Class 11 Physics

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.

Doppler Effect In Sound and Light  Question and Answers

Question 1. Each of the two men A and B is carrying a source of sound of frequency. If A approaches B with a velocity u,

1. How many beats per second will be heard by A and
2. How many beats per second will be heard by B? (Velocityofsound =c)

Answer:

Apparent frequency, n’ = \(\frac{c+u_o}{c-u_s} \times n\)

The distance between A and B is decreasing. So, the velocity of the listener u₀ and that of the source u_s, are both positive.

1. When A is the listener, velocity of the listener, u₀ = u; velocity of the source B, u_s = 0

So, \(n_A=\frac{c+u}{c} \times n\)

Evidently, n_A > n

∴ Number of beats per second

= \(n_A-n=\left(\frac{c+u}{c}-1\right) \times n=\frac{u}{c} n .\)

2. When B is the listener, velocity of the listener, u₀ = 0; velocity of the source A, u_s = u

So, \(n_B=\frac{c}{c-u} \times n\)

Evidently, \(n_B>n\)

∴ Number of beats per second

= \(n_B-n=\left(\frac{c}{c-u}-1\right) \times n\)=\(\frac{u}{c-u} n\)

Question 2. A car Is approaching a hill at a high speed. At that time, if the horn of the car is blown, the driver hears the echo sharper than the original sound. Explain the reason.
Answer:

Answer: The original sound after being reflected from the hill approaches the car and the driver listens to the echo. So, in this case, the listener is moving towards the source of the echo. So, due to the Doppler effect, the echo is of a higher apparent frequency. Thus, it appears to be sharper than the original sound to the driver.

Question 3. Certain characteristic wavelength in the light from a galaxy has a longer wavelength compared to that from a terrestrial source. Is the galaxy approaching or receding?
Answer:

The galaxy is receding. It can be concluded from the increase in wavelength, i.e., decrease in frequency due to the Doppler effect, that the distance between the source of light (the galaxy) and the observer (the earth) is gradually increasing. Hence, the galaxy is receding.

Question 4. Show that the apparent frequency f’ of a source of sound moving with a speed vs towards a stationary receiver is \(f^{\prime}=\frac{f c}{c-v_s}\), where c is the velocity of sound and f is the frequency.
Answer:

If the source of sound approaches the stationary listener with velocity v_s, the number of sound waves f produced per second occupies the distance c – v_s.

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{c-v_s}{f}\)

So, apparent frequency, f’ = \(\frac{c}{\lambda^{\prime}}=\frac{f c}{c-v_s}\)

Question 5. Two sources, each emitting a sound of wavelength A, are kept at a fixed distance. How many beats will be heard by a listener moving with a velocity u along the line joining the two sources?
Answer:

If n is the frequency and v is the velocity of sound, the apparent frequency to a listener in motion for a stationary source,

n’ = \(\frac{v+u}{v} \times n\)

The velocities of the listener for the two sources in question are -u and +u.

So, \(n_1^{\prime}=\frac{v-u}{v} \times n \text { and } n_2^{\prime}=\frac{v+u}{v} \times n\)

∴ Number of beats per second

= \(n_2^i-n_1^{\prime}=\frac{2 u}{v} \times n=\frac{2 u}{\frac{v}{n}}=\frac{2 u}{\lambda} .\)

Example 6. A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency of 2f. If v is the velocity of sound, what will be the velocity of the car?
Answer:

If u_s and u₀ are the velocities of the source of sound and the listener respectively,

the frequency of the echo, f’ = \(\frac{v+u_o}{v-u_s} \times f\)

Here, u_s = u₀ = u(say) and f’ = 2f

∴ 2f = \(\frac{v+u}{v-u} \times f \text { or, } v+u=2(v-u) \text { or, } 3 u=v \text { or, } u=\frac{v}{3} \text {. }\)

Example 7. What should be the velocity of a source of sound so that the apparent frequency to a listener will be half the actual frequency of the source? The velocity of sound in air = v.
Answer:

If n is the actual frequency, apparent frequency = \(\frac{n}{2}\).

Here the listener is stationary, i.e., u₀ = 0

Again decrease of frequency means that the source is receding from the listener. So the velocity of the source us is negative.

Therefore, from the relation n’ = \(\frac{v+u_o}{v-u_s} \times n\), we have

⇒ \(\frac{n}{2}=\frac{v}{v+u_s} \times n \text { or, } v+u_s=2 v \text { or, } u_s=v\)

i.e., the source is receding from the listener with the velocity of sound.

Question 8. What should be the velocity of a source of sound so that the apparent frequency to a listener will be twice the actual frequency of the source? The velocity of sound in air = v.
Answer:

Apparent Frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case n’ = 2n. The listener is stationary; so u₀ = 0.

The apparent frequency is higher, i.e., the source is approaching the listener. So, the velocity of the source u_s is positive. Therefore,

2n = \(\frac{v}{\nu-u_s} \times n \text { or, } v=2\left(\nu-u_s\right) \text { or, } u_s=\frac{v}{2}\)

i.e., the source is approaching the listener with half the velocity of sound.

Question 9. What should be the velocity of a listener so that the apparent frequency of the sound coming from a stationary source to him will be twice the actual frequency? The velocity of sound in air = v.
Answer:

Apparent frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case velocity of the source, u_s= 0.

Since the apparent frequency is higher, the listener is approaching the source, i.e., the velocity of the listener u₀ is positive.

Again, n’ = 2n.

∴ 2n = \(\frac{v+u_o}{v} \times n \text { or, } 2 v=v+u_o \text { or, } u_o=v\)

i.e., the listener is approaching the stationary source with the velocity of sound.

Question 10. Doppler effect gives an idea of a continuously expanding universe—explain.
Answer:

Light coming from the distant star can be analysed with the spectrometer. The experiment shows that the wavelength of a spectral line for a light source situated for away from Earth is greater than that for the same light source on Earth i.e., frequency is comparatively low.

From this apparent decrease in frequency, known as the redshift in the Doppler effect, we conclude that all distant stars are receding from Earth which indicates that the Universe is continuously expanding.

Question 11. A listener moving with constant velocity passes a stationary source. Draw a graph to show the change of apparent frequency of the source to the listener with time. The actual frequency of the source is n.
Answer:

While approaching the stationary source apparent frequency will be,

n’ = \(\frac{V+u_0}{V} n\)…(1)

[velocity of sound in air = V, velocity of listener = u₀]

While receding the stationary source apparent frequency will be,

n” = \(\frac{V-u_0}{V} n\)…(2)

from equations (1) and (2) it is clear that nf and n” remain constant with time and also n’ > n> n”. The change of apparent frequency with time is shown below.

Here OS denotes the time taken by the listener to pass by the source. After that, the listener continually moves away from the source.

Question 12. Both of a sound source and a listener are approaching each other with the same speed \(\frac{c}{10}\) (speed of sound in air =c). What will be the percentage of apparent increase or decrease in frequency of sound?
Answer:

Apparent frequency,

n’ = \(\frac{c+u_0}{c-u_S} n\)

So, the apparent increase in frequency,

n’ – n = \(\frac{u_0+u_S}{c-u_S} n\)

∴ Percentage of change in frequency

= \(\frac{n^{\prime}-n}{n} \times 100=\frac{u_0+u_S}{c-u_S} \times 100\)

= \(\frac{\frac{c}{10}+\frac{c}{10}}{c-\frac{c}{100}} \times 100=\frac{2}{9} \times 100=22.2 \%\)

Question 13. A band of music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v_m. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.
Answer:

Two separate sounds will be heard by the motorist, one is direct from the band and the other is the echo from the wall.

Apparent frequency of direct sound, \(f_1=f\left(\frac{v+v_m}{v+v_b}\right)\)

Apparent frequency of the echo, \(f_2=f\left(\frac{v+v_m}{v-v_b}\right)\)

Therefore, the beat frequency heard by the motorist,

n = \(f_2-f_1=f\left(v+v_m\right) \cdot\left[\frac{1}{v-v_b}-\frac{1}{v+v_b}\right]\)

= \(\frac{2 f v_b\left(v+v_m\right)}{v^2-v_b^2}\)

Question 14. Why Doppler effect is clearly realised in the case of sound but not in the case of light waves?
Answer:

Our auditory system is more sensitive to realise a small change in audible frequency. On the other hand, our visual system is not so sensitive to detect such a small change in the visible frequency of light. In our daily life, the magnitude of our relative velocity of us with a light source on earth is not high enough to observe such a noticeable change in frequency, i.e., the Doppler effect.

Doppler Effect In Sound  and Light Multiple Choice Questions

Question 1. A source of sound with a frequency of 256 Hz is moving with a velocity of v towards a wall and an observer is stationary between the source and the wall. When the observer is between the source and the wall

1. He will hear beats
2. He will hear no beats
3. He will not get any sound
4. He will get the sound of the same frequency

Answer: 2. He will hear no beats

Question 2. The frequency of a progressive wave may change due to

1. Reflection
2. Refraction
3. Interference
4. Doppler effect

Answer: 4. Doppler effect

Question 3. A train blowing a whistle of frequency 1000 Hz is moving with a uniform velocity from west to east. The apparent frequency of the sound of the whistle to a stationary listener is 990 Hz. The position of the listener relative to the train is

1. On the north
2. On the south
3. On the east
4. On the west

Answer: 4. On the west

Question 4. A source of sound and a listener are moving in the same direction with the same velocity. If the actual frequency of sound is 200 Hz, the apparent frequency of the sound to the listener is

1. 200 Hz
2. Less than 200 Hz
3. Greater than 200 Hz
4. None of these

Answer: 1. 200 Hz

Question 5. A bus is moving towards a huge wall with a velocity of 5 m · s^-1. The driver sounds a horn of frequency 200 Hz. The beat frequency heard by the passenger will be

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

Answer: 2. 6

Question 6. A motor car sounding a horn is approaching a large reflector. If the frequency of the horn is 1000 Hz, the frequency of the echo to the driver will be

1. 1000 Hz
2. Less than 1000 Hz
3. Greater than 1000 Hz
4. None of these

Answer: 3. Greater than 1000 Hz

Question 7. An observer standing on a railway crossing receives frequencies of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. The speed of the sound in air is 300 m · s^-1. The velocity of the train (in m · s^-1)

1. 60
2. 30
3. 90
4. 70

Answer: 2. 30

Question 8. A whistle producing sound waves of frequency 9500 Hz and above is approaching a stationary person with speed v m · s^-1. The velocity of sound in air is 300 m · s^-1. If the person can hear frequencies upto a maximum of 10000 Hz, the maximum value of v up to which he can hear the whistle is

1. 15√2 m · s^-1
2. 15/√2 m · s^-1
3. 15 m · s^-1
4. 30 m · s^-1

Answer: 3. 15 m-s^-1

Question 9. What do you understand by the red shift of a star or of a galaxy of stars?

