair of solid and liquid.
• If the angle of contact is less than 90°, then the liquid is said to wet the solid, and it rises in a capillary tube. But if the angle of contact is more than 90°, then the liquid does not wet the solid, and falls in a capillary tube.

Jurin’s law: The rise or fall of a liquid in a capillary tube is inversely proportional to the radius of the tube.

Viscosity And Surface Tension Useful Relations For Solving Numerical Problems

For two adjacent layers of a flowing liquid, if the opposing force acting is F, the area of the liquid surface is A and the velocity gradient is \(\frac{d v}{d x}\), then the coefficient of viscosity, \(\eta=\frac{F}{A \frac{d y}{d x}}\)

If Reynolds number is N, velocity of fluid flow is v, characteristic length of the fluid is l, the coefficient of viscosity is η and density of the fluid is ρ, then N = \(\frac{e l v}{\eta}\).

If a small sphere of radius r falls through a medium having a coefficient of viscosity η with a terminal velocity v, then the opposing force acting on the sphere due to viscosity is F = 6πηrv and the terminal velocity of the sphere is,

v = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

where ρ and σ are the densities of the material of the sphere and the material of the medium respectively.

If the cross-sectional area at any place of a tube of flow is a and the velocity of the fluid at that place is v, then the equation of continuity is expressed as, vα = constant.

Bernoulli’s theorem: \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}\) = constant,

where \(\frac{v^2}{2 g}\) is the velocity head, h is the elevation head, and \(\frac{p}{\rho g}\) is the pressure head.

Torricelli’s theorem: The velocity of efflux of a liquid through an orifice situated at a depth h of its container is, v = √2gh

Surface tension = \(\frac{\text { tangential force }}{\text { length }}\)

Work done to increase the area of a liquid surface by unity at constant temperature = surface energy stored per unit area = surface tension.

If the radius of a capillary tube is r, the density’ of liquid is ρ the angle of contact of the liquid with respect to the material of the tube is θ and the surface tension of the liquid is T, then the rise of the liquid in that capillary tube is

h = \(\frac{2 T \cos \theta}{r \rho g}\)

The excess pressure inside a spherical drop or bubble \(p=\frac{2 T}{r}\)

where T = surface tension and r = radius of curvature

The excess pressure inside a spherical soap bubble \(p=\frac{4 T}{r}\)

where T = surface tension and r = radius of curvature.

Viscosity And Surface Tension Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1.

Question 2.

Statement 1: The viscosity of liquid increases with rise in temperature.

Statement 2: Viscosity of a liquid is the property of the liquid by virtue of which it opposes the relative motion amongst its different layers.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: All the rain drops hit the surface of the earth with the same constant velocity.

Statement 2: An object falling through a viscous medium eventually attains a terminal velocity.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 4.

Statement 1: Air flows from a small bubble to a large bubble when they are connected to each other by a capillary tube.

Statement 2: The excess pressure because of surface tension inside a spherical bubble decreases as its radius increases.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 5.

Statement 1: When height of the tube is less than the rise in liquid in a capillary tube, the liquid does not overflow.

Statement 2: Product of radius of meniscus and height of liquid In the capillary tube always remains constant.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 6.

Statement 1: It is easier to spray water in which some soap is dissolved.

Statement 2: Soap is easier to spread.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 7.

Statement 1: A needle placed carefully on the surface of water may float, whereas a ball of the same material will always sink.

Statement 2: The buoyancy of an object depends both on the material and shape of the object.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 8.

Statement 1: A large force is required to draw apart normally two glass plates enclosing a thin water film.

Statement 2: Water works as glue and sticks two glass plates.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 9.

Statement 1: Tiny drops of liquid resist deforming forces better than bigger drop.

Statement 2: Excess pressure inside a drop is directly proportional to surface tension.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 10.

Statement 1: The uplift of the wing of an aircraft moving horizontally is caused by a pressure difference between the upper and lower faces of the wing.

Statement 2: The velocity of air moving along the upper surface is higher than that along the lower surface.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

 

Expansion Of Gases Introduction

Gases expand on heating and contract on cooling, like solids and liquids. Like liquids, gases too do not have any fixed shape and therefore the linear or the surface expansion of a gas is irrelevant.

Only the change in volume with the change in temperature is of importance. Other characteristic features of gaseous expansions, compared to those of solids and liquids, are discussed below.

1. The coefficient of volume expansion for a gas has a much higher value than that for solids and liquids. For a certain rise in temperature, the expansion of the container is negligible compared to the expansion of the gas in it. Hence, the apparent expansion of gas is practically the same as real expansion and is usually not reckoned separately unless a very high accuracy is required.
2. Unlike solids and liquids, the volume of a fixed mass of a gas is considerably affected due to any change in pressure. So, the effects of both temperature and pressure have to be studied in connection with the expansion or contraction of a gas. While studying the effect of one, the other is usually kept constant.
3. The rate of expansion or contraction, due to the change in either pressure or temperature, does not differ for different gases. Unlike solids and liquids, the coefficient of expansion is the same for all gases.

The state of a fixed mass of gas is therefore determined by the parameters

1. Volume,
2. Pressure and
3. Temperature.

The rules, that govern the change of one parameter with the change of another keeping the third one constant, are called gas laws.

1. It should be mentioned here that the pressure of a gas means the pressure exerted by the gas. In equilibrium (i.e., when each parameter of the gas is a constant) the pressure exerted by the gas and the pressure applied on the gas are equal.
2. The gas which follows Boyle’s law and Charles’ law at any temperature and pressure is called an ideal gas. In reality, an ideal gas does not exist. However, the ideal gas concept provides a very useful tool for the analysis of real gases.

 Expansion Of Gases Charles Law

The relationship between the volume and temperature of a fixed mass of gas at constant pressure was investigated experimentally by the French scientist Charles in 1787.

