Class 11 Physics

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.

WBBSE Class 11 Superposition of Waves Notes

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

Short Answer Questions on Superposition of Waves

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.

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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°.

Understanding Superposition Principle in Waves

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).

Real-Life Examples of Wave Superposition

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 =  …….

Constructive and Destructive Interference Explained

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

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

Mathematical Representation of Wave Superposition

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.

Applications of Superposition of Waves

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, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}, \cdots\), the amplitude is  A’ = ±2A = maximum. These antinodes of the stationary’ wave.
2. At x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \cdots\), 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

WBBSE Class 11 Vibration of Strings Overview

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.

Vibrating Air Columns: Key Concepts

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.

Fundamental Frequency of a Vibrating String

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, \(n_1=\frac{V}{\lambda_1}=3 \cdot \frac{V}{4 l}\)

Applications of String Vibration in Musical Instruments

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.

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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

Harmonics in Vibrating Strings

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:

Mathematical Formulas for Vibration of Strings

Given, the velocity of sound in air = 320 m · s^-1

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:

Given, the velocity of sound in air =332 m · s^-1

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:

Given

The length of an organ pipe, closed at one end is 90 cm.

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:

Given

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

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:

Given, the velocity of sound in air = 350 m s^-1.

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:

Given

A 20 cm long closed pipe emits a tone of frequency 400Hz.

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:

Given

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,

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:

Given

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.

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

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

Comparison of Open and Closed Air Columns

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:

Given

In winter, the frequency of a tone emitted at 10°C by an open organ tube is 400 Hz

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:

Given

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.

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.

Factors Affecting Frequency of Vibrating Strings

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.

Understanding Vibrating Strings in Physics

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)

WBBSE Class 11 Vibration of Strings and Air Columns Notes

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)

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Fundamental Frequency of a Vibrating String

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.

Harmonics in Vibrating Strings

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.

Mathematical problems for Vibration Frequencies

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

Air Column Vibrations in Musical Instruments problems

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

WBBSE Class 11 Superposition of Waves Overview

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.

Principle Of Superposition

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

Definition of Superposition of Waves for Class 11

Superposition Of Waves

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

Examples of Superposition of Waves in Real Life

• 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

Mathematical Representation of Superposition of Waves

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:

Given

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.

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.

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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:

Given

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.

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:

Given

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.

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:

Given

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.

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:

Given

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.

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

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

WBBSE Class 11 Revision Notes on Superposition

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:

Given

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.

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:

Given

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.

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:

Given

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.

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:

Given

Two waves from 10 beats in 3s in a gas, The wavelength are  1m and 1.01m, respectively.

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:

Given

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,

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:

Given

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,

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:

Given

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.

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:

Given

Two progressive waves y₁ = 4 sin 500πt and y₂ = 2 sin506π t are supported.

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:

Given

Three transverse progressive waves are x₁ = Acos(kx-ωt), x₂  = Acos(kx + ωt) , x₃ = Acos (ky-ωt).

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

Superposition Of Waves Long Answer Type Questions

Question 1. When you sing in the shower why is the sound heard so good?
Answer:

It is because of superposition. The sound waves you produce in the air bounce off the walls and interfere with each other in a way that produces a sweet sound.

Question 2. Which from an of a string the tone clamped will at be both absent ends, in the when note it is emitted strode at a distance of one-third of its length from one end?
Answer:

The stretched string AB is struck at the point C, where AC = \(\frac{1}{3}\)AB. As a result, an antinode will be formed at C; a node can never be formed there. So the modes of vibration, as shown in the figure, will be absent. This means that the absent ones would correspond to the 3rd, 6th 9th,… harmonics.

Question 3. What would be the change in the fundamental frequency of a stringed instrument if

1. The length of the string is doubled,
2. The tension is doubled,
3. The diameter is doubled?

Answer:

1. The frequency is inversely proportional to the length of the string. So, if the length is doubled, the fundamental frequency would be reduced to half.
2. Frequency is proportional to the square root of the tension. So, if the tension is doubled, the fundamental frequency would be √2. or 1.414 times of its initial value.
3. If the diameter is doubled, the area of the cross-section becomes 22 or 4 times the initial value. If the materials of the string is the same, the mass per unit length also becomes 4 times its original value. The frequency is inversely proportional to the square root of the mass per unit length. So, in this case, the frequency would be \(\frac{1}{\sqrt{4}} \text { or } \frac{1}{2}\) times, i.e., half the initial value.