1. They are gradually receding from the earth
2. They are gradually approaching the earth
3. They are stationary
4. None of the above

Answer: 1. They are gradually receding from the earth

Question 10. When a source of light and an observer approach each other, the apparent change of the wavelength of light is called

1. Redshift
2. Violet shift
3. Blueshift
4. Black shift

Answer: 3. Blueshift

In this type of questions, more than one options are correct.

Question 11. State in which of the following cases, an observer will not see any Doppler effect?

1. Both the source and observer remain stationary but a wind blows.
2. The observer remains stationary but the source moves in the same direction and with the same speed as the wind.
3. The source remains stationary but the observer and the wind have the same speed away from the source.
4. The source and the observer move directly against the wind but both with the same speed.

Answer:

1. Both the source and observer remain stationary but a wind blows.

4. The source and the observer move directly against the wind but both with the same speed.

Question 12. Consider a source of sound S and an observer P. The sound source is of frequency n₀. The frequency observed by P is found to be n₁ if P approaches S at speed v and S is stationary, n₂ if S approaches P at speed v and P is stationary and n₃ if each of P and S has speed \(\frac{v}{2}\) towards one another. Which of the following conclusions is correct?

1. n₁ = n₂ = n₃
2. n₁< n₂
3. n₃ > n₀
4. n₃ lies between n₁ and n₂

Answer:

2. n₁< n₂

3. n₃>n₀

4. n₃ lies between n₁ and n₂

Question 13. An observer A is moving directly towards a stationary sound source while another observer B is moving away from the source with the same velocity. Which of the following conclusions are correct?

1. The average of frequencies recorded by A and B is equal to the natural frequency of the source.
2. The wavelength of waves received by A is less than that of waves received by B.
3. The wavelength of waves received by two observers will be the same.
4. Both observers will observe the wave travelling at same speed.

Answer:

1. The average of frequencies recorded by A and B is equal to natural frequency of the source.

3. The wavelength of waves received by two observers will be the same.

Question 14. Two cars, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these cars blows a whistle of frequency f₁. An observer in the other car hears the frequency of the whistle to be f₂. The speed of sound in still air is v. Correct statement(s) is are:

1. If the wind blows from the observer to the source, f₂ > f₁
2. If the wind blows from the source to observer, f₂>f₁
3. If the wind blows from the observer to the source, f₂<f₁
4. If the wind blows from the source to the observer, f₂<f₁

Answer:

1. If the wind blows from the observer to the source, f₂ > f₁
2. If the wind blows from the source to the observer, f₂>f₁
3. If the wind blows from the observer to the source, f₂<f₁

 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

 

Nature Of Vibration Long Answer Type Questions

Question 1. Why does an empty container emit a louder sound than a water-filled container when they are struck?
Answer:

When we strike the container, its wall vibrates. These vibrations produce forced vibrations in the air or water inside the container.

• If the container is filled with water, it becomes heavy and the amplitude of vibration becomes small. But if the container is empty, it is comparatively lighter and the amplitude of vibration is large.
• As the loudness of sound is proportional to the square of the amplitude of vibration, the empty container emits a louder sound.

Question 2. Why do buildings get demolished by earthquakes?
Answer:

An earthquake induces forced vibration in the walls of buildings. For intense earthquakes, the amplitudes of the forced vibrations become very high, which in many cases cross the elastic limit of the constituent materials of the buildings. So many buildings get demolished.

Question 3. A vibrating tuning fork is held at the mouth of a cylindrical tube. The tube is dipped into water. It is found that when the level of water rises to a definite height, a sound of large intensity is heard. Explain the reason behind it.
Answer:

The empty cylindrical tube is filled with air. The vibrating tuning fork produces forced vibration in the air column of the tube. As a result of the forced vibration of the air column, an additional sound is produced.

• The frequency of the sound produced depends on the length of the air column. By dipping the tube into water, the length of the air column can be changed.
• So, when the level of water rises to a particular height, the frequency of the air column in the tube and that of the tuning fork become equal. Then resonance takes place and a sound of large intensity is heard.

Question 4. In the presence of a resonant body, the sound produced by a body Is intensified. Is the principle of conservation of energy violated here?
Answer:

The principle of conservation of energy is not violated here.

• Suppose, a tuning fork is set into vibration. The loudness of the sound produced is not very large. If the vibrating fork is held on a hollow wooden box, the sound is intensified. In this case, the air inside the box acts as a resonant body.
• But it is found that the sound of the tuning fork lasts for a long time in the first case. But in the second case, the sound of the tuning fork is short-lived. The total dissipation of energy in both cases is the same.
• In the first case, the rate of dissipation of energy is low but continues for a long time. But in the second case, the rate of dissipation of the same amount of energy is comparatively high, and therefore the sound dies down in a short time. It does not violate the principle of conservation of energy.

Nature Of Vibration Conclusion

If the effect of forces other than the restoring force acting on a vibrating body is negligible, then its vibration is called free or natural vibration.

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.

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.

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.

Nature Of Vibration Very Short Answer Type Questions

Question 1.Under the influence of which force the oscillation of a pendulum gradually dies out?
Answer: Resistive force

Question 2. Which force acts on a body during its free vibration?
Answer: Restoring force

Question 3. Which force acts on a body that vibrates freely?
Answer: Only a restoring

Question 4. A man with a wristwatch falls freely from a tall building. Will the watch give the correct time?
Answer: Yes, independent of g

Question 5. Which quantity of vibration gradually decreases during damped vibration?
Answer: Amplitude

Question 6.If a resistive force acts on a vibrating body, then its amplitude of vibration gradually ________
Answer: Decreases

Question 7. What phenomenon will occur if the frequency of free vibration of a vibrating body becomes equal to the frequency of an external periodic force?
Answer: Resonance

Question 8. Which characteristic of sound increases during resonance?
Answer: Intensity

Question 9. A special case of ______ vibration is resonance.
Answer: Forced

Question 10. Which type of vibration is the vibration of the diaphragm of a microphone during a speech?
Answer: Forced

Question 11. What kind of external force has to act on a vibrating body, to occur forced vibration?
Answer: Periodic

Nature Of Vibration 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 vibrating body always moves to and fro about an equilibrium position.

Statement 2: Due to inertia of motion a vibrating body does not stop at its equilibrium position.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: The sound gets amplified when a vibrating tuning fork is made to touch the surface of a table.

Statement 2: The dimension of the table is more than that of the tuning fork.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.

Nature Of Vibration Match The Columns

Question 1.

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

Nature Of Vibration Integer Answer Type Questions

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

1. The amplitudes of the first two oscillations of a damped pendulum are 9.0 cm and 3.0 cm respectively. What will be the amplitude (in cm) of the pendulum in its third oscillation?
Answer: 1

Question 2. After 50 complete oscillations a pendulum’s amplitude becomes 1/3 of its initial vibration. If its amplitude becomes \(\frac{1}{n^3}\) of its initial vibration after 150 complete oscillations, then find the value of n.
Answer: 3

 

Nature Of Vibration Multiple Choice Questions And Answers

Question 1. In the case of free vibration of a body, the quantity that remains constant is

1. Velocity
2. Acceleration
3. Time period
4. Phase

Answer: 3. Velocity

Question 2. During vibration, the restoring force is

1. Directly proportional to the displacement
2. Directly proportional to the velocity
3. Directly proportional to the kinetic energy
4. Directly proportional to the potential energy

Answer: 1. Directly proportional to the displacement

Question 3. During the damped oscillation of a body, the force that acts is

1. Only the restoring force
2. Only the resistive force
3. The restoring force along with the resistive force
4. The restoring force along with the resistive force and the external periodic force

Answer: 3. The restoring force along with the resistive force

Question 4. During damped vibration, the quantity which gradually decreases is

1. Velocity
2. Phase
3. Frequency
4. Amplitude

Answer: 4. Amplitude

Question 5. In case of resonance, the characteristic property which is the same for free vibration of the body and the external periodic force is

1. Amplitude
2. Phase
3. Velocity
4. Frequency

Answer: 4. Frequency

Question 6. During the forced vibration of a particle

1. Only the restoring force acts on the particle
2. Both the restoring force and a dissipative force act on the particle
3. Both the restoring force and an external periodic force act on the particle
4. The restoring force, a dissipative force, and an external periodic force act on the particle

Answer: 4. The restoring force, a dissipative force, and an external periodic force act on the particle

Question 7. During forced vibration, the frequency of the external periodic force is

1. Equal to
2. Less than
3. Greater than
4. Equal to, or greater than, or less than the frequency of free vibration of the body.

Answer: 4. Equal to, or greater than, or less than the frequency of free vibration of the body.

Question 8. During forced vibration, if the frequency of free vibration of a body is equal to the frequency of the external periodic force, then the phenomenon that occurs is called

1. Beats
2. Interference
3. Resonance
4. Reverberation

Answer: 3. Resonance

Question 9. If a vibrating tuning fork is held at the open end of a tube closed at one end, then for a particular length of the tube an intense sound is heard. This phenomenon is known as

1. Beats
2. Stationary wave
3. Interference
4. Resonance

Answer: 4. 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

Superposition Of Waves Vibration Air Columns

Musical instruments like flute, organ, etc., are played by vibrating the air columns enclosed within them. Similarly, sound can be produced by blowing the air columns in whistles, thin pipes, etc.

Such an instrument, whatever be its shape, can be considered, as a pipe or tube enclosing an air column within it. In general, there are two types of pipes.

1. A pipe having one end open and the other closed is called a closed pipe;
2. A pipe having both ends open is called an open pipe.

• In the case of air columns, a closed-end in a column of air is analogous to the fixed end of a vibrating string. That is, at the closed end of an air column, the air is not free to undergo movement.
• Thus it is assumed the nodal position a standing wave, conversely, the air is free to undergo its back-and-forth longitudinal motion at the open end of the air column, and as such, the standing wave pattern will depict antinodes at the open ends of air columns.
• Now, we may hold a vibrating tuning fork or we may whiff at the open end B of a closed or an open pipe. Then a longitudinal sound wave propagates along the pipe towards the end A. This wave is reflected at A.

When this reflected wave comes back towards B, the incident and the reflected waves superpose. As a result, a stationary wave is generated within the pipe. This stationary wave produces the musical note emitted by instruments like flute, organ, etc.