• He concluded that at constant pressure a fixed volume of all gases would expand by the same amount for an equal rise in temperature. Gay Lussac arrived at almost the same result in 1802.
• He found that the coefficient of volume expansion of all gases has the same value if pressure is kept constant. In 1842 Regnault showed that the experimental value of this coefficient is 1/273 per degree Celsius.
• Charles’ law is enunciated by combining the experimental results of Gay Lussac and Regnault.

Statement: When pressure is kept constant, the volume of a fixed mass of gas increases or decreases by 1/273 th part of its volume at 0°C, for each degree Celsius rise or fall in temperature.

Mathematical expression: Let V₀ be the volume of a mass m of a gas at 0°C. As per Charles’ law, the volume at 1°C will be

⇒ \(V_1=V_0+\frac{V_0}{273}=V_0\left(1+\frac{1}{273}\right)\)

The volume of the gas at \(2^{\circ} \mathrm{C}\),

⇒ \(V_2=V_0+\frac{2 V_0}{273}=V_0\left(1+\frac{2}{273}\right)\)

∴ The volume of the gas at \(t^{\circ} \mathrm{C}\),

⇒ \(V_t=V_0+\frac{V_0 t}{273}=V_0\left(1+\frac{t}{273}\right)\)….(1)

Similarly, if temperature is decreased by t°C, i.e., at a temperature -t°C, the volume of the gas becomes \(V_{-t}=V_0\left(1-\frac{t}{273}\right)\)

Therefore, at constant pressure volume of a fixed mass of gas changes linearly with its temperature. So, at fixed pressure, a graph plotted between the volume of a gas of a fixed mass and its temperature gives a straight line (AB).

Expansion Of Gases Pressure Law

The law that relates the change in pressure of a fixed mass of a gas at a fixed volume, with change in temperature is called pressure law or Regnault’s law.

Statement: When volume is kept constant, the pressure of a fixed mass of gas increases or decreases by1/273th part of its pressure at 0°C, for each degree centigrade rise or fall in temperature.

Mathematical expression: Let p₀ be the pressure of a fixed mass of a gas at 0°C. The pressure is p_t when the temperature is raised to t °C.

Therefore the pressure of the gas at 1°C, \(p_1=p_0+\frac{p_0}{273}=p_0,\left(1+\frac{1}{273}\right)\)

and the pressure of the gas at t°C, \(p_t=p_0+\frac{p_0 t}{273}=p_0\left(1+\frac{t}{273}\right)\)

Similarly, the pressure of the gas at -t°C, \(p_{-t}=p_0-\frac{p_0 t}{273}=p_0\left(1-\frac{t}{273}\right)\)

Hence the change in pressure of a fixed mass of gas is linearly related to the change in temperature at constant volume.

So, at constant volume, a graph plotted between the pressure of a gas of fixed mass and its temperature gives a straight line.

The increase in either the volume or the pressure of a gas with a rise in temperature is loosely termed as thermal expansion, although an increase in pressure is not actually an expansion.

Expansion Of Gases – Alternative Forms Of Charles Law And Pressure Law

Charles’ law: Suppose a fixed mass of a gas. at constant pressure, has volume V₀ at 0°C, at V₁ at t₁°C and V₂ at t₂°C. From Charles’ law.

⇒ \(V_1=V_0\left(1+\frac{t_1}{273}\right)=V_0\left(\frac{273+t_1}{273}\right)=\frac{V_0}{273} \cdot T_1\)

[where T₁ = t₁ + 273]

Obviously, the temperature t₁ in the Celsius scale is the same as the temperature T₁ K in the Kelvin scale.

Similarly, \(V_2=\frac{V_0}{273} \cdot T_2\)

where \(T_2=t_2+273\)

Here, \(t_2{ }^{\circ} \mathrm{C}=T_2 \mathrm{~K}\).

∴ \(\frac{V_1}{V_2}=\frac{T_1}{T_2}=\) constant

or, \(V \propto T\) at constant pressure.

Hence, Charles’ law can also be stated as the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in an absolute scale.

Pressure law: Suppose a fixed mass of a gas at constant volume has pressure p₀ at 0°C, p₁ at t₁ °C. and p₂ at t₂ °C. Hence, from pressure law.

⇒ \(p_1=p_0\left(1+\frac{t_1}{273}\right)=\frac{p_0}{273}\left(273+t_1\right)=\frac{p_0}{273} \times T_1\)

where \(T_1=t_1+273\)

It is dear that the temperature t°C is the same as the temperature T₁K.

Similarly, \(p_2=\frac{p_0}{273} \times T_2\) [where T₂ = t₂ + 273]

Here, t₂ °C = T₂ K.

∴ \(\frac{p_1}{p_2}=\frac{T_1}{T_2} \quad \text { or, } p \propto T\), at constant volume.

Hence, pressure law can be expressed as. the pressure of a fixed mass of a gas at constant volume is directly proportional to its temperature in an absolute scale.

Expansion Of Gases- Combination Of Boyles Law And Charles

Ideal or Perfect gas: The gases that obey Bodes and Charles’ law at any temperature are called ideal gases.

The equation obtained by combining Boyles’s law and Charles’ law is called the equation of the state of an ideal gas.

From Bovle’s law, \(V \propto \frac{1}{p}\) when T is constant.

From Charles’ law, V∝T when p is constant.

∴ \(V \propto \frac{T}{p}\) when both p and T vary

or, pV= kT ….(1)

where k is a constant whose value depends on the units of p, V, T, and the mass of the gas. If a gas of fixed mass occupies a volume V₁ at pressure p₁ and temperature T₁, and after a general change, a volume V₂, at pressure p₂, and temperature T₂, then

⇒ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)….(2)

It is known that the physical property of an ideal gas of fixed mass depends on its pressure, volume, and temperature.

Hence, the equation pV = kT is called the equation of state of an ideal gas.

Note that no real gas totally follows this equation of state.

The deviation, though small at ordinary temperature and pressure, becomes significant at high pressure and at low- temperature.