Question 4. Compare the fundamental frequencies of the tones emitted by two pipes of equal length, one open at both ends and the other closed at one end.
Answer:

Let V be the velocity of sound in air and l, be the length of each of the pipes.

∴ The frequency of the fundamental tone emitted by the pipe open at both ends, \(n_1=\frac{V}{2 l}\)

and the frequency of the fundamental tone emitted by the pipe closed at one end, \( n_2=\frac {V}{4 l}.\)

∴ \(\frac{n_1}{n_2}=\frac{\frac{V}{2 l}}{\frac{V}{4 l}}=2 \quad \text { or, } n_1=2 n_2\)

So, the fundamental frequency for a pipe open at both ends is twice that of a pipe of the same length but closed at one end.

WBBSE Class 11 Superposition of Waves Q&A

Example 5. A closed pipe and an open pipe emit fundamental tones of the same frequency. Find out the ratio of their lengths.
Answer:

Let the length of the closed pipe and the open pipe be l₁ and l₂, respectively. V is the velocity of sound in air.

For the closed pipe, the fundamental frequency is \(n_1=\frac{V}{4 l_1}\); for the open pipe, it is \(n_2=\frac{V}{2 l_2}\)

In the problem, \(n_1=n_2 \text { or, } \frac{V}{4 l_1}=\frac{V}{2 l_2} \text { or, } \frac{l_1}{l_2}=\frac{1}{2}\)

Therefore, the ratio of their lengths is  1: 2

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Question 6. Would the frequencies of the tones emitted from a closed or an open pipe change If the temperature of the air column in the pipe increases?
Answer:

The frequencies of the tones are proportional to the velocity of sound in air, which increases with the increase of temperature. So, as the temperature increases, the frequencies of the tones emitted from the air column in the pipe would also increase.

Question 7. How would the fundamental frequency emitted from an organ pipe, open at both ends, change, If

1. An open end is suddenly closed,
2. The length of the pipe is increased,
3. The diameter of the pipe is increased?
4. What would happen if the air is blown heavily through an open end?

Answer:

1. If an open end is suddenly closed, the pipe open at both ends becomes a closed pipe of equal length. In this case, the fundamental frequency would be halved.
2. The frequency of the emitted fundamental tone is inversely proportional to the length of the pipe. So, the frequency would decrease with the increase in the length of the pipe.
3. The end error of a pipe increases with the increase in diameter. As a result, the effective length of the pipe increases. The fundamental frequency is inversely proportional to the length of the pipe; so the frequency would decrease due to this increase in effective length.
4. A heavier blow would increase the loudness of the emitted tones. Moreover, the hither (Le, 2nd, 3rd, 4th,…) harmonics would be. formed more easily.

Question 8. Why is the musical sound emitted from an open pipe more pleasant than that emitted from a closed pipe?
Answer:

A dosed pipe can emit the fundamental rone and its odd harmonics. But an open pipe can emit both the odd and the even harmonics as well as the fundamental tone. So, a note emitted from, an open pipe contains a larger number of constituent tones.

The quality of a note depends on the number of overtones present in it. Store the number of overtones present in a note, the more pleasant it sounds. This results in a higher quality of sound emitted from an open pipe.

Question 9. A stretched vibrating string is touched at a distance of 1/3 rd of its length from one end. What would happen to the musical sound emitted?
Answer:

The point of touch would be a nodal point. So only the 3rd, 6th. 9th…. harmonics, which have a node at 1/3 would be emitted. No other harmonics have a node at 1/3. So, all of them would be suppressed.

Mathematical Problems in Superposition of Waves

Question 10. Two tuning forks, vibrating simultaneously, produce 6 beats per second. The first of them has a frequency of 312 Hz. Some amount of wax is added to one arm of the second tuning fork; the number of beats per second reduces to 3. Find out the frequency of this second tuning fork. Is it possible to increase the beat frequency to 6 per second by adding some more wax to die second tuning fork?
Answer:

Initially, the number of beats per second is 6. So the frequency of the 2nd fork is either 312-*-6 = 318 Hz or 312 – 6 = 306 Hz. After putting the wax, the beat frequency is 3 per second.