Characteristics Of The Stationary Waves In An Air Column:

1. Since sound waves are longitudinal waves, the stationary waves generated in an air column are longitudinal stationary waves. We may recall that the waves in a vibrating stretched string are transverse stationary waves.
2. The air particles at the closed end (end A) of a closed pipe cannot vibrate at all. So, a node is generated at the closed end. On the other hand, the air particles at the open end (end B) of a closed pipe or at both the open ends of an open pipe, can vibrate with maximum amplitude. So, an antinode is generated at each open end.
3. The distance between two consecutive nodes or between two consecutive antinodes is \(\frac{\lambda}{2}\); here A = wavelength of the sound wave. Likewise, the distance between a node and its adjacent antinode = \(\frac{\lambda}{4}\).

Closed Pipe: Let the length of a closed pipe be l and the velocity of sound be V. In a closed pipe, incident waves, reflected waves from closed ends and the stationary wave produced are longitudinal waves. But for convenience, the waves in the pipe are drawn.

Fundamental Tone In A Dosed Pipe: For the simplest stationary wave in a closed pipe, the node is at the closed end and the antinode is at the open end. There is no other node or antinode between them. So, the tone produced in the pipe is called the fundamental tone or the fundamental.

If λ₀ is the wavelength of the sound wave, the length of the pipe, l = the distance between a node and its adjacent antinode = \(\frac{\lambda_0}{4}\)

or, λ₀ = 41.

So, the frequency of the emitted fundamental tone, \(n_0=\frac{V}{\lambda_0}=\frac{V}{4 l}\)…..(1)

This fundamental tone is also known as the 1st harmonic. Equation (1) implies that n₀ increases when l decreases. As a result, the pitch of the emitted tone will be higher for shorter pipes.

Overtones In A Closed Pipe: When there is another node-antinode pair between the node at the closed end and the antinode at the open end, the next higher tone or the 1st overtone is produced.

If λ₁ is the wavelength of the sound wave, the length of the pipe, l = distance between the node at the closed end and the second antinode from it = \(\frac{3 \lambda_1}{4}\)

∴ \(\lambda_1=\frac{4 l}{3}\)

So, the frequency of the emitted tone, \(n_1=\frac{V}{\lambda_1}=3 \cdot \frac{V}{4 l}\)

Clearly, n₁ = 3n₀ …(2)

This means that the frequency of the 1st overtone is three times that of the fundamental. For this reason, the 1st overtone is called the 3rd harmonic.

Similarly, for two node-antinode pairs between the open and the closed ends, the 2nd overtone is produced. The same calculations show that n₂ = 5 n₀; the frequency of the 2nd overtone is five times that of the fundamental.

So, it is the 5th harmonic. Clearly, the next overtones will have frequencies, n₃ = 7n₀, n₄ = 9n₀, ………., n_p = (2p+ 1)n₀, etc.

The following two features are evident in this case:

1. The frequency of each overtone is an integral multiple of the fundamental frequency. So, each overtone is a harmonic.
2. The harmonics with frequencies \(2 n_0, 4 n_0, 6 n_0, \cdots\) are absent. So, a closed pipe can emit the fundamental tone and its odd harmonics only.

Open Pipe: Let the length of an open pipe be l and the velocity of sound be V. In an open pipe, incident waves, reflected waves from open ends and the stationary wave produced are longitudinal waves. But, for convenience, the waves in the pipe are drawn in as transverse waves.

Fundamental Tone In An Open Pipe: For the simplest stationary wave in an open pipe, two antinodes are formed at the two ends. There is only one node just at the middle of the pipe. This is called the fundamental tone.

If λ₀ is the wavelength of the sound wave, the length of the pipe, l = the distance between two consecutive antinodes = \(\frac{\lambda_0}{2}\)

or, λ₀ = 21

So, the frequency of the emitted fundamental tone, \(n_1=\frac{V}{\lambda_1}=\frac{V}{l}=2 \cdot \frac{V}{2 l}\)…(2)

This fundamental tone is also known as the 1st harmonic.

Overtones In On Open Pipe: The 1st overtone is formed when two nodes lie between the two antinodes of the two ends and a third antinode is at the mid-point.

If λ₁ is the wavelength of the sound wave, the length of the pipe, l = distance between three consecutive antinodes = λ₁

i.e., λ₁ = 1

So, the frequency of the emitted tone, \(n_1=\frac{V}{\lambda_1}=\frac{V}{l}=2\cdot \frac{V}{2 l}\)

Clearly, \(n_1=2 n_0\)

This means that the frequency of the 1st overtone is twice that of the fundamental. For this reason, the 1st overtone is called the 2nd harmonic.

Similarly, for three nodes between the antinodes at the open ends, the 2nd overtone is produced. The same calculations show that n₂ = 3n₀; the frequency of the 2nd overtone is 3 times that of the fundamental frequency. So, it is the 3rd harmonic. Clearly, the next overtones will have frequencies, \(n_3=4 n_0, n_4=5 n_0, n_5=6 n_0, \cdots, n_p=(p+1) n_0 \text {, etc. }\).

The Following Two Features Are Evident In This Case:

1. The frequency of each overtone is an integral multiple of the fundamental frequency. So, each overtone is a harmonic.
2. An open pipe can emit the fundamental tone and all its even and odd harmonics. No harmonic is absent in this case.

A closed pipe does not emit the even harmonics, but all the even and odd harmonics are emitted from an open pipe. So, in general, notes emitted from an open pipe have a greater number of overtones.

• As a result, this note is of a higher quality and is more pleasant to hear as compared to that emitted from a closed pipe. For this reason, open pipes are used in all good-quality flutes, organs, etc.
• The fundamental frequency is inversely proportional to the length of a closed or open pipe. In musical instruments like flute, some holes are made along its length. A good instrumentalist can open or close the holes according to the need and can change the effective length of the pipe. In this way, he can change and control the frequency of the emitted tones.

Comparison Of Frequencies Of The Tones Emitted By Closed And Open Pipes Of Equal Length: Let l be the length of each of a closed and an open pipe and V be the velocity of sound in air.

Comparison Of Fundamental Frequencies: For the fundamental tone emitted by the closed pipe, a node is formed at the closed end and an antinode at the open end.

There is no other node or antinode between them.

If λ₀ is the length of the stationary wave,

l = distance from a node to its adjacent antinode = \(\frac{\lambda_0}{4}\)

or, λ₀ = 4l

So, the fundamental frequency, \(n_0=\frac{V}{\lambda_0}=\frac{V}{4 l}\)…..(1)

For the fundamental tone emitted by an open pipe, two antinodes are formed at the two open ends. There is only a single node in between them.

If λ’₀ is the length of the stationary wave, l = distance between two consecutive antinodes = \(\frac{\lambda_0^{\prime}}{2}\)

or, λ’₀ = 2l

• So, the fundamental frequency, \(n_0^{\prime}=\frac{V}{\lambda_0^{\prime}}=\frac{V}{2 l}\)….(2)
• Comparing the relations (1) and (2), we get \(\frac{n_0}{n_0^{\prime}}=\frac{\frac{V}{4 l}}{\frac{V}{2 l}}=\frac{1}{2} \quad \text { or, } n_0^{\prime}=2 n_0\)…(3)
• So, the fundamental frequency of an open pipe is twice that of a closed pipe of the same length. We know that the pitch of a musical sound increases with its frequency. While whiffing at one end of an open pipe, if the opposite end is suddenly closed with a finger, the pipe becomes a closed one.
• As a result, the pitch of the sound falls abruptly. Conversely, when the closed end of a closed pipe is suddenly opened, the sound emitted from the pipe becomes abruptly sharper.

Comparison Of The Frequencies Of Overtones: In the note emitted from a closed pipe, the frequencies of the fundamental and the overtones are \(n_0, 3 n_{o^{\prime}} 5 n_0, \ldots\), etc.

• On the other hand, the fundamental and the overtone frequencies in a note emitted from an open pipe are \(n_0^{\prime}, 2 n_0^{\prime}, 3 n_0^{\prime}, \cdots\), etc. If the closed pipe and the open pipe are of equal length, then n’₀ = 2n₀. So for the open pipe, the frequencies are \(2 n_0, 4 n_0, 6 n_0, \cdots, \text { etc. }\).
• So, it is seen that if one end of a pipe can be opened and closed as required, it can emit the fundamental tone as well as all the even and odd harmonics.

End Error In Closed And Open Pipes: Some amount of air is trapped inside a closed or an open pipe. Accurate experiments show that a small layer of air in close vicinity outside an open end also behaves like a trapped layer. So, this layer should also be considered as a part of the air column enclosed in the pipe.

• As a result, the antinode of a stationary wave is formed not exactly at the open end; rather it is formed at a point slightly outside every open end. This means that the effective length of the pipe is slightly greater than its actual length.
• This increase in the effective length, due to the open end of a closed pipe or due to both the open ends of an open pipe, is called the end error. If c is the magnitude of the end error for an open end of a pipe of length l then, effective length of a closed pipe, L = l+c; effective length of an open pipe, L = 1 + 2c. The relations of Sections 4.6.1 and 4.6.2 should be modified accordingly.

So, the fundamental frequency for a closed pipe, \(n_0=\frac{V}{4 L}=\frac{V}{4(l+c)}\)

and the fundamental frequency for an open pipe, [lat6ex]n_0=\frac{V}{2 L}=\frac{V}{2(l+2 c)}[/latex]

• This is known as end correction.
• Scientists Helmholtz and Rayleigh experimentally showed that, c = 0.58 ≈ 0.6 r; where r = radius of the tube.
• This means that narrower tubes correspond to fewer end errors. The theory also proposes that the end error increases with the wavelength of the sound emitted.
• For higher harmonics, the frequencies are very high and the wavelengths are very low. In that case, the end error becomes negligible.

Effect of Different Physical Quantities on the Frequencies of Air Columns: Considering the end errors, the fundamental frequencies of the notes emitted by a closed and an open pipe are, respectively,

⇒ \(n_0=\frac{V}{4(l+0.6 r)} \text { and } n_0^{\prime}=\frac{V}{2(l+2 \times 0.6 r)}\)

where, l = length of the pipe, r = radius of the pipe and V = velocity of sound in air.

These expressions show that the frequency becomes higher when

1. l is less,
2. r is less,
3. V is higher due to rise in temperature and
4. V is higher due to an increase in humidity.

The pitch of the emitted sound increases with the increase in frequency. So, the pitch becomes higher for,

1. Shorter pipes,
2. Narrower pipes,
3. The higher temperature of air and
4. Higher humidity in the air.

Comparison Between The Vibrations Of Air Columns In A Closed And An Open Pipe:

Unit 10 Oscillation And Waves Chapter 4 Superposition Of Waves Vibration Air Columns Numerical Examples

Example 1. Find out the frequency of the first overtone emitted by a1.25 m long organ pipe closed at one end. Given, the velocity of sound in air = 320 m · s^-1.
Solution:

For the 1st overtone, a node-antinode pair is formed in between a node at the closed end and an antinode at the open end.