Expansion Of Gases Absolute Scale Of Temperature

From Charles’ law, If a fixed mass of gas at constant pressure occupies a volume V₀ at 0°C, then at t°C its volume will be \(V_t=V_0\left(1+\frac{t}{273}\right)\)

Now if the temperature is reduced to t = -273°C

⇒ \(V_t=V_0\left(1-\frac{273}{273}\right)=0\), i.e., the volume of the gas vanishes at -273°C temperature.

Similarly from pressure law, if a fixed mass of a gas at constant volume has a pressure p at 0°C, then at t°C its pressure will be \(p_t=p_0\left(1+\frac{t}{273}\right) \text {. }\)

Now if the temperature is reduced to t = -273 °C, \(p_t=p_0\left(1-\frac{273}{273}\right)=0\), i.e., the pressure of the gas vanishes at -273°C temperature.

So, at -273 °C, the volume and pressure of a fixed mass of gas have worked out to be zero.

If the temperature could be lowered below -273 °C, the pressure and volume would have been negative, which is meaningless. Hence, the lowest conceivable temperature in this universe is -273 °C.

Therefore -273 °C is called the absolute zero of temperature, or simply, absolute zero. More precisely, the value of absolute zero is -273.15°C.

Definition: The temperature at which the volume and the pressure of a gas reduce to zero is called the absolute zero of temperature. This temperature is the lowest temperature in reality.

Taking absolute zero i.e., -273°C as zero, a new scale of temperature was developed by Lord Kelvin, and it is called the absolute scale of temperature or Kelvin scale after the inventor.

In this scale, the temperature reading is denoted by A (absolute) or K (Kelvin). Each degree in this scale is taken equal to a degree in the Celsius scale. Hence, the scale of temperature where -273°C is taken as zero (0) and each degree is equal to a degree in Celsius scale, is called the absolute scale or Kelvin scale.

The freezing point of water (0°C) in an absolute scale is 273 K and the boiling point of water (100°C) in this scale is 373 K.

If any temperature is represented by t°C in the Celsius scale and by TK on the Kelvin scale, then, T = t + 273.

In the Kelvin scale, a temperature may be zero or positive. A negative Kelvin temperature does not exist.

0° K In Fahrenheit scale: in the relation \(\frac{C}{5}=\frac{F-32}{9}\), putting C = -273°C (0K) we have, \(\frac{-273}{5}=\frac{F-32}{9}\) or, F = – 459.4°F.

Incidentally, gas laws are valid as long as the matter remains in a gaseous state. All gases get liquified much before they attain the absolute zero temperature. No gas can be cooled to the temperature of absolute zero, and as such, zero volume or zero pressure of a gas is never realized in practice.

Absolute scale of temperature—why it is so named: The selection of 0 degrees in Celsius and Fahrenheit scale has no scientific reason behind it. The freezing point of water is taken arbitrarily as 0°C and the boiling point as 100°C in the Celsius scale.

• But the selection of 0 degrees in absolute scale is scientific as it denotes the temperature at which the volume and pressure of any ideal gas would reduce to zero. This is the lowest conceivable temperature in this universe.
• In addition, the scale does not depend on the nature of the gas. Hence, it is justified to call it the absolute scale of temperature.

Expansion Of Gases Universal gas Constant

The equation of state for an ideal gas is pV = kT. The value of the constant k depends on the mass of the gas used. Keeping pressure (p) and temperature (T) constant, if the mass of a gas is doubled, the volume is also doubled.

• So k is doubled. We can conclude that k is directly proportional to the mass of the gas.
• When one gram-molecule or one mole of an ideal gas is taken, the constant k is written as R. As per Avogadro’s law, at the same temperature and pressure, volume of 1 gram-molecule of any gas is the same.
• Therefore, the value of the constant R, for all 1-gram-molecule ideal gases, is the same. Hence, R is called the universal gas constant or molar gas constant.

Hence, for 1 gram-molecule of any ideal gas,

pV = RT ….(1) f the volume of n gram-molecule of gas is V, then the volume of 1 gram-molecule of gas, = V/n

Then from equation (1) we get, \(p \cdot \frac{V}{n}=R T \quad \text { or, } p V=n R T\)….(2)

Obviously, for mg of a gas of molecular weight M the number of moles, n = m/n

∴ pV = \(\frac{m}{M} R T\)….(3)

Comparing equation(3) with the gas equation pV = kT, k = \(\frac{m}{M} R T\)

Therefore for 1 gram of gas, k = \(\frac{R}{M}\) = r (say), r is also a constant and is called the specific gas constant.

As the molecular weights of different gases are different, the magnitude of this constant is also different for different gases.

Magnitude of universal or molar gas constant: From equation (1) above, the relation R = \(\frac{p V}{T}\) is valid for any ideal gas at any temperature and pressure.

Hence, R is also equal to \(\frac{p_0 V_0}{T_0}\) where p₀ = 76 cm Hg
(standard pressure) = 76 x 13.596 x 980.665 dyn · cm^-2, T₀ (standard temperature) = 0 + 273 = 273 K and V₀ = the volume at STP of 1 mol of a gas = 22.4 litre, as per Avogadro’s law.

∴ R = \(\left(76 \times 13.596 \times 980.665 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\right) \frac{\times 22.4 \times 1000 \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1}}{273 \mathrm{~K}}\)

= 8.314 x 10⁷ dyn · cm · mol^-1 · K^-1

= 8.314 x 10⁷ erg · mol”1 · K^-1

= 8.314 J · mol^-1 · K^-1

Magnitude of specific gas constant: Value of r for agasis r= R/M.