• So the changed frequency of the 2nd fork is, either 312÷3=315Hz or 312 – 3 = 309 Hz. But the addition of wax reduces the frequency of the 2nd fork; so it cannot change from 306 Hz to 309 Hz; the actual change is from 318 Hz to 315 Hz. This means that the actual frequency of the 2nd tuning fork is 318 Hz.
• By adding a sufficient amount of wax on the 2nd fork, its frequency can be reduced from 318 Hz to 306 Hz. Then again the beat frequency would be 312 – 306 = 6 per second.

Question 11. Two Identical wires of equal lengths are stretched In such a way that their simultaneous vibrations produce 6 beats per second. The tension In one of the wires is changed slightly and it is observed that the beat frequency remains the same. How is it possible?
Answer:

The frequency of a stretched string is proportional to the square root of the tension. If the fundamental frequency in the 1st wire is more than that in the 2nd wire, the tension in the 1st ware (T₁) > the tension in the 2nd wire (T₂).

Then, n₁ = n₂ ÷ 6, where n₁ and n₂ are the fundamental frequencies of the two wires respectively. Now, the tension T₂ is increased gradually until the fundamental frequency of the 2nd wire changes to n₂‘ = n₂ ÷ 12. Under these circumstances, the frequency difference becomes, \(n_2^{\prime}-n_1=\left(n_2+12\right)-\left(n_2+6\right)=6\)

This means that 6 beats would be heard again per second.

Question 12. Three tuning forks of frequencies n-x,n, and n + x are vibrated simultaneously. If the amplitudes of vibration are equal, show that the forks would form beats.
Answer:

Let A be the amplitude of the vibration of each tuning fork.

Then, the equations of the waves are y₁ = Asin2π(n- x)t, y₂ = Asin2πnt if and y₃ = Asin2π(n + x)t.

∴ The equation of the resultant wave is

y = \(y_1+y_2+y_3\)

= \(A[\sin 2 \pi(n-x) t+\sin 2 \pi n t+\sin 2 \pi(n+x) t]\)

= \(A[\{\sin 2 \pi(n+x) t+\sin 2 \pi(n-x) t\}+\sin 2 \pi n t]\)

= \(A[2 \sin 2 \pi n t \cos 2 \pi x t+\sin 2 \pi n t]\)

= \(A(1+2 \cos 2 \pi x t) \sin 2 \pi n t\)

∴ The amplitude of the resultant wave = A(1 + 2 cos2πxt)

This amplitude clearly depends on time. As a result, the intensity of the emitted sound would also increase and decrease periodically with time [intensity ∝ (amplitude)²]. Thus beats are produced. So, beats can be produced not only due to the superposition of two waves but also due to the superposition of more than two waves.

Short Notes on Wave Superposition Concepts

Question 13. Three sources emitting sound waves of the same amplitude have frequencies 400 Hz, 401 Hz, and 402 Hz, respectively. Find out the number of beats heard per second.
Answer:

⇒ \(y_1=A \sin (2 \pi \cdot 400 t) ; y_2=A \sin (2 \pi \cdot 401 t)\)

⇒ \(y_3=A \sin (2 \pi \cdot 402 t)\)

∴ y = \(y_1+y_2+y_3\)

= \(A[\sin (2 \pi \cdot 400 t)+\sin (2 \pi \cdot 401 t)+\sin (2 \pi \cdot 402 t)\} \)

= \(A\left[\sin (2 \pi \cdot 401 t)+2 \sin \left\{2 \pi \frac{402+400}{2} t\right\}\right.\)

. \(\left.\cdot \cos \left\{2 \pi \frac{402-400}{2} t\right\}\right]\)

= \(A[\sin (2 \pi \cdot 401 t)+2 \sin (2 \pi \cdot 401 t) \cos 2 \pi t]\)

= \(A(1+2 \cos 2 \pi t) \sin (2 \pi \cdot 401 t)\)

Here, the amplitude of the resultant wave, \(A^{\prime}=A(1+\cos 2 \pi t).\)

This is maximum when \(\cos 2 \pi t=1=\cos 2 n \pi[n=0,1,2,3, \cdots]\)

or, t = \(n=0,1,2, \cdots\)

Then, the time interval between two consecutive maxima is Is; so the beat frequency = 1 per second.