So, the distance between the node at the closed end and the 2nd antinode from it = \(\frac{3 \lambda}{4}\) = l = length of the pipe.

∴ \(\lambda=\frac{4 l}{3}\)

So, the frequecny of this 1st overtone is \(n_1=\frac{V}{\lambda}=\frac{3 V}{4 l}=\frac{3 \times 320}{4 \times 1.25}=192 \mathrm{~Hz}\)

Example 2. Find out the frequencies of the fundamental and its nearest harmonic emitted by a 1m long closed organ pipe. Given, the velocity of sound in air =332 m · s^-1.
Solution:

Fundamental frequency, \(n_0=\frac{V}{4 l}=\frac{332}{4 \times 1}=83 \mathrm{~Hz}\)

A closed organ pipe can produce the odd harmonics only. So the frequency of the next harmonic,

n₁ = 3n₀ = 3 x 83 = 249 Hz.

Example 3. The length of an organ pipe, closed at one end is 90 cm. Find out the frequency of the harmonic next to the fundamental. Velocity of sound in air = 300 m · s¹
Answer:

A closed pipe emits the odd harmonics only. So, the frequency of the harmonic next to the fundamental is

n₁ = 3 x fundamental frequency

= \(3 \cdot \frac{V}{4 l}=3 \times \frac{(300 \times 100)}{4 \times 90}=250 \mathrm{~Hz} .\)

Example 4. The length of an organ pipe open at two ends is twice that of another organ pipe closed at one end. If the fundamental frequency of the open pipe is 100 Hz, find out the frequency of the 3rd harmonic emitted by the closed pipe.
Solution:

The fundamental frequency of the open pipe, \(n_0^{\prime}=\frac{V}{2 l}; \text { so, } V=2 \ln n_0^{\prime}\); so, V = 2ln₀‘ [V = velocity of sound in air, l = length of the open pipe]

The length of the closed pipe is \(\frac{l}{2}\). So the frequency of the 3rd harmonic is

n = 3x fundamental frequency

= \(3 n_0=3 \cdot \frac{V}{4\left(\frac{l}{2}\right)}=\frac{3 \cdot 2 \mathrm{~V}}{4 l}\)

= \(\frac{3 \times 2 \times 2 l n_0^{\prime}}{4 l}=3 n_0^{\prime}\)

= \(3 \times 100=300 \mathrm{~Hz} .\)

Example 5. Find out the fundamental frequency of a 125 cm long organ pipe closed at one end. Given, the velocity of sound in air = 350 m s^-1.
Solution:

If the length of the closed pipe is l and the wavelength of the fundamental is λ,

l = \(\frac{\lambda}{4}\) or, λ= 4l

If V is the velocity of sound in air, fundamental frequency, \(n_0=\frac{V}{4 l}=\frac{350 \times 100}{4 \times 125}=70 \mathrm{~Hz}.\)

Example 6. A 20 cm long closed pipe emits a tone of frequency 400Hz. Find out the length of an open pipe emitting a tone of frequency 600 Hz at the same
Solution:

The fundamental frequency for the dosed pipe, \(n_0=\frac{V}{4 l}\) [l = length of closed pipe]

The fundamental frequency for the open pipe, \(n_0^{\prime}=\frac{V}{2 L}\)

[L = length of open pipe]

∴ \(\frac{n_0}{n_0^{\prime}}=\frac{V}{4 l} \cdot \frac{2 L}{V}=\frac{1}{2} \cdot \frac{L}{l}\)

or, \(L=2 l \cdot \frac{n_0}{n_0^{\prime}}=2 \times 20 \times \frac{400}{600}=26.67 \mathrm{~cm}\)

Example 7. The length of an open organ pipe is twice the length of a closed organ pipe. If the fundamental frequency of the open pipe is 100 Hz, what is the frequency of the third harmonic of the closed pipe?
Solution:

Let the length of the closed organ pipe = l

Length of the open organ pipe =2l.

Frequency of the fundamental of the open pipe, \(n_0=\frac{V}{2 \cdot 2 l}=\frac{V}{4 l}\) [V= velocity of sound in air]

or, l = \(\frac{V}{4 n_0}=\frac{V}{4 \times 100}=\frac{V}{400}\)

Frequency of the fundamental of the closed pipe, \(n_c=\frac{V}{4 l}=\frac{V}{4} \cdot \frac{400}{V}=100 \mathrm{~Hz}\)

Only the odd harmonics are produced from a closed pipe; so the frequency of the third harmonic = 100 x (3 x 2- 1) = 500 Hz.

Question 8. One end of an open pipe is suddenly closed. It is the closed pipe is 100 Hz higher than the fundamental frequency of the open pipe. Find out this fundamental frequency when both ends are open.
Solution:

If V is the tire velocity of sound and l is the length of the pipe, the fundamental frequencies in the open and closed conditions are respectively,

⇒ \(n_0^{\prime}=\frac{V}{2 l} \text { and } n_0=\frac{V}{4 l}=\frac{n_0^{\prime}}{2}\)

The frequency of the 3rd harmonic for the closed pipe,

n = \(3 n_0=\frac{3}{2} n_0^{\prime}\)

∴ \(n-n_0^{\prime}=100 \text { or, } \frac{3}{2} n_0^{\prime}-n_0^{\prime}=100 \text { or, } n_0^{\prime}=200 \mathrm{~Hz}\).

Question 9. In winter, the frequency of a tone emitted at 10°C by an open organ tube is 400 Hz. What will be the frequency of this tone at 40°C In summer?
Solution:

Let V₀ be the velocity of sound in air at 0°C.

So, the velocity at 10°C , V₁ = V₀(1 + 0.00183 x 10) = 1.0183 V₀;

The velodty at 40°C, V2 = VQ(1 + 0.00183 x 40) = 1.0732 V₀

The frequency of a tone is proportional to the velocity of sound.

∴ \(\frac{n_1}{n_2}=\frac{V_1}{V_2}\)

or, \(n_2=n_1 \cdot \frac{V_2}{V_1}=400 \times \frac{1.0732 V_0}{1.0183 V_0}=421.57 \mathrm{~Hz}\).

Example 10. A 1 m long uniform cylindrical container Is closed by two thin vibrating membranes A and B at Its two ends. A third thin vibrating membrane C divides the container Into two equal parts. The parts AC and BC arc are filled with hydrogen and oxygen gases, respectively. The membranes A and B are vibrated with the same frequency. Find out the minimum frequency of this vibration so that a node is formed at C. Given, velocities of sound In hydrogen and oxygen gases arc 1100 m – s1 and 300 m · s^-1, respectively.
Solution:

AC = BC = 0.5 m = l (say).

If membranes A and B are vibrated, antinodes are produced at points A and B. Given, a node is produced at point C.

For an antinode at A and its adjacent node at C, the frequency is \(n_H=\frac{V_H}{4 l}=\frac{1100}{4 \times 0.5}=550 \mathrm{~Hz}\)

Similarly, for the part BC \(n_O=\frac{V_O}{4 l}=\frac{300}{4 \times 0.5}=150 \mathrm{~Hz}\)

Now, \(\frac{n_H}{n_O}=\frac{550}{150}=\frac{11}{3} or, 3 n_H=11 n_O\)

So, the frequency of the 3rd harmonic in AC matches exactly with that of the 11th harmonic in BC.

∴ The minimum frequency = 3 x 550 = 1650 Hz.

Example 11. Find out the coefficient of linear expansion of the material of an open pipe so that the frequency of any tone emitted from it does not vary with temperature.
Solution:

The fundamental frequencies at 0°C and t°C are respectively,

⇒ \(n_1=\frac{V_0}{2 l_0} \text { and } n_2=\frac{V_t}{2 l_t}\)

Here, V₀ = velocity of sound at 0°C, l₀ = length of the pipe at 0°C, V_t = velocity of sound at t°C, l_t = length of the pipe at t°C.

According to the question, n₁ = n₂

So, \(\frac{V_0}{2 l_0}=\frac{V_t}{2 l_t}\)

or, \(V_0 l_t=V_t l_0 \quad \text { or, } V_0 l_0(1+\alpha t)=V_0(1+0.00183 t) l_0 \)

(\(\alpha=\text { coefficient of linear expansion }\))

or, \(\alpha=0.00183^{\circ} \mathrm{C}^{-1} \text {. }\)

Determination of the Velocity of Sound in Air by Resonant Air Column: Consider a uniform glass tube P which has its upper end open. The tube is partially filled with water; so the glass tube effectively behaves as a pipe closed at one end.

• The tube P is connected to a big water container T through a rubber tube R. The length of the air column in the closed pipe P can be increased or decreased according to the need by raising or lowering the container T. The length of the air column can also be measured from a scale marked on the tube P.
• Now, a tuning fork F vibrating with a frequency n is held over the open end of the tube P. As a result, forced vibration is produced in the air column of P. The air column gradually rises from a very low value until the tuning fork and the air column are in unison.

Under these circumstances, the frequencies become equal and a loud sound is heard due to resonance. This corresponds to the fundamental tone emitted from the tube P. So, the fundamental frequency of the pipe is also n. If l₁ is the length of the resonant air column, then neglecting the end error we get,

⇒ \(l_1=\frac{\lambda}{4} \quad \text { or, } \lambda=4 l_1\)

So, \(n=\frac{V}{\lambda}=\frac{V}{4 l_1} \quad or, \quad V=4 n l_1\)…(1)

Here, λ is the wavelength of the sound emitted from the tuning fork and V is the velocity of sound in air.

So, V can be calculated from equation (1) by knowing the frequency n of the tuning fork and by measuring the length l₁ of the air column in P.

End Correction: Due to the end error, equation (1) cannot provide the accurate value of V. If c is the end error corresponding to the open upper end of P, equation (1) is modified as

V = \(4 n\left(l_1+c\right) \quad \text { or, } \quad l_1+c=\frac{V}{4 n}\)…(2)

Now, the length of the air column in P can be gradually increased further until the next overtone, which is the 3rd harmonic, comes in unison with the tuning fork. Here again, a loud sound is heard due to resonance. If l₂ is the length of the air column in this case, taking the end error into consideration we have,

n = \(3 \cdot \frac{V}{4\left(l_2+c\right)} \quad \text { or, } \quad l_2+c=\frac{3 V}{4 n}\)…(3)

Subtracting (2) from (3), \(l_2-l_1=\frac{V}{2 n} \quad \text { or, } \quad V=2 n\left(l_2-l_1\right)\)…(4)

Equation (4) has been obtained by eliminating the end error c effectively. So it provides the accurate value of the velocity of sound V in air.