1. For hydrogen, molecular weight = 2,

r = \(frac{R}{2}=\frac{8.314 \times 10^7}{2}\)

= \(4.16 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

2. For oxygen, molecular weight = 32

r = \(\frac{R}{32}=\frac{8.314 \times 10^7}{32}\)

= \(0.26 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

3. For nitrogen, molecular weight = 28

r = \(\frac{R}{28}=\frac{8.314 \times 10^7}{28}\)

= \(0.297 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

4. For carbon dioxide, molecular weight = 44

r = \(\frac{R}{44}=\frac{8.314 \times 10^7}{44}\)

= \(0.189 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Universal gas Constant Numerical Examples

Example 1. The mass of 1 litre of hydrogen at STP is 0.0896 g. Calculate the value of R from this data.
Solution:

Volume of 0.0896 g of hydrogen at STP = 1000 cm³.

Hence, a volume of 2 g or 1 mol of hydrogen at STP = \(\frac{1000 \times 2}{0.0896}=22321.4 \mathrm{~cm}^3.\)

Standard pressure = 76x 13.6×981 dyn · cm^-2; standard temperature = 0°C = 273 K

∴ R = \(\frac{p V}{T}=\frac{76 \times 13.6 \times 981 \times 22321.4}{273} \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

= \(8.29 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Example 2. The mass of 3.76 litre of oxygen at 2 standard atmo¬sphere pressure and 20°C is 10 g. Find the value of R
Solution:

Given, p = 2 standard atmosphere = 2 x 76 x 13.6 x 981 dyn · cm^-2,

V = 3.76 x 10³ cm³, T = 273 + 20 = 293 K and n = 10/32 mol, [where 32 is the molecular weight of oxygen]

∴ pV=nRT

R = \(\frac{p V}{n T}=\frac{2 \times 76 \times 13.6 \times 981 \times 3.76 \times 10^3 \times 32}{10 \times 293}\)

= \(8.33 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Example 3. The density of air at STP = 1.293 g · L^-1 and that of mercury = 13.6 g ·  cm^-3. Find the value of the gas constant for 1 g of air.
Solution:

If k is the constant for 1 g of air, then k = pV/T

According to the problem, 1.293 g of air occupies a volume of 1000 cm³

∴ 1g of air occupies a volume of \(\frac{1000}{1.293} \mathrm{~cm}^3\)

Here, p = \(76 \times 13.6 \times 981 \mathrm{dyn} \cdot \mathrm{cm}^{-1}, T=273 \mathrm{~K}\)

∴ k = \(76 \times 13.6 \times 981 \times \frac{1000}{1.293} \times \frac{1}{273}\)

= \(0.287 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Example 4. The masses, volumes, and pressures of two samples of oxygen and hydrogen gases are equal. Find the ratio of their absolute temperatures.
Solution:

Let the volume of each gas = V, the pressure of each gas = p, and the absolute temperatures of the gases be T₁ and T₂ respectively.

If the mass of each sample is m g, then the number of moles of the gases are respectively, \(n_1=\frac{m}{32} \quad \text { and } n_2=\frac{m}{2} \text {. }\)

As per the equation pV = nRT, for oxygen gas, \(p V=\frac{m}{32} R T_1\) and for hydrogen gas, \(p V=\frac{m}{3} R T_2\)

∴ \(\frac{m}{32} R T_1=\frac{m}{2} R T_2 \quad \text { or, } \quad \frac{T_1}{T_2}=\frac{32}{2}=16: 1\)

 Expansion Of Gases Conclusion

Boyle’s law: At constant temperature, the volume (V) of a fixed mass of gas is inversely proportional to the pressure (p) of the gas.

i.e \(V \propto \frac{1}{p}\) or, pV = constant.

Charles’ law: At constant pressure, the volume of a fixed mass of gas increases or decreases by 1/273 of its volume at 0°C per degree Celsius rise or fall in temperature.

Pressure law: At constant volume, the pressure of a fixed mass of gas increases or decreases by 1/273 of its pressure at 0°C per degree Celsius rise or fall in temperature.

Volume coefficient: The volume coefficient of a fixed mass of a gas at a constant pressure is the increment of its volume per unit volume when its temperature is raised by 1°C from 0°C.

Pressure coefficient: The pressure coefficient of a fixed mass of a gas at a constant volume is the increment of its pressure per unit pressure when its temperature is raised by 1°C from 0°C.

• The volume coefficient of all gases is equal.
• The pressure coefficients of all gases are equal.
• In the case of any ideal gas, the volume coefficient and the pressure coefficient are equal, i.e., \(\gamma_p=\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}=\gamma_v.\)
• While determining the volume and pressure coefficients of a gas, we always refer to the initial volume or pressure at 0°C.
• The gases which obey Boyle’s and Charles’ laws at all temperatures are known as ideal gases. But in actual practice no real gas is ideal.

Absolute zero temperature: The temperature at which the volume and the pressure of an ideal gas theoretically become zero is called the absolute zero temperature. This is the lowest possible temperature in reality.

The scale of temperature in which the temperature -273°C is taken as zero and each degree interval is equal to a degree interval in the Celsius scale is called the absolute scale of temperature or Kelvin scale.

Absolute zero temperature = 0K = -273°C = 459.4°F.

Universal or molar gas constant.

R =8.314 x 10⁷ erg · mol^-1  K^-1

= 8.314 J ⋅ mol^-1 K^-1

At constant pressure, the density of a gas is inversely proportional to its absolute temperature.

At constant temperature, the density of a gas is directly proportional to its pressure.