Question 14. A 100 cm long stretched string is struck at a point 25 cm from one of its ends. Which of the overtones would be absent in the emitted note?
Answer:

At the striking point, an antinode would be formed. So, the overtones having a node at the point of 25 cm would not be formed. As 25 cm = \(\frac{100 \mathrm{~cm}}{4}=\frac{1}{4}\)(l = length of the stretched string), the absent overtones would be the 4th, 8th, 12th, …. harmonics.

Question 15. In a particular vibrating mode of a stretched string of length l clamped at both ends, n nodes are formed. What is the wavelength of the stationary wave formed in this mode?
Answer:

Both endpoints are nodes. So, the number of loops formed between the n nodes = n-1. Every loop has a length = \(\frac{\lambda}{2} \text {. So, } \frac{l}{n-1}=\frac{\lambda}{2} \text { or, } \lambda=\frac{2 l}{n-1} \text {. }\)

Question 16. The equation of two progressive waves superposing on a string is y₁ = Asin[k(x- ct)] and y₂ = Asin[k(x+ ct)]. What is the distance between two consecutive nodes?
Answer:

Here, \(y_1 \text { or } y_2=A \sin [k(x \mp c t)]\)

The general form of the two waves, y = \(A \sin \frac{2 \pi}{\lambda}(x \mp V t)\)

So, k = \(\frac{2 \pi}{\lambda} \text { or, } \lambda=\frac{2 \pi}{k}\)

∴ The distance between two consecutive nodes = \(\frac{\lambda}{2}=\frac{\pi}{k}\).

Real-Life Examples of Superimposed Waves

Question 17. Two waves represented as y₁ = A₁ sin ωt and y₂ = A₂ cosωt superpose at a point in space. Find out the amplitude of the resultant wave at that point.
Answer:

⇒ \(y_1=A_1 \sin \omega t ; y_2=A_2 \cos \omega t=A_2 \sin \left(\omega t+\frac{\pi}{2}\right) \text {. }\)

So, the phase difference between the two waves = \(\frac{\pi}{2}\).

Then, the amplitude of the resultant wave is A = \(\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \frac{\pi}{2}}=\sqrt{A_1^2+A_2^2} .\)

Question 18. The length of a stretched string between two rigid supports is 40 cm. What would be the maximum length of a stationary wave that can be formed in the string?
Answer:

Let the maximum wavelength that can be formed for the given string be λ. For the stationary wave of maximum length, only two nodes are present and they are at the two ends of the string. There is only one antinode between them. So the string vibrates in a single loop. Then the length of this loop = \(\frac{\lambda}{2}\) = 40 cm; so, λ = 80 cm.

Question 19. The equation of a transverse progressive wave is y = 0.02sin(x+40t) m. Find out the tension In a wire of linear density 10^-4 kg · m^-1, if the wave travels along it.
Answer:

Comparing the given equation with the standard form y = A sin(ωt+ kx), we have k = 1 and ω = 40.

So, the wave velocity, V = \(\frac{\omega}{k}=\frac{40}{1}=40 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

If T = tension and m = linear density of the wire,

V = \(\sqrt{\frac{T}{m}} \text { or, } T=V^2 m=(40)^2 \times 10^{-4}=0.16 \mathrm{~N}\)

Question 20. The superposition of two progressive waves produces a stationary wave represented as y = Acos(0.01x)sin(100t) m. What is the velocity I of the two-component waves?
Answer:

Two progressive waves represented as y₁ = A sin(ωt- kx) and y₂ = A sin(ωt+ kx) superpose to form a stationary wave y = y₁ + y₂ = 2A cos kx sin ωt.

Comparing the given equation with this standard form, we have k = 0.01 m^-1 and ω = 100 Hz. So the wave velocity,

V = \(\frac{\omega}{k}=\frac{100}{0.01}=10^4 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

Step-by-Step Solutions to Superposition Questions

Example 21. A uniform wire has length I and area of cross-section α and the density of its material is ρ. If the wire is stretched with a tension T, what would be the velocity of a transverse wave traveling along It?
Answer:

Volume of unit length of the wire = unit length x area of cross-section = 1 x α = α; then linear density = mass per unit length, m = ρα.