Estimation Of End Error: From equations (2) and (3), \(3\left(l_1+c\right)=\frac{3 V}{4 n}=l_2+c\)

or, \(3 l_1+3 c=l_2+c \quad \text { or, } 2 c=l_2-3 l_1\)

or, c = \(\frac{1}{2}\left(l_2-3 l_1\right)\)…(5)

So, the end error can be estimated by measuring the values of l₁ and l₂. It is observed that c is nearly equal to 0.6r, where r is the radius of the tube P.

Superposition Of Waves Transverse Vibration In A String

Stretched string: A stretched string is a thin metal wire clamped between two rigid supports (A and B) with tension.

Such a string is the source of sound emitted by musical instruments like sitar, violin, piano, etc. To generate vibrations in the string, it is initially disturbed in a direction normal to its length.

This transverse disturbance is initiated usually by one of the three following methods:

1. Plucking the wire at any point—this method is used for sitar, guitar, etc.
2. Striking the wire at any point—this method is used to play a piano.
3. Bowing on the wire at any point—this method is used to play a violin.

The vibration in a stretched string has two salient features:

1. When a stretched string is displaced from its equilibrium position, two identical progressive waves are produced. These two waves travel towards the two ends of the wire. After getting reflected from the two ends, they again travel towards the opposite ends.
   • Thus, the two waves are reflected again and again from both ends. Hence, a stationary wave is generated in the string. This is known as a stationary wave in a stretched string.
   • As the string is rigidly clamped at the two ends, two nodal points are formed at the ends. One or more than one loops may be formed between them. Only one loop is shown there is no node other than A and B, and only one antinode is formed at the midpoint.
2. When the stretched string is vibrated in one or more than one loop, every point on the wire vibrates in a direction normal to the length of the wire. So, it is a transverse vibration.

Formation Of Stationary Waves In A Stretched String: Let a uniform string of length l be stretched by a tension T along the x-axis, with its ends rigidly fixed at the end x = 0 and x = l. Suppose a transverse wave is produced in the string travels along the positive x-axis and gets reflected at the fixed end, x = l.

The incident and reflected waves can be written as, y₁ = a sin (ωt- kx) and y₂ = – a sin(ωt+ kx) The equation of the resultant wave due to the superposition of the incident and the reflected progressive waves will be

y = \(y_1+y_2 =a \sin (\omega t-k x)-a \sin (\omega t+k x)\)

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

= \(-2 a \cos \omega t \sin k x\)… (1)

(Using the relation sinA – sinB = \(2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}\))

Since the supports A and B are rigid, nodes will be formed at these two points. So, y = 0 at x = 0 and y = 0 at x = l

[The coordinates of the supports A and B are x = 0 and x = l respectively, l = length of the string]

Now putting x = l and y = 0 in equation (1) we have

0 = -2a cosωt sinkl

So, for any value of t, sin kl = 0

i.e., kl = pπ, where p = 0, 1, 2, 3…..

Now if p = 0, then y = -2a cosωt sin kx

= \(-2 a \cos \omega t \sin \frac{p \pi}{l} x=0\)

i.e., for any value of t and x, y = 0.

This equation represents a vibrationless stationary string. This is not our subject of discussion.

So the significant values of p are 1, 2, 3,…..

Now, \(\frac{p \pi}{l} \quad \text { or, } \frac{2 \pi}{\lambda}=\frac{p \pi}{l} \quad \text { or, } \lambda=\frac{2 l}{p}\)…..(2)

Different Modes Of The Stationary Waves:

1. If p = 1, from equation (2), \(\lambda=2 l \text { or, } l=\frac{\lambda}{2}\). This means that two consecutive nodes must be apart by a distance equal to the length of the string, i.e., the string should vibrate in one segment.
2. If p = 2, from equation (2), \(\lambda=\frac{2 l}{2}=l \text { or, } l=\lambda\).

This means that between two nodes at the two ends of the string, there is another node. So the string must vibrate in two segments.

Generally, for the values of p = 1, 2, 3, 4,…… the stationary waves are formed in the stretched string and the string vibrates in one, two, three, four …. segments respectively. In each case, there are two nodes at the two ends of the string. Other nodes are formed according to the number of segments.

Laws Of Transverse Vibration In A Stretched String: Let a thin, uniform, and flexible wire be clamped with a tension between two rigid supports. Here, l = length of the wire, M = mass of the wire, r = radius of the wire,ρ = density of the material of the wire, and T = tension along the wire.

So, the area of cross-section = πr²;

volume of unit length of the wire = πr² · l = πr²

∴ Mass per unit length of the wire, m = \(\pi r^2 \rho=\frac{M}{l}\). This mass per unit length is often called the linear density of the wire. It is a characteristic property of the wire used because for a given wire, the linear density is fixed it is independent of the tension applied and the total length of the wire.

As proposed by the French scientist Marin Mersenne, the frequency (n) of transverse vibration in a stretched string varies with the relevant properties of the string according to the following laws:

Law Of Length: If the tension and the mass per unit length remain constant, the frequency of transverse vibration is inversely proportional to the length of the string, i.e., n ∝ \(\frac{1}{l}\), when T and m are constants.

Law Of Tension: If the length and the mass per unit length remain constant, the frequency of transverse vibration is proportional to the square root of the tension in the string, i.e., n ∝ √T, when l and m are constants.

Law Of Mass: If the length and the tension in the wire remain constant, the frequency of transverse vibration is inversely proportional to the square root of the mass per unit length of the string,

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

The combination of the three laws gives, \(n \propto \frac{1}{l} \sqrt{\frac{T}{m}} \quad \text { or, } n=\text { constant } \times \frac{1}{l} \sqrt{\frac{T}{m}}\)

When the string vibrates in a single loop, the fundamental tone is emitted. In that case, the value of the above constant is \(\frac{1}{2}\). Then the frequency of the fundamental tone is \(n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}\)….(1)

It Is To Be Noted That,

1. The frequency of transverse vibration of the wire does not depend upon the total mass; it depends on mass per unit length of the wire.
2. The frequency of the wire does not depend upon the extent of stretching or the strength with which the wire is hit, causing the wire to vibrate.

Now, m = \(\pi r^2 \rho \text { or, } \sqrt{m}=\sqrt{\pi} \cdot r \cdot \sqrt{\rho} \text {. So, the law } n \propto \frac{1}{\sqrt{m}}\) means that \(n \propto \frac{1}{r \sqrt{\rho}}\).

This leads to two supplementary laws on transverse vibration in a string:

Law Of Radius: If the material of the string remains the same, the density remains constant. Thus the frequency of transverse vibration is inversely proportional to the radius of the string when the length and the tension in the string are constant;

i. e., n ∝ \(\frac{l}{r}\), when l, T, and ρ are constants.}{r}[/latex]

Law Of Density: For strings of the same radius, but made of different materials, the frequency of transverse vibration is inversely proportional to the square root of the density of the material, if the length and the tension in the string remain constant;

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

Fundamental Tone And Overtones In A Stretched String: Let us consider the vibration of a stretched string in a single loop. Here, there are two nodes at the two ends, A and B, of the string and only one antinode at the mid-point.

If λ is the wavelength of the stationary wave produced, the distance between two consecutive nodes.

= \(\frac{\lambda}{2}\) = length of the wire (l).

or, λ = 2l

If V is the wave velocity, the frequency is \(n_1=\frac{V}{\lambda}=\frac{V}{2 l}\)…(1)

Now, if the string vibrates in two loops, a node is formed at the mid-point (C) in addition to the nodes A and B at the two ends. So, the distance between three consecutive nodes = λ = length of the wire (l).

Therefore, the frequency is \(n_2=\frac{V}{\lambda}=\frac{V}{l}=2 \cdot \frac{V}{2 l}=2 n_1\)

Similarly, the frequency becomes \(3 n_1, 4 n_1, 5 n_1, \ldots\), vibrations in three, four, five,… loops, respectively.

The above discussions show that the lowest frequency for all the tones that can be emitted from a vibrating string is n₁. So the tone of frequency n₁ is the fundamental tone or the 1st harmonic. This means that only the fundamental tone is emitted when a stretched string vibrates in n single loop.

It is the fundamental tone that is produced by the string. But the string is making all those other possible vibrations too, all at the same time, so that the actual vibration of the string is pretty complex.

The tones with higher frequencies \(2 n_1, 3 n_1, 4 n_1, \ldots\), etc,, are the overtones. Again, the frequency of each overtone is a simple multiple of the fundamental frequency; so each overtone is a harmonic, i.e., the frequency of 2nd harmonic = 2n₁, the frequency of 3rd harmonic = 3n₁,… the frequency of p-th harmonic = pn₁ and so on.

We cannot hear the harmonics as separate notes. It may be noted that a stretched string may emit all tire even and odd harmonics. They are what give the string its rich, musical, string-like sound timbre. (The sound of a single frequency alone is mechanical, uninteresting, and unmusical sound)

If the above equation (1) is compared with equation (1), we get, V  \(=\sqrt{\frac{T}{m}}\)

where, T = tension in the string, m = mass per unit length of the string, and V = velocity of the transverse wave in the string.

The frequencies of the fundamental and the different overtones, in terms of the tension (T), the length of the wire (l), and the mass of the wire per unit length (m), are

⇒ \(n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}; n_2=2 n_1=\frac{2}{2 l} \sqrt{\frac{T}{m}} ;\)

∴ \(n_p=\frac{p}{2 l} \sqrt{\frac{T}{m}}\)….(3)

where, p = number of loops in the vibration; n_p is called the frequency of the p-th harmonic.

In principle, a vibrating string can emit all the overtones.

• But, in reality, the presence of the overtones depends on the point of initial disturbance. For example, let the mid-point of the string be plucked to initiate the vibrations. Clearly, this point will have the maximum amplitude of vibration and will become an antinode.
• So, a node at the mid-point is never obtained by plucking the mid-point. Accordingly, the harmonics having a node at the mid-point will be absent.
• These absent harmonics are the 2nd, 4th 6th harmonics. So, a vibrating string generates only the odd harmonics when it is initially plucked at the mid-point.

Initial plucking or striking at a point on the string generates an antinode at that point. At the same time, if some other point is touched loosely, then a node is formed there.

• A metal bar is used for this loose touch while playing a guitar. In this way, a good instrumentalist can control the musical sound emitted by forming nodes and antinodes at points as per his desire.
• It is possible to vibrate every wire in sitar, piano, etc. in such a way that the same wire can emit different tones simultaneously. Shows an example where the fundamental and the 2nd harmonic are being produced simultaneously. The string is vibrating in two loops to produce the 2nd harmonic.
• At the same time, the 2-loop formation is also vibrating in a single loop, so that the fundamental tone is emitted.
• We know that the quality or timbre of a musical sound depends on the number of overtones in the emitted sound and also on the relations of the overtones with the fundamental. Accordingly, it is evident that instruments like sitar, violin, piano, etc., can emit notes of very rich quality.