Expansion Of Gases Useful Relations For Solving Numerical Problems

If the volume of a definite mass of a gas is V and its pressure is p, then according to Boyle’s law pV = constant

Let at constant pressure, the volume of some definite mass of a gas at 0°C be V₀. According to Charles’ law the final volume of the gas due to riC rise or fall in temperature,

⇒ \(V_t=V_0\left(1 \pm \frac{t}{273}\right)\)

Let at constant volume, the pressure of some definite mass of a gas at 0°C be p₀. According to pressure law. the final pressure of the gas due to t°C rise or fall in temperature,

⇒ \(p_t=p_0\left(1 \pm \frac{t}{273}\right)\)

Volume coefficient of a gas at constant pressure, \(\gamma_p=\frac{V_t-V_0}{V_0 \cdot t}\)

Pressure coefficient of a gas at constant volume, \(\gamma_v=\frac{p_t-p_0}{p_0 \cdot t}\)

⇒ \(\gamma_p=\gamma_v=\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\)

T = t + 273 lf any temperature is represented by t and T in Celsius and Kelvin scales respectively]

For n mol of an ideal gas, pV = nRT = \(\frac{m}{M} R T\) [where m and M are the mass and molecular weight of the gas respectively]

If ρ is the density of an ideal gas, then \(\frac{p}{\rho T}=\frac{R}{M}\) [symbols have usual meanings]

Expansion Of Gases Very Short Answer Type Questions

Question 1. At what Celsius temperature, does the volume of a gas become zero according to Charles’ law?
Answer: -273°C

Question 2. How does the volume of a definite mass of gas change with pressure at constant temperature?
Answer: Inversely proportional

Question 3. How does the volume of a definite mass of gas change with its absolute temperature at constant pressure?
Answer: Directly proportional

Question 4. At constant volume, the pressure of a definite mass of gas is directly proportional to its absolute temperature. Is the statement true or false?
Answer: True

Question 5. What is the value of the volume coefficient of a gas?
Answer: 1/273 °C^-1

Question 6. What is the value of the pressure coefficient of a gas?
Answer: 1/273 °C^-1

Expansion Of Gases Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2 Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: Equal masses of helium and oxygen gases are given equal quantities of heat. There will be a greater rise in the temperature of helium compared to that of oxygen.

Statement 2: The molecular weight of oxygen is more than the molecular weight of helium.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: In the upper part of the atmosphere, the temperature of air is of the order of 1000 K, even then it is quite cold there.

Statement 2: Molecular density at high altitudes is low.

Answer:  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: Shows the V – T graphs of a certain mass of an ideal gas at two pressures p₁ and p₂. It follows from the graph that p₁ is greater than p₂.

Statement 2: The slope of V- T graph for an ideal gas is directly proportional to pressure.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 4.

Statement 1: V-T graph in a process is a rectangular hyperbola. Then p-T graph in the same process will be a parabola.

Statement 2: If the V-T graph is a rectangular hyperbola, with an increase in T, the volume will decrease and hence, pressure will increase.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The size of a hydrogen balloon increases as it rises in the air.

Statement 2: The material of the balloon can be easily stretched.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Expansion Of Gases Match Column 1 With Column 2

Question 1. The V-T graph of two gases A and B is shown.

Answer: 1. D, 2. D, 3. D, 4. D

Expansion Of Gases Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and the temperature at the surface is 40°C. The atmospheric pressure is 76 cm of Hg and g = 980 cm · s^-2.

1. What is the pressure at the bottom of the lake?

1. 1279325 dyn • cm^-2
2. 1359943 dyn • cm^-2
3. 1257928 dyn • cm^-2
4. 1378174 dyn • cm^-2

Answer:  1. 1279325 dyn • cm^-2

2. What is the temperature at the bottom of the lake?

1. 9.77°C
2. 10.37°C
3. 11.31°C
4. 11.67°C

Answer: 2. 10.37°C

Question 2. A fixed thermally conducting cylinder has a radius of R and a height of L₀. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L, from the top surface as shown. The atmospheric pressure is p₀.

1. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be

1. \(\frac{p_0}{2}+\frac{M g}{\pi R^2}\)
2. \(\frac{p_0}{2}\)
3. \(p_0\)
4. \(\frac{p_0}{2}-\frac{M g}{\pi R^2}\)

Answer: 1. \(\frac{p_0}{2}+\frac{M g}{\pi R^2}\)

2. When the piston is at a distance 2L from the top the hole at the top is sealed. The piston is then released, at a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

1. \(\left(\frac{2 p_0 \pi R^2}{\pi R^2 p_0+M g}\right)(2 L)\)
2. \(\left(\frac{p_0 \pi R^2-M g}{\pi R^2 p_0}\right)(2 L)\)
3. \(\left(\frac{p_0 \pi R^2+M g}{\pi R^2 p_0}\right)(2 L)\)
4. \(\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)\)

Answer: 4. \(\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)\)

3. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown.

The density of the water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

1. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0\)
2. \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0\)
3. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)
4. \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0\)

Answer: 3. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)

Expansion Of Gases Integer Type Questions And Answers

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. An ideal gas is heated from 27 °C to 627°C at constant pressure. If the initial volume was 3 m³, then what would be the final volume (in m³) of gas?
Answer: 9

Question 2. An air bubble of diameter 1 cm is formed at a depth of 238 ft in a lake. What will be the diameter (in cm) of the bubble when it reaches the free surface? Given that the temperature from top to bottom in the lake is the same and the height of a water barometer is 34 ft.
Answer: 2

Question 3. The volume of some air saturated with water vapor is 80 cm³ at a pressure of 74 cm Hg. Keeping the temperature fixed if pressure is made 146 cmHg then the volume becomes halved. What will be the pressure (in cm Hg) of water vapor then?
Answer: 2

Question 4. Find the percentage increase in the pressure when air enclosed at 30 °C is raised to 57 °C at a constant volume.
Answer: 9

Expansion Of Gases Short Answer Type Questions And Answers

Question 1. The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 x 10⁵ Pa. If n_i and n_f are the number of molecules in the room before and after heating, then n_f – n_i will be

1. 1.61 x 10²³
2. 1.38 x 10²³
3. 2.5 x 10²⁵
4. -2.5 x 10²⁵

Answer:

If n₁ and n₂ are the number of moles at the initial and final temperature inside the room respectively, then

⇒ \(n_1 =\frac{p_1 V_1}{R T_1}=\frac{10^5 \times 30}{8.31 \times(273+17)}\)

= \(1.245 \times 10^3\)

⇒ \(n_2 =\frac{p_2 V_2}{R T_2}=\frac{p_1 V_1}{R T_2}=\frac{10^5 \times 30}{8.31 \times(273+27)}\)

= \(1.203 \times 10^3\)

Therefore \(n_f-n_1=\left(n_2-n_1\right) \times 6.023 \times 10^{23}\)

= \(-2.5 \times 10^{25}\)

The option 4 is correct.