∴ Velocity of a transverse wave, V = \(\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\rho \alpha}}\)

 Superposition Of Waves Multiple Choice Question And Answers

Question 1. A wave y= asin(ωt- kx) being superposed with another wave produces a node at x = 0. The equation of the second wave should be

1. y = asin(ωt+kx)
2. y = -asin(ωt+kx)
3. y= asin(ωt-kx)
4. y = -asin(ωt- kx)

Answer: 2. y = -asin(ωt+kx)

Question 2. At the instant when all the particles in the medium of a stationary wave are at their equilibrium positions the

1. Kinetic Energy Becomes Zero
2. Potential Energy Becomes Zero
3. Net Energy Becomes Zero
4. None Of These Becomes Zero

Answer: 2. Potential Energy Becomes Zero

Question 3. Which of the following remains constant in case of vibration of the particles in a stationary wave?

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

Answer:  3. Amplitude

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. When two identical sound waves superpose at a point with a phase difference of 180°,

1. The point will be a node
2. The intensity of the sound will increase at that point
3. The point will be an antinode
4. Beats will be heard

Answer: 1. The point will be a node

Question 5. The equation of a progressive wave is y = A cos (kx-ωt). This wave superposes with another wave and produces a stationary wave by creating a node at x = 0. Which of the following equations is correct for the other wave?

1. Asin(kx+ωt)
2. -Asin(kx-ωt)
3. -Acos(kx+ωt)
4. -Asin(kx-ωt)

Answer: 3. -Acos(kx+ωt)

WBBSE Class 11 Superposition of Waves MCQs

Question 6. A fundamental tone of frequency n₁ is produced when a string stretched at both ends is vibrated. The frequency of the fundamental tone changes to n₂ when the tension in the string is doubled. What is the ratio of n₁ and n₂?

1. 1: 2
2. 2: 1
3. 1:√2
4. √2: 1

Answer: 3. 1: √2

Question 7. The waves generated due to the vibration of an air column in a pipe open at one end or a pipe open at both ends are

1. Transverse progressive waves
2. Transverse stationary waves
3. Longitudinal progressive waves
4. Longitudinal stationary waves

Answer: Longitudinal stationary waves

Question 8. When a fundamental tone is produced from a pipe of length l, open at both ends, the wavelength of the stationary wave is

1. \(\frac{l}{2}\)
2. l
3. 2l
4. 4l

Answer: 3. 2l

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Question 9. If the frequency of the fundamental tone emitted from a pipe closed at one end is 200 Hz, then the frequencies of the first three overtones will be

1. 400 Hz, 600 Hz, 800 Hz
2. 600 Hz, 1000 Hz, 1400 Hz
3. 400 Hz, 800 Hz, 1200 Hz
4. 600 Hz, 800 Hz, 1000 Hz

Answer: 2. 600 Hz, 1000 Hz, 1400 Hz

Question 10. A fundamental tone is emitted from a pipe closed at one end. If the closed end is suddenly opened,

1. The pitch of the tone decreases
2. The intensity of the tone decreases
3. The pitch of the tone increases
4. The intensity of the tone increases

Answer: 3. The pitch of the tone increases

Constructive and Destructive Interference MCQs

Question 11. When a fundamental tone is produced from a pipe of length l, closed at one end, the wavelength of the stationary wave is

1. \(\frac{l}{2}\)
2. l
3. 2l
4. 4l

Answer: 4. 4l

Question 12. Due to the end error in stationary waves produced in a closed or an open pipe

1. Antinodes are not formed at the closed end
2. Antinodes are formed inside the pipe very close to its open end.
3. Nodes are formed slightly outside the open end of the pipe
4. None of these

Answer: 3. Nodes are formed slightly outside the open end of the pipe

Question 13. When the temperature of the air in a closed or so open pipe increases, the frequency of stationary wave will

1. Remain the same
2. Increase
3. Decrease
4. Depending on other factors also

Answer: 2. Increase

Question 14. The frequencies of the fundamental tones in a dosed and an open, pipe are the state The talks of the lengths of the two pipes is

1. 1:1
2. 1:2
3. 2:1
4. 4:1

Answer: 2. 1:2

Question 15. While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, he measures the column length to be x cm for the second resonance. Then,

1. 18 > x
2. x > 54
3. 54 > x > 36
4. 36 > x > 18

Answer: 2. x > 54

Question 16. A student is performing the experiment of the resonance column. The diameter of the column tube is 4cm The frequency of the turning fork is 512HZ. The air temperature is  38C in which the speed of sound is 336m/s. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is