Superposition Of Waves – Transverse Vibration In A String Numerical Examples

Example 1. Two stretched wires made of the same material have lengths, diameters, and tensions, each in the ratio 1:2. The first wire emits a fundamental tone of frequency 200 Hz. What is the fundamental frequency of the 2nd wire?
Solution:

If d is the diameter of a wire, the mass per unit length is m = \(\frac{\pi d^2}{4} \rho\)

where ρ = density of the material of the wire. It is the same for the two wires.

Now, the fundamental frequency,

n = \(\frac{1}{2} l \sqrt{\frac{T}{m}}=\frac{1}{2 l} \sqrt{\frac{T \cdot 4}{\pi d^2 \rho}}=\frac{1}{l d} \sqrt{\frac{T}{\pi \rho}}\)

So, for the two wires, \(\frac{n_1}{n_2}=\frac{l_2}{l_1} \cdot \frac{d_2}{d_1} \cdot \sqrt{\frac{T_1}{T_2}}\)

or, \(\quad n_2=n_1 \cdot \frac{l_1}{l_2} \cdot \frac{d_1}{d_2} \cdot \sqrt{\frac{T_2}{T_1}}=200 \times \frac{1}{2} \times \frac{1}{2} \times \sqrt{\frac{2}{1}}\)

[Here, \(n_1=200\)]

∴ \(n_2=50 \sqrt{2}=70.7 \mathrm{~Hz} \text {. }\)

Example 2. The lengths of two wires made of the same material are in the ratio 2:3. Their diameters are equal and the fundamental of the shorter wire is one octave higher than that of the longer wire. Find the ratio
Solution:

The two wires are of the same material and the diameters are equal. So, the mass per unit length m is the same.

Here, the ratio 2 : 3 implies that the 1st wire is shorter

∴ \(n_1=2 n_2 \text { or, } \frac{n_1}{n_2}=\frac{2}{1} \text {. }\)

The fundamental frequency, n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}\)

So, for the two wires, \(\frac{n_1}{n_2}=\frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}\)

or, \(\frac{T_1}{T_2}=\left(\frac{n_1}{n_2}\right)^2 \cdot\left(\frac{l_1}{l_2}\right)^2=\left(\frac{2}{1}\right)^2 \cdot\left(\frac{2}{3}\right)^2=\frac{16}{9}\)

i.e., the ratio between the tensions is 16:9.

Example 3. A wire of density 9 g · cm^-3 Is elongated by 0.05 cm when stretched between two clamps 100 cm apart Find out the lowest frequency of transverse vibration in the wire, Given, Young’s modulus of the material of the wire = 9 x10¹¹ dyn · cm^-2.
Solution:

Length of the wire, L = 100 cm; elongation, l = 0.05 cm; density of the material, ρ = 9 g · cm^-3; Young’s modulus of the material,

Y = 9 x 10¹¹ dyn · cm^-2.

Let α = area of the cross-section of the wire

∴ Mass per unit length, m = αρ

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

longitudinal strain = \(\frac{\text { elongation }}{\text { original length }}=\frac{l}{L} .\)

∴ Y = \(\frac{\text { longitudinal stress }}{\text { longitudinal strain }}=\frac{T / \alpha}{l / L}=\frac{T L}{\alpha l}\)

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

The lowest frequency of transverse vibration = fundamental frequency,

n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}=\frac{1}{2 L} \sqrt{\frac{Y \alpha l}{L} \cdot \frac{1}{\alpha \rho}}=\frac{1}{2 \times 100} \sqrt{\frac{\left(9 \times 10^{11}\right) \times 0.05}{100 \times 9}}\)

= 35.36 Hz

Example 4. A stationary wave having 5 loops is generated in a 10 m long wire. What is the frequency, if the wave velocity is 20 m · s^-1?
Solution:

If λ is the wavelength, length of each loop = \(\frac{\lambda}{2}\)

∴ Length of 5 loops = 5 • \(\frac{\lambda}{2}\) = 10 m or, \(\lambda=\frac{10 \times 2}{5}\) = 4m

So, frequency, n = \(\frac{V}{\lambda}=\frac{20}{4}=5 \mathrm{~Hz}\) = 5 Hz.

Example 5. A uniform wire of length 12 m and mass 6 kg is suspended from a rigid support A mass of 2 kg is attached to the lower free end. A transverse wave of length 0.06 m is generated at the lower end of the wire. What is its wavelength when the wave reaches the upper end?
Solution:

The velocity of transverse vibration in a stretched wire, V = \(\sqrt{\frac{T}{m}}\), where T = tension in the wire and m = mass per unit length = constant, for the uniform wire.

If V₁ and V₂ are the velocities at the lower and upper ends, respectively, \(\frac{V_1}{V_2}\)=\(\sqrt{\frac{T_1}{T_2}}\)

Here, T₁ = tension at the lower end = weight of the suspended mass = 2 x 9.8 N;

T₂ = tension at the upper end = weight of the wire and the suspended mass = (2 + 6) x 9.8 = 8 x 9.8 N

The entire wire will emit a sound of a fixed frequency; so the wavelength A is proportional to the wave velocity V.

∴ \(\frac{\lambda_1}{\lambda_2}=\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(\lambda_2=\lambda_1 \sqrt{\frac{T_2}{T_1}}=0.06 \times \sqrt{\frac{8 \times 9.8}{2 \times 9.8}}=0.12 \mathrm{~m}\).

Example 6. A wire is stretched with negligible tension at 30°C between two rigid supports. Find the velocity Of the transverse wave in the wire at 20°C. Given, the coefficient of linear expansion, Young’s modulus, and the density of the material of the wire are α = 18 x 10^{-6 °}C^-1, Y = 12 x 10¹¹ dyn • cm^-2, and ρ = 6 g · cm^-3, respectively.
Solution:

Decrease in temperature, θ = 30-20 = 10°C; due to this decrement, the wire will tend to contract. As a result, thermal stress will be developed.

This thermal stress = Yαθ;

So the tension in the wire, T = Yαθ A, where A = area of the cross-section of the wire.

Again, mass per unit length of the wire,

m = A · 1 · ρ = Aρ

∴ The velocity of the transverse wave in the wire is

V = \(\sqrt{\frac{T}{m}}=\sqrt{\frac{Y \alpha \theta A}{A \rho}}\)

= \(\sqrt{\frac{Y \alpha \theta}{\rho}}=\sqrt{\frac{\left(12 \times 10^{11}\right) \times\left(18 \times 10^{-6}\right) \times 10}{6}}\)

= \(6000 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Superposition Of Waves Beats

When two progressive waves with equal or nearly equal amplitudes, but with a slight difference in then- frequencies, move in the same direction and superpose in a region of space, the amplitude and intensity of the resultant wave increase and decrease periodically. This phenomenon is known as beats.

Beats Experiment: Let two tuning forks of the same frequency be kept on a hollow box. If the two forks are vibrated simultaneously, nothing special is observed in the emitted sound. Now, one arm of one of the forks is waxed slightly. Thus the frequency of this fork decreases slightly.

The two forks are vibrated again simultaneously keeping them close to each other. It is observed that the loudness of the emitted sound rises and falls periodically. This phenomenon is known as beats.

Characteristics Of Beats:

1. Two progressive waves with equal or nearly equal amplitudes, but differing slightly in frequencies are allowed to superpose. The resultant amplitude rises and falls periodically and beats are formed.
2. Rise and fall of resultant amplitude result in rise and fall of intensity. This corresponds to periodic increments and decreases of the loudness for sound waves and the brightness for light waves.
3. If the resultant intensity goes to maximum (or minimum) n times per second, then n is called the number of beats per second or beat frequency. It is equal to the difference in the frequencies of the component waves, i.e., if n₁ and n₂ are the component frequencies (n₁>n₂), the beat frequency is n = n₁-n₂.
4. The persistence of hearing of the human ear is \(\frac{1}{10}\)s. This means that if more than one sound of the same type comes to our ear within \(\frac{1}{10}\)s, we cannot distinguish them. So, if the beat frequency is more than 10 Hz, we cannot feel the effect of rise and fall of the intensity. We may conclude that to feel the effect of beats, the two superposing waves should have a frequency difference of less than 10 Hz.
   • For example, if two sound waves of frequencies 200 Hz and 205 Hz superpose, the beat frequency is (205 – 200) = 5; that can be easily perceived.
5. For sound waves, two sources having a frequency difference of less than 10 Hz are often realized in practice. But for light waves, individual frequencies are of the order of 10¹⁵ Hz in the visible range. So, a frequency difference of less than 10 Hz is practically impossible to observe. For this reason, the formation of beats is a phenomenon useful for sound waves but has no physical significance for light waves.

The above discussions provide the precise definition of beats:

Beats Definition: The periodic rise and fall of the loudness of the resultant sound wave, produced by the superposition of two progressive sound waves of equal amplitude but of slightly different frequencies, is called beats.

Graphical Representation Of Beat Formation: Let two sound waves of nearly equal amplitudes be incident on each other with the initial phase. For example, we take two progressive waves of frequencies 10 Hz and 8 Hz.

• They superpose in a region of space with equal amplitude and with the same initial phase. Shows the displacements y₁ and y₂ of the two waves A and B, respectively. The resultant displacement is shown by C. From the principle of superposition, the algebraic sum (y₁ + y₂) denotes the resultant displacement y at any instant.
• When t = 0, \(\frac{1}{2}\)s and 1 s, the two waves are in the same phase. So, the resultant y is the maximum. On the other hand, at t = \(\frac{1}{4}\)s and \(\frac{3}{4}\)s, the two waves are in opposite phase and the resultant y is nearly zero, i.e., minimum.

• So in a span of 1 s, rise or fall in the intensity of sound occurs twice, i.e., two beats are formed per second.
• This means that \(\frac{1}{2}\)s time is required to produce a single beat. Moreover, shows that the time interval between two maximum amplitudes = \(\frac{1}{2}\) – 0 = \(\frac{1}{2}\)s; the time interval between two minimum amplitudes = \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{2}\)s.

So, it may be concluded that a beat is formed between two consecutive maxima or between two consecutive minima of the resultant amplitude.