Question 2. Explain why air pressure in a car increases during driving.
Answer:

The temperature of air inside the increases due to motion. We know from Charles’ law, p ∝ T, therefore air pressure inside the the increases during driving.

Calorimetry Introduction

An important characteristic of heat must be noted from the very beginning. Unlike the quantities volume, pressure, temperature, etc., heat is not related to any equilibrium state of a body. Heat manifests itself only when a body undergoes a transition from one state to another.

Statements like ‘heat of a body’, ‘initial heat’, and ‘final heat’ have no physical meaning. Physically significant are the quantities like ‘heat gained’, or ‘heat lost’ by a body during any process. In this sense, heat is regarded as an energy in transit.

The rise or fall in temperature of a body implies the addition or extraction of heat from it. If two objects are in contact with each other then there will be heat flow between them till both attain an equilibrium temperature.

In such situations a need arises to know the quantity of heat transferred from one body to another, or, from an object to its surroundings. The study of problems that involve heat exchange between different bodies is called calorimetry.

Calorimetry Factors Affecting Heat Transfer

We know that it takes longer to boil some water than to warm it. Also, boiling water takes longer to cool down than lukewarm water does. So for rise and fall of different amount of temperature, different amounts of heat are absorbed or released.

• Again, increasing the amount of water to be boiled, increases the time taken to boil it too. From this we can conclude that the amounts cf heat absorbed (or released) by the bodies of same material but different mass for a fixed rise (or fall) in temperature depend on their masses.
• When a piece of iron and some water of the same mass are heated, the piece of iron heats up faster than the water.
• Clearly, the material of an object plays a role in determining the amount of heat absorbed (or released) by the object for a certain rise (or fall) in temperature.
• From these observations, it can be inferred that, the amount of heat absorbed or released by a body depends on the change of temperature of the body, mass of the body, and material of the body, provided that the material does not undergo any phase change.

Temperature: Amount of heat gained or lost by a body of fixed mass is directly proportional to the rise or fall in its temperature.

Mass: Amount of heat gained or lost by a body, for a fixed rise or fall in temperature, is directly proportional to its mass.

Material: Amount of heat gained or lost by a body depends on the material of the body.

When a body of mass m receives heat H, the rise in temperature is t,
H ∝ mt or, H = smt……(1)

where s is specific heat whose value depends on the material of the body.

Therefore, H (the heat absorbed or released by a body) = m (mass of the body) x s (specific heat of the material of the body) x t (rise or fall in temperature)
It is to be noted that t in equation (1) is the change in temperature of the body.

So only the heat gained or lost can be measured by this equation.

Calorimetry Units Of Heat

Some useful useful units for the measurement of heat are defined as follows:

Calorie: It is the unit of heat in CGS system. The amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C is called 1 calorie (1 cal).

• Practically, the heat required to raise the temperature of 1 g of pure water from 0°C to PC is not the same as the heat required to raise Its temperature, for example, from 45T to 46°C though the rise in temperature is the same.
• Hence, quantity of heat differs at the different ranges in the scale of temperature.
• So, most acceptable definition of calorie is, 1 cal = 1/100 X heat required to raise the temperature of 1 g of pure water from 0°C to 100°C. It is often known as mean calorie.
• Its value is equal to the amount of heat required to raise the temperature of 1 g of pure water from 14.5°C to 15.5°C. So it is sometimes called 15°C calorie.

Kilocalorie or kilogram calorie: The quantity of heat required to raise the temperature of 1 kg of pure water from 14.5°C to 15.5°C, is called 1 kilocalorie.

As 1 kg = 1000 g, 1 kilocalorie (kcal) = 1000 calorie (cal).

Joule: This is the unit of heat in SI. 1 cal = 4.2 J.

Calorimetry Specific Heat Or Specific Heat Capacity

It has been stated that heat gained (or lost) by a body depends on its mass, the rise (or fall) in temperature and the material of the body.

• Let us take two bodies of same mass but made of different materials. If we try to change the temperature of both the bodies by the same amount, we will find that the heat required is different for them.
• Therefore it is obvious, the quantity of heat required also depends on a special property. This property is known as specific heat or specific heat capacity of the body. It depends on the material of the body.
• It is obvious that different materials possess different specific heat capacities.

Specific Heat Or Specific Heat Capacity Definition: Specific heat capacity is the quantity of heat required to raise the temperature of unit mass of a substance by unit amount.

Definition of specific heat capacity in CGS system: It is the quantity of heat required to raise the temperature of 1 g of a substance through 1 °C. In this system the unit of specific heat is cal · g^-1 · °C^-1.

For example, ‘the specific heat capacity of copper is 0.093 cal · g^-1 · °C^-1 means that to raise the temperature of 1 g of copper by 1 °C, the heat required is 0.093 cal.

Definition of specific heat capacity in SI: The quantity of heat required to raise the temperature of 1 kg of a substance through 1 K is known as specific heat capacity of the substance.

The unit of specific heat in this system is J · kg^-1 · K^-1

In SI, the specific heat capacity of water is 4186 J · kg^-1 · K^-1.

Relation between CGS and SI unit of specific heat capacity: \(1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}=\frac{4.186 \mathrm{~J}}{10^{-3} \mathrm{~kg} \times 1 \mathrm{~K}}=4186 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

Specific heat capacities of a few solids and liquids

 

Calorimetry Fundamentals Principal Of Calorimetry

When two bodies at different temperatures are brought in contact, heat is transferred from the body at higher temperature to that at lower temperature. So, the former starts to cool down whereas the latter starts to warm up.