1. 14 cm
2. 15.2 cm
3. 16.4cm
4. 17.6cm

Answer: 2. 15.2 cm

Mathematical Problems for Wave Superposition

Question 17. A student is performing an experiment using a resonance column and a turning fort of frequency 244 s^-1. He is told that the air in the tube has been replaced by another (assume the fear fee column remains filled with gas). If the minimum height at which resonance occurs is (0.350 ± 0.005) m, the gas in tube is (using information: √167RT = \(640 \mathrm{~J}^{1 / 2} \cdot \mathrm{mol}^{-1 / 2}; \sqrt{140 R T}=590 \mathrm{~J}^{1 / 2} \cdot \mathrm{mol}\). The molar masses M in grams are given in fee options. Take the value of \(\sqrt{\frac{10}{M}}\) for each gas as given there)

1. Neon \(\left(M=20, \sqrt{\frac{10}{20}}=\frac{7}{10}\right)\)
2. Nitrogen \(\left(M=28, \sqrt{\frac{10}{28}}=\frac{3}{5}\right)\)
3. Oxygen \(\left(M=32, \sqrt{\frac{10}{32}}=\frac{9}{16}\right)\)
4. Argon \(\left(M=36, \sqrt{\frac{10}{36}}=\frac{17}{32}\right.\)

Answer: 4. Argon \(\left(M=36, \sqrt{\frac{10}{36}}=\frac{17}{32}\right.\)

Question 18. What should be fee beat frequency so that it is not audible to the human ear?

1. Greater than 10 Hz
2. 10 Hz
3. 5 Hz
4. Less than 5 Hz

Answer: 1. Greater than 10 Hz

Question 19. If n₁ and n₂ are fee frequencies of two sound waves, the frequency of the beat produced due to the superposition of these waves will be

1. \(n_1-n_2\)
2. \(n_1+n_2\)
3. \(\frac{n_1+n_2}{2}\)
4. \(2\left(n_1-n_2\right)\)

Answer: 1. \(n_1-n_2\)

Question 20. Beats are not observed for light waves, because

1. There is no difference in the velocities of the two light waves
2. It is impossible to stabilize the frequency difference of two light waves below 10 Hz
3. It is impossible to keep the intensities of two light waves equal or nearly equal
4. None of these

Answer: 2. It is impossible to stabilize the frequency difference of two light waves below 10 Hz

Question 21. When two waves y₁ = A sin2008πt and y₂ = A sin2008πt are superposed, the number of beats produced per second is

1. 0
2. 1
3. 4
4. 8

Answer: 3. 4

Question 22. A set of 25 tuning fork is arranged in a series of decreasing frequencies such that each fork gives 3 beats with the succeeding one. The first fork is the octave of the last. Frequency of fee 10th fork is

1. 120Hz
2. 117Hz
3. 110Hz
4. 89Hz

Answer: 2. 117Hz

Short Answer Questions Related to Wave Superposition

Question 23. The displacement of a particle in a medium at an instant due to the effect of more than one wave is

1. Not dependent on the displacements due to the individual waves
2. Equal to the vector sum of fee displacements due to fee individual waves
3. Equal to fee displacement due to any one of the waves
4. Random due to the effect of all the waves

Answer: 2. Equal to fee vector sum of fee displacements due to fee individual waves

Question 24. A sound wave of frequency 500 Hz is advancing along the positive x-axis with a speed of 300 m · s^-1. The phase difference between the two points x₁ and x₂ is 60°. The least distance between those two points is

1. 1 mm
2. 1cm
3. 10cm
4. 1 m

Answer: 3. 10cm

Question 25. When two progressive waves y₁= 4sin(2x-6t) and \(y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)\) are superposed, the amplitude of the resultant wave is

1. 5
2. 6
3. \(\frac{5}{3}\)
4. \(\frac{1}{2}\)

Answer: 1. 5

Question 26. A hollow pipe of length 0.8 m is closed at one end. At its open end, a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 m • s^-1, the mass of the string is

1. 5g
2. 10 g
3. 20 g
4. 40 g

Answer: 2. 10 g

Question 27. A sound of 20dB is more intense than a sound of 10 dB by

1. 100
2. \(\frac{1}{10}\)
3. 10
4. 0.01

Answer: 3. 10

In this type of question, more than one option is correct.