Mathematical Analysis Of Beats: Let us consider two sound waves having frequencies n₁ and n₂ (n₁> n₂) traveling through a medium. They have equal amplitude A and equal initial phase. The displacements of the two waves at any point is \(y_1=A \sin 2 \pi n_1 t \text { and } y_2=A \sin 2 \pi n_2 t\)

According to the principle of superposition, the resultant displacement is

y = \(y_1+y_2=A\left(\sin 2 \pi n_1 t+\sin 2 \pi n_2 t\right)\)

= \(2 A \sin \left(2 \pi \frac{n_1+n_2}{2} t\right) \cos \left(2 \pi \frac{n_1-n_2}{2} t\right)\)

or, y = \(A^{\prime} \sin 2 \pi n t\)….(1)

where n = \(\frac{n_1+n_2}{2}\)…(2)

and \(A^{\prime}=2 A \cos 2 \pi \frac{\left(n_1-n_2\right) t}{2}\)…(3)

The factor sin2πnt in equation (1) implies that the resultant wave is a sound wave with frequency n = \(\frac{n_1+n_2}{2}\) = average of the component frequencies.

Besides, equation (3) shows that the amplitude of the resultant wave (A’) is not a constant, rather it varies with time t. As the intensity of sound is proportional to the square of the amplitude, this intensity also varies with time. The amplitude changes with time between a maximum and a minimum. As a result, the intensity of sound periodically increases or decreases. This is known as beats.

Beat Frequency:

1. If \(\cos 2 \pi \frac{\left(n_1-n_2\right) t}{2}=0\), the intensity of sound becomes zero, i.e., A’ = 0.

In this condition, no sound is heard.

Here, \(\cos 2 \pi \frac{\left(n_1-n_2\right) t}{2}=0\)

This corresponds to \(2 \pi \frac{\left(n_1-n_2\right) t}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \cdots\)

or, t = \(\frac{1}{2\left(n_1-n_2\right)}, \frac{3}{2\left(n_1-n_2\right)}, \frac{5}{2\left(n_1-n_2\right)}, \cdots\)

So, the time interval between two consecutive minima is

\(t_0 =\frac{3}{2\left(n_1-n_2\right)}-\frac{1}{2\left(n_1-n_2\right)}=\frac{1}{n_1-n_2}\)

= \(\frac{5}{2\left(n_1-n_2\right)}-\frac{3}{2\left(n_1-n_2\right)}\)

i.e, the number of minima per second = \(\frac{1}{t_0}=n_1-n_2\).

2. If \(\cos 2 \pi \frac{\left(n_1-n_2\right) t}{2}= \pm 1\), the intensity of sound becomes maximum, i.e., A’ = ±2A (maximum amplitude). In condition, aloud sound is heard.

Here, \(\cos 2 \pi \frac{\left(n_1-n_2\right) t}{2}= \pm 1\)

This corresponds to \(2 \pi \frac{\left(n_1-n_2\right) t}{2}=0, \pi, 2 \pi, \cdots\)

or, t=0, \(\frac{1}{n_1-n_2}, \frac{2}{n_1-n_2}, \frac{3}{n_1-n_2}, \cdots\)

The time interval between two consecutive maxima is \(t_0^{\prime}=\frac{1}{n_1-n_2}-0=\frac{1}{n_1-n_2}=\frac{2}{n_1-n_2}-\frac{1}{n_1-n_2} ;\)

i.e., the number of maxima per second

= \(\frac{1}{t_0^{\prime}}=n_1-n_2\)

So, the number of beats per second or the beat frequency = n_{1 ∼} n₂ = magnitude of difference in frequencies of the two superposed waves.

It is to be noted that if the amplitudes A and B of the two superposed waves are slightly different, then for the resultant wave, maximum amplitude = (A+B) and minimum amplitude = (A-B)≠0

In this case, the sound intensity never falls to zero, that is, absolute silence is never achieved. Rather, a feeble sound is heard between every two loud maxima.

Conditions For The Recognition Of Beats: The beat frequency must be less than 10 Hz, i.e., more than one beat must not be formed within a time span of \(\frac{1}{10}\)s. Otherwise, beats cannot be separately recognized by the ear and a continuous sound is heard. This is known as a beat note.

The amplitudes and intensities of die two superposed waves must be equal or nearly equal. If they are widely different, the intensity variation between the maxima and the minima of the resultant wave is much less, and it becomes hard to recognize the resultant beats.

Application Of Beats:

Determination Of An Unknown Frequency: To determine the unknown frequency of a source of sound, a few tuning forks of known frequencies are taken. Let it be the unknown frequency of a source of sound and n₁, n₂, n₃, ….. be the known frequencies of the standard tuning forks.

At first, die tuning forks, one by one, are vibrated simultaneously with the source of sound to detect whether recognizable beats are formed or not. Suppose, a particular tuning fork of frequency nf is able to form such beats. If the beat frequency is N, then

Either, n-n_i = N i.e., n = n_i + N;

or, n₁ – n = N i.e., n = n_i – N

• For example, if a tuning fork of frequency 200 Hz forms 4 beats per second, the unknown frequency of the source will be either 200  4 = 204 Hz or 200 – 4 = 196 Hz
• Now, the correct value between 204 Hz and 196 Hz is to be ascertained. For this purpose, a small drop of wax is put or a small piece of paper is pasted on one arm of the tuning fork.
• This results in a slight decrease in the frequency of the fork. So the beat frequency will also change slightly, it may be less or more than its previous value.
• If the number of beats is reduced, the unknown frequency will be 196 Hz. On the other band, if the number of beats is increased, the unknown frequency wall he 204 Hz.

Tuning Of A Musical Instrument: A musical instrument can be tuned with die help of beats. Let a particular string of a musical instrument be brought to unison with a standard source of musical sound. The string and the standard source are vibrated simultaneously.

The length and the tension of the string are gradually varied until beats are formed and recognized. It indicates that the frequency of the string is now within 10 Hz of the frequency of the standard source. Then a little more variation of the length and the tension would bring the string tuned exactly with the source. No more formation of beats would confirm this tuning.

Superposition Of Waves Beats Numerical Examples

Example 1. When two tuning forks are vibrated simultaneously, 4 beats are heard per second. The second fork is waxed slightly and then 6 beats are heard per second when both are vibrated again simultaneously. Find out the frequency of the second fork. Given, that the frequency of the first fork is 510 Hz.
Solution:

Let n₂ be the frequency of the 2nd tuning fork.

We know, the number of beats per second = difference in frequencies of the two tuning forks.

If n₂ > 510 Hz , then n₂ -510 = 4 or, n₂ = 514 Hz. When the 2nd fork is waxed, its frequency becomes less than 514 Hz. Then the number of beats per second decreases. So, 6 beats will not be formed per second.

If n₂ < 510 Hz , then 510- n₂ = 4 or, n₂ = 506 Hz . The frequency decreases further for waxing. So, 6 beats can be formed per second. This means that the value n₂ = 506 Hz matches with the given problem.

Example 2. 2 of length 78 cm and 80 cm of a sonometer are kept at the same tension. A tuning fork produces 4 beats per second with each of them. Find out the frequency of the tuning fork.
Solution:

Fundamental frequency of the 78 cm long wire, \(n_1=\frac{1}{2 \times 78} \sqrt{\frac{T}{m}}\)

Fundamental frequency of the 80 cm long wire, \(n_2=\frac{1}{2 \times 80} \sqrt{\frac{T}{m}}\)

∴ \(\frac{n_1}{n_2}=\frac{80}{78} ; \text { clearly, } n_1>n_2\)

If n is the frequency of the tuning fork, then \(n_1-n=4 \text { and } n-n_2=4\)

∴ \(n_1-n_2=8 \text { or, } \frac{n_1}{n_2}-1=\frac{8}{n_2}\)

or, \(\frac{80}{78}-1=\frac{8}{n_2} \text { or, } \frac{2}{78}=\frac{8}{n_2} \text { or, } n_2=312 \mathrm{~Hz}\)

∴ n = \(312+4=316 \mathrm{~Hz}\).

Example 3. A 75 cm long stretched string is tuned with a tuning fork. If the length of the string is reduced by 3 cm, it produces 6 beats with the tuning fork per second. Find out the frequency of the timing fork.
Solution:

Frequency of the tuning fork, n = fundamental

Frequency of the 75 cm long string = \(\frac{1}{2 \times 75} \sqrt{\frac{T}{m}}\)

Again, the fundamental frequency of the (75 – 3) cm or 72 cm long string, \(n^{\prime}=\frac{1}{2 \times 72} \sqrt{\frac{T}{m}}\)

∴ \(\frac{n}{n^{\prime}}=\frac{72}{75} ; \text { clearly, } n<n^{\prime}\) .

As 6 beats are produced per second, we have,

n’ – n = 6

or, \(\frac{n^{\prime}}{n}-1=\frac{6}{n}\)

or, \(\frac{75}{72}-1=\frac{6}{n} \text { or, } \frac{3}{72}=\frac{6}{n} \text { or, } n=144 \mathrm{~Hz}\).

Question 4. A tuning fork of unknown frequency produces 5 beats per second with another tuning fork. The second fork can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when a small amount of wax is dropped on the first fork. Find out the frequency of the first tuning fork. Given, the speed of sound in air = 320 m · s^-1.
Solution:

Length of the organ pipe =40 cm = 0.4 m

So, its fundamental frequency = \(\frac{V}{4 l}=\frac{320}{4 \times 0.4}=200 \mathrm{~Hz}\)

∴ The frequency of the 2nd tuning fork, n₂ = 200 Hz.

Clearly, the frequency of the 1st tuning fork is n₁  = (200+ 5) Hz or, (200-5) Hz;

The wax would decrease the frequency of the 1st fork. This decreased frequency must be closer to n₂ = 200 Hz, as the beat frequency also decreases. So, the frequency of the 1st tuning fork is n₁ = 205 Hz.

Example 5. Two tuning forks A and B, when vibrated simultaneously, produce 5 beats per second. 40 cm and 40.5 cm lengths of a sonometer wire, kept in the same tension, are tuned with the forks A and B, respectively. Find out the frequencies of the two tuning forks.
Solution:

According to the question, \(n_A=\frac{1}{2 \times 40} \sqrt{\frac{T}{m}} \text { and } n_B=\frac{1}{2 \times 40.5} \sqrt{\frac{T}{m}}\)

∴ \(\frac{n_A}{n_B}=\frac{40.5}{40}=\frac{81}{80} ; \text { clearly, } n_A>n_B\)

As the number of beats per second is 5, we have, \(n_A-n_B=5 \quad \text { or, } \frac{n_A}{n_B}-1=\frac{5}{n_B} \quad \text { or, } \frac{81}{80}-1=\frac{5}{n_B}\)

or, \(\frac{1}{80}=\frac{5}{n_B} \quad \text { or, } n_B=400 \mathrm{~Hz}\)

∴ \(n_A=n_B+5=400+5=405 \mathrm{~Hz}\)

Example 6. Two wires are tied on a sonometer. The tensions, lengths, diameters, and densities of the materials of the two wires are in the ratio 8: 1, 36: 35, 4: 1, and 1:2, respectively. Find out the number of beats produced per second when the two wires are vibrated simultaneously. Given, the frequency of the wires are vibrated simultaneously. The frequency of the wire emitting a tone of higher pitch is 360 Hz.
Solution:

Mass per unit length of a wire, m = \(\frac{\pi d^2}{4} \rho\), where d = diameter of the wire and ρ = density of the material.