The flow of heat continues till both reach the same temperature. The state where both the bodies are at the same temperature is called thermal equilibrium.

Now we know that heat is a form of energy. According to the law of conservation of energy, energy can neither be created nor be destroyed.

So during such heat transfers, if heat does not transform into another form of energy (or, if no other form of energy transforms into heat energy) then, heat released by a body = heat absorbed by another body

This is the fundamental principle of calorimetry. This is also true for any number of bodies brought in contact with one another. In that case, we can write down the principle of calorimetry as: heat released by the hot bodies = heat absorbed by the cold bodies.

Calorimetry Calorimeter

A calorimeter is a cylindrical metallic container mostly used in calorimetric experiments which is generally made of copper. The container is partly filled with a proper liquid called a calorimetric substance.

• A stirrer (S) made of copper is used to stir the liquid and a thermometer (T) is used to measure the temperature of the liquid. Both are immersed in the liquid.
• In a calorimetric experiment when an object is dipped into the liquid of the calorimeter there must be a difference in temperature between the object and the liquid. After some time the object, the calorimeter, and the liquid inside the calorimeter reach thermal equilibrium.

During this interval of time if

1. No heat exchange occurs between the calorimeter and the surrounding,
2. No exothermic or endothermic reaction (physical or chemical) happens inside the calorimeter and
3. The object does not dissolve in the liquid, then according to the fundamental principle of calorimetry, heat released by the hot objects = heat absorbed by the cold objects.

There will be error in the experimental result if the above three conditions are not satisfied. Heat exchange may take place between the calorimeter and the environment by means of

1. Conduction,
2. Convection and
3. Radiation.

A calorimeter is so constructed that heat exchange by any of the three processes can be minimised.

Calorimetric Substance

Water Advantages:

1. For the same rise or fall in temperature, heat gained or lost by water is more than that gained or lost by the same mass of other solids or liquids of lower specific heat.
2. Water is commonly used as a calorimetric substance because water is easily available.

Water Disadvantages:

1. Again due to high specific heat of water, for absorption of same amount of heat, rise in the temperature of water is less than that of other liquids of lower specific heat. Now if the increase of temperature is low then there is high chance of error in reading.
2. The boiling point of water is 100°C. If an object of temperature more than 100°C is mixed with water, a sudden vaporisation occurs. Due to these reasons water is not useful as a calorimetric substance.

Aniline: Recendy aniline has been recognized as the best calorimetric substance. Aniline is present in pure form. Its specific heat is low (0.62 cal · g^-1 • °C^-1) and boiling point is high (183.9°C).

 

Calorimetry Conclusion

If the state of a body remains unchanged, heat gained or lost by the body depends on

1. Rise or fall in temperature of the body,
2. Mass of the body and
3. Nature of the material of the body.

Heat absorbed or released by a body only for a change in its temperature can be measured. Total heat contained in a body at a particular temperature cannot be measured.

• The amount of heat required by lg of pure water for the rise of its temperature from 14.5°C tol5.5°C is called 1 cal.
• The amount of heat required by lg of pure water for the rise in its temperature from 0° to 100°, divided by 100, gives the value of mean calorie.
• The SI unit of heat is joule. 1 cal = 4.2 J.
• Specific heat is the quantity of heat required to raise the temperature of unit mass of a substance through 1 degree.

According to this definition, units of specific heat: cal · g^-1 · °C^-1 (in CGS system) and J · kg^-1 · K^-1 (in SI).

Thermal capacity of a body is defined as the quantity of heat required to raise its temperature by unity.

Unit of thermal capacity: cal • °C^-1 (in CGS system) and J · K^-1 (in SI).

Water equivalent of a given body is the mass of water for which the rise in temperature is the same as that for the body, when the amount of heat supplied is the same for both.

Unit of water equivalent: g (in CGS system) and kg (in SI).

During heat transfer, if heat energy is not converted into any other form of energy (or no other form of energy is converted into heat energy), then
heat lost by one body = heat gained by another body.

This is the basic principle of calorimetry.

Calorimetry Useful Relations For Solving Numerical Problems

If no change of state occurs, then the heat gained or lost by a body, H = mst.

[where m = mass of the body, s = specific heat, t = change in temperature]

Thermal capacity of a body = ms = H/t

Thermal capacity per unit mass of a body = specific heat of the material of the body

Water equivalent of a body, W = \(\frac{m s}{s_w}\)

According to the basic principle of calorimetry, for more than two bodies at different temperatures in contact, heat lost by hot bodies = heat gained by cold bodies.

In CGS system, specific heat of water = 1 cal · g^-1 · °C^-1, and in SI, the specific heat of water = 4200 J · kg^-1 · ^-1.

 

Calorimetry Very Short Answer Type Questions

Question 1.Which substance, used in our everyday life, has the highest specific heat?
Answer: Water

Question 2. For a body of mass m and specific heat capacity s, how much heat is required to increase its temperature by t?
Answer: mst

Question 3. There are two spheres of the same radius. One is solid but the other is hollow. If both the spheres are heated to the same temperature and then allowed to cool down, which sphere will cool faster?
Answer: Hollow

Question 4. What is the CGS unit of thermal capacity?
Answer: cal • °C^-1

Question 5. What is the CGS unit of water equivalent?
Answer: g

Question 6. Which substance is treated as the best calorimetric substance nowadays?
Answer: Aniline

Calorimetry Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: A gas have a unique value of specific heat.

Statement 2: Specific heat is defined as the amount of heat required to raise the temperature of unit mass of the substance through unit degree.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: When a body of mass M loses heat, the time rate of fall of temperature for given amount of loss of heat is inversely proportional to mass.

Statement 2: ΔQ = MsΔT where, ΔQ = amount of heat, s = specific heat, andΔT = decrease in temperature.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: Specific heat capacity is the cause of formation of land and sea breeze.