∴ Fundamental frequency, n = \(\frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{2 l} \sqrt{\frac{4 T}{\pi d^2 \rho}}=\frac{1}{l d} \sqrt{\frac{T}{\pi \rho}}\)

So, for the two wires, \(\frac{n_1}{n_2}=\frac{l_2}{l_1} \cdot \frac{d_2}{d_1} \cdot \sqrt{\frac{T_1}{T_2} \cdot \frac{\rho_2}{\rho_1}}=\frac{35}{36} \times \frac{1}{4} \times \sqrt{\frac{8}{1} \times \frac{2}{1}}=\frac{35}{36}\)

Clearly, n₂ >n₁ so the pitch of the tone emitted by the 2nd wire is higher.

Here, n₂ = 360 Hz

∴ \(\frac{n_1}{360}=\frac{35}{36} \quad \text { or, } n_1=350 \mathrm{~Hz}\)

∴ Number of beats per second =360-350 = 10.

Example 7. Two tuning forks A and B, when vibrated simultaneously, produce 5 beats per second. The forks produce resonances respectively with 36 cm and 37 cm long air columns of a tube closed at one end. What are the frequencies of the tuning forks?
Solution:

If is the velocity of sound air, the fundamental frequency of a tube of length l closed at one end, is

n = \(\frac{V}{4 l}\)

∴ \(n_A=\frac{V}{4 \times 36} ; n_B=\frac{V}{4 \times 37}\)

So, \(\frac{n_A}{n_B}=\frac{37}{36}\) ; clearly, \(n_A>n_B\).

As 5 beats are produced per second, we have \(n_A-n_B=5\)

or, \(\frac{n_A}{n_B}-1=\frac{5}{n_B} \quad \text { or, } \frac{37}{36}-1=\frac{5}{n_B} \quad \text { or, } \frac{1}{36}=\frac{5}{n_B} \\
\text { or, } n_B=180 \mathrm{~Hz}\)

∴ \(n_A=n_B+5=180+5=185 \mathrm{~Hz} .\)

Example 8. A diver sends an audio signal from some depth underwater. This signal produces 5 beats per second with the note emitted by a 20 cm long pipe whose one end is closed. Find out the frequency and the wavelength of the audio signal inside water. Given, the velocity of sound in air = 360 m · s^-1 and that in water = 1500 m · s^-1.
Solution:

The fundamental frequency of the 20 cm long pipe closed at one end, \(n_0=\frac{V}{4 l}=\frac{360}{4 \times(20 \times 0.01)}=450 \mathrm{~Hz}\)

As 5 beats are produced per second, the frequency of the audio signal is either, (450- 5) = 445 Hz or (450 + 5) = 455Hz

The frequency does not change due to refraction from water to air. So the wavelength inside water is

Either \(\frac{1500}{445}=3.37 \mathrm{~m} \quad \text { or, } \frac{1500}{455}=3.30 \mathrm{~m}\)

Example 9. Two waves from 10 beats in 3s in a gas, The wavelength are  1m and 1.01m, respectively. Find out the velocity of sound in the gas.
Solution:

Let V be the velocity of sound in the gas.

As 10 beats are formed in 3 s, the beat frequency = \(\frac{10}{3}\)Hz.

The frequency of the 1st wave, \(n_1=\frac{V}{\lambda}=\frac{V}{1} \mathrm{~Hz} \text {; }\)

The frequency of the 2nd wave, \(n_2=\frac{V}{\lambda_2}=\frac{V}{1.01} \mathrm{~Hz}\)

Here, \(n_1>n_2\)

∴ \(n_1-n_2=\frac{10}{3} \text { or, } \frac{V}{1}-\frac{V}{1.01}=\frac{10}{3}\)

or, V = \(336.7 \mathrm{~m} \cdot \mathrm{s}^{-1}\) .

Example 10. 24 tuning forks lire Otttutged lit the ascending writer of their frequencies. Itaeh fork produces 4 between per second with Ha Immediately preceding fork, The tell fork emits tut octave to Hint emitted by the first one, Find out the frequencies of the first the the last tuning forks.
Solution:

Let the frequencies In ascending order of the 24 tuning forks be n₁, n₂,……… n₂₄. frequencies tell that

\(\begin{gathered}
n_2-n_1=4 \\
n_3-n_2=4 \\
\ldots \ldots \ldots \ldots \ldots \ldots \\
n_{24}-n_{23}=4 \\
\hline \text { (on addition) } n_{24}-n_1=23 \times 4=92
\end{gathered}\)

 

As the 24th tuning fork emits an octave to that emitted by the first one, n₂₄ = 2n₁

∴ 2n₁ – n₁ = 92 or, n₁ = 92 Hz.

∴ n₂₄ = 2n₁ = 2 x 92 = 104 Hz.

Example 11. A wire of length 25 cm and muss 2.5 g Is stretched with a fixed tension. The length of a pipe dosed at one end Is 40 cm. During vibrations, the first overtone of the wire produces 8 beats per second with the fundamental emitted by the pipe. The number of beats reduces with the decrease In tension In the wire. If the velocity of sound In air Is 320 m · s^-1, find the tension in the wire.
Solution:

The fundamental frequency of the pipe closed at one end, \(n_1=\frac{V}{4 l}=\frac{(320 \times 100)}{4 \times 40}=200 \mathrm{~Hz}\)

The mass per unit length of the wire,m = \(\frac{2.5}{25}=0.1 \mathrm{~g} \cdot \mathrm{cm}^{-1}\)

The 1st overtone is the 2nd harmonic of the wire; the frequency is \(n_2=2 \times \frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{25} \sqrt{\frac{T}{0.1}}=\frac{\sqrt{10 T}}{25} \mathrm{~Hz}\)

Clearly, n₂ reduces with the decrease in tension T; so (n₂ – n₁) also decreases. For this reason, the beat frequency reduces. This means, n₂ > n₁.

As B beats are produced per second, we get n₂  – n₁ = 8

or, n₂ = n₁ + 8  = 200 + 8 = 208 Hz.

∴ 208 = \(\frac{\sqrt{107}}{25} \text { or, } 10 T=(208 \times 25)^2\)

or, T = \(\frac{(200 \times 25)^2}{10}=27.04 \times 10^5 \mathrm{dyn}=27,04 \mathrm{~N} .\)

Example 12. A sonometer wire Is stretched by hanging a 10cm high brass cylinder vertically from one of Its ends. The wire resonates with a tuning fork of frequency 250 Hz. Now the cylinder Is partially Immersed In water. If the wire and the tuning fork are vibrated simultaneously, 4 beats are heard per second. Calculate the length of the portion of the cylinder that was immersed In water. Given, the density of brass = 0,5 g · cm^-3.
Solution:

The apparent loss of weight of the partially Immersed cylinder reduces the tension m the wire.

So, the frequency of the vibrating wire decreases. When the cylinder is partially immersed In water, the fundamental frequency of the wire is \(n_2=n_1-4=256-4=252 \mathrm{~Hz}\)

Now, \(\frac{n_1}{n_2}=\sqrt{\frac{T_1}{T_2}} or, \frac{256}{252}=\sqrt{\frac{T_1}{T_2}}\)

Here, T₁ = weight of the cylinder

= 10αρg = 10α x 0.5 x g

T₂ = weight of the partially immersed cylinder

= weight of the cylinder in the air – the weight of the water displaced

= 10α x 8.5 x g – lα x 1 x g

where α = area of the cross-section of the cylinder

l = height of the cylinder immersed in water

∴ \(\frac{256}{252}=\sqrt{\frac{10 \alpha \times 8.5 \times g}{10 \alpha \times 8.5 \times g-l \alpha \times 1 \times g}}=\sqrt{\frac{85}{85-l}}\)

or, \(\frac{85}{85-l}=\left(\frac{256}{252}\right)^2 \text { or, } 85-l=85 \times\left(\frac{252}{256}\right)^2\)

or, \(l=85 \times\left[1-\left(\frac{252}{256}\right)^2\right]=2.64 \mathrm{~cm}\)

Example 13. Two progressive waves y₁ = 4 sin 500πt and y₂ = 2 sin506π t are supported. Find the number of beats produced in one minute.
Solution:

Comparing the given equations with the standard equation y = Asinωt,

We have, \(\omega_1=500 \pi \text { or, } n_1=\frac{\omega_1}{2 \pi}=\frac{500 \pi}{2 \pi}=250 \mathrm{~Hz} ;\)

and \(\omega_2=506 \pi \text { or, } n_2=\frac{\omega_2}{2 \pi}=\frac{506 \pi}{2 \pi}=253 \mathrm{~Hz}\)

∴ Number of beats per second = difference in frequencies =253-250 =3

∴ A Number of beats produced In one minute = 3 x 60 = 100

Example 14. Three transverse progressive waves are x₁ = Acos(kx-ωt), x₂  = Acos(kx + ωt) , x₃ = Acos (ky-ωt). How may these be superposed to generate

1. A stationary wave,
2. A wave propagating In a direction inclined at an angle of 45° with both the positive x and y-axes? In each case, find out the positions where the resultant intensity would always be zero.

Solution:

1. The first and the second waves are two identical but oppositely directed waves. So, they would generate a stationary wave.

The equation of the resultant wave would be

z = \(z_1+z_2 =A[\cos (k x-\omega t)+\cos (k x+\omega t)]\)

= \(2 A \cos k x \cdot \cos \omega t\)

The resultant intensity is zero, where 2Acos kx = 0

∴ coskx = 0 or, x = \(\frac{(2 n+1) \pi}{2 k}\{n=0,1,2,3, \ldots\}\)

2. The first wave directed along the positive x-axis and the third wave directed along the positive y-axis are identical waves. So the resultant wave propagates in a direction that Is Inclined at 45″ with both the x and the yaxes. The equation of the resultant wave would be

z = \(z_1+z_3=A[\cos (k x-\omega t)+\cos (k y-\omega t)]\)

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

Tire resultant intensity is zero, where \(2 A \cos \frac{k(x-y)}{2}=0\)

∴ \(\cos \frac{k(x-y)}{2}=0\)

or, \(\frac{k(x-y)}{2}=\frac{(2 n+1) \pi}{2}[n=0,1,2,3, \ldots]\)

or, \(x-y=\frac{(2 n+1) \pi}{k}\) .