Statement 2: The specific heat of water is more than land.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 4.

Statement 1: Specific heat of a body is always greater than its thermal capacity.

Statement 2: Thermal capacity is the heat required for raising temperature of a body by unity.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 5.

Statement 1: Two bodies at different temperatures, if brought in thermal contact do not necessarily settle to the mean temperature.

Statement 2: The two bodies may have different thermal capacities.

Answer: 1. Statement 1: Two bodies at different temperatures, if brought in thermal contact do not necessarily settle to the mean temperature.

 Calorimetry Match Column 1 And Column 2

Question 1. Three liquids A, B, and C having the same specific heat and masses m, 2 m, and 3 m have temperatures 20°C, 40°C, and 60°C respectively. Temperature of the mixture when

Answer: 1. E, 2. A, 3. D, 4. B

Calorimetry Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A vacuum flask contains 0.4 kg liquid paraffin whose temperature can be increased at the rate of 1°C per minute using an immersion heater of power 12.3 W. When a heater of 19.2 W is used to heat 0.5 kg of liquid paraffin, in the same flask, the rate of rise of temperature is 1.2°C per minute.

1. What is the specific heat of the paraffin?

1. 0.529cal · g^-1 · °C^-1
2. 0.429 cal · g^-1 · °C^-1
3. 0.472cal · g^-1 · °C^-1
4. 0.62.3 cal · g^-1 °C^-1

Answer: 1. 0.529 cal · g^-1 · °C^-1

2. What is the thermal capacity of the material of the flask?

1. 17.67cal · °C^-1
2. 16.85 cal · °C^-1
3. 17.01cal · °C^-1
4. 16.63 cal · °C^-1

Answer: 3. 17.01 cal · °C^-1

Question 2. When a block of metal of specific heat 0.1 caI · g^-1 · °C^-1 and weighing 110 g is heated to 100°C and then quickly transferred to a calorimeter containing 200 g of a liquid at 10°C, the resulting temperature is 1°C. On repeating the experiment with 400 g of the same liquid in the same calorimeter at same initial temperature, the resulting temperature is 14.5°C.

1. Find the specific heat of the liquid.

1. 0.42 cal · g^-1 · °C^-1
2. 0.52 cal · g^-1 · °C^-1
3. 0.48 cal · g-1 · °C^-1
4. 0.62 cal · g^-1 · °C^-1

Answer: 3. 0.48 cal · g^-1 · °C^-1

2. Find the water equivalent of calorimeter.

1. 15.5 g
2. 16.6 g
3. 17.3 g
4. 18.2 g

Answer: 2. 16.6 g

Calorimetry Integer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. 50 g of copper is heated to increase its temperature by 10°C. If the same quantity is given to 10 g of water, what will be the rise in its temperature (in °C)?
Answer: 5

Question 2. The water of volume 2 L in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J · s^-1. In now much time (in minutes) temperature will rise from 27 °C to 75°C? [Given specific heat of water is 4.2 kJ • kg·^-1 ]
Answer: 8

Question 3. If 0.5 kg of coal on burning raises the temperature of 50 L of water from 20 °C to 90 °C, then the heat of combustion of coal is n x 10⁶ cal · kg^-1. Find the value of n.
Answer: 7

Question 4. When x gram of steam is mixed with y grams of ice at 0°C, we obtain (x + y) grams of water at 100°C. Find the value of y/x
Answer: 3

Calorimetry Short Answer Type Questions

Question 1. A 10 Watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the tempera¬ture of water and the container rises by 3°K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of oil. The same heater now raises the tem¬perature of the container-oil system by 2°K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 J • kg^-1 • K^-1, the specific heat of oil in the same unit is equal to

1. 1.50 x 10³
2. 2.55 x 10³
3. 3.00 x 10³
4. 5.10 x 10³

Answer:

In first case the temperature of water and the container is increased by 3K in 15 min or (15 x 60)s by heating.

So, 10 x (15 x 60) = (0.5 x 4200 x 3) + (Wx 3)

[W = water equivalent]

or, \(W=\frac{9000-6300}{3}=900 \mathrm{~kg}\)

In second case, the heater increases the temperature of the oil and the container by 2K in 20 min or (20 x 60)s.

So, 10 x (20×60) = (2 x s x 2) + (W x 2)

[s = specific heat of oil]

or, s = \(\frac{12000-2 W}{4}=\frac{12000-1800}{4}=\frac{10200}{4}\)

= 2550 = 2.55 x 10³ J • kg^-1 · K^-1

The option 2 is correct.

Question 2. Three bodies of the same material and having masses m, m, and 3m are at temperatures 40°C, 50°C, and 60°C respectively. If the bodies are brought in the thermal contact, the final temperature will be

1. 45°C
2. 54°C
3. 52°C
4. 48°C

Answer:

If the final temperature is t°C,

ms(t-40) + ms(t-50) + 3ms(t-60) = 0

or, (t-40) + (t-50) + 3(t-60) = 0

or, 5t-270 = 0

∴ t = 54°C

The option 2 is correct.

Question 3. A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by: (given- room temperature = 30°C, specific heat of copper= 0.1 cal/g °C)

1. 800°C
2. 885°C
3. 1250°C
4. 825°C

Answer:

From the fundamental principle of calorimetry, heat released by copper ball = heat absorbed by calorimeter and water

or, 100 x0.1(t-75) = 100 x 0.1 x (75-30) + 170 x 1 x (75-30)

or, t = 885°C

The option 2 is correct.

Question 4. Explain why the coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
Answer:

This is due to the fact that heat absorbed by a substance is directly proportional to the specific heat of the substance.

Question 5. Explain why two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
Answer:

If two bodies, at different temperatures, come in thermal contact, heat flows from the body at higher temperature to the body at lower temperature till the temperature becomes same. The final temperature can be mean temperature, i.e., (T₁ + T₂)/2, only when both the bodies have equal thermal capacities.