Question 28. For a certain transverse standing wave on a long string, an antinode is formed at x = 0, and next to it, a node is formed at x = 0.10 m. The displacement y(t) of the string particle at x = 0 is shown.

1. The transverse displacement of the particle at x = 0.05 m and t = 0.05 s is -2√2 cm
2. Transverse displacement of the particle at x = 0.04 m and t = 0.025 s is -2 √2 cm
3. The speed of the traveling waves that interface to produce this standing wave is 2 m • s^-1
4. The transverse velocity of the string particle at x = \(\frac{1}{15}\)m and t = 0.1 s is 20π cm· s^-1

Answer:

1. The transverse displacement of the particle at x = 0.05 m and t = 0.05 s is -2√2 cm

3. The speed of the traveling waves that interface to produce this standing wave is 2 m • s^-1

4. The transverse velocity of the string particle at x = \(\frac{1}{15}\)m and t = 0.1 s is 20π cm· s^-1

Question 29. Following are equations of four waves

1. \(y_1=a \sin \omega\left(t-\frac{x}{v}\right)\)
2. \(y_2=a \cos \omega\left(t+\frac{x}{v}\right)\)
3. \(z_1=a \sin \omega\left(t-\frac{x}{v}\right)\)
4. \(z_2=a \cos \omega\left(t+\frac{x}{v}\right)\)

Which of the following statements is correct?

1. On superposition of waves (1) and (2), a traveling wave having amplitude a√2 will be formed
2. Superposition of waves (2) and (3) is not possible
3. On superposition of (1) and (2), a stationary wave having amplitude a√2 will be formed
4. On superposition of (3) and (4), a transverse stationary wave will be formed

Answer:

1. On the superposition of waves (1) and (2), a traveling wave having amplitude a√2 will be formed

4. On superposition of (3) and (4), a transverse stationary wave will be formed

Question 30. One end of a taut string of length 3m along the x-axis is fixed at x = 0. The speed of the waves in the string is 100 m · s^-1. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. Obtain the possible waveforms of this stationary wave.

1. \(y(t)=A \sin \frac{\pi x}{6} \cos \frac{50 \pi t}{3}\)
2. \(y(t)=A \sin \frac{\pi x}{3} \cos \frac{100 \pi t}{3}\)
3. \(y(t)=A \sin \frac{5 \pi x}{6} \cos \frac{250 \pi t}{3}\)
4. \(y(t)=A \sin \frac{5 \pi x}{2} \cos 250 \pi t\)

Answer:

1. \(y(t)=A \sin \frac{\pi x}{6} \cos \frac{50 \pi t}{3}\)

3. \(y(t)=A \sin \frac{5 \pi x}{6} \cos \frac{250 \pi t}{3}\)

4. \(y(t)=A \sin \frac{5 \pi x}{2} \cos 250 \pi t\)

Wave Motion Introduction

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

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

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

Understanding Wave Motion in Physics

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

Wave Motion Mechanical Waves

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

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

WBBSE Class 11 Wave Motion Notes

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

Characteristics Of A Mechanical Wave:

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

Wave Motion Transverse And Longitudinal Waves

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

There Are Two Types Of Simple Harmonic Waves

1. Transverse wave and
2. Longitudinal waves

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

Examples Of Transverse Waves:

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

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

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

Types of Waves: Transverse and Longitudinal

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

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

WBBSE Class 11 Wave Motion Notes

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

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

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

Characteristics of Mechanical Waves

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

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

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

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

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

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

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

Difference Between Transverse And Longitudinal Waves:

Wave Motion Other Classifications Of Waves

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

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

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

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

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

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

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Wave Motion Some Physical Terms Related To Waves

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

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

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

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

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

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

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

Applications of Wave Motion in Real Life

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

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

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

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

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

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

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

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

Relations Among The Physical Quantities

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

So, according to the definition of frequency,

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

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

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

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

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

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

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

i.e., wave velocity = frequency x wavelength

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

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

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

Wave Motion Numerical Examples

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

Wavelength, λ = 20 cm;

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

If n is the frequency, V = nλ

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

So, 5 waves are produced per second.

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

Wavelength of radio wave, λ = 300 m.

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

So, the velocity of radio waves,

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

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

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

Here V = 320 m · s^-1; n = 400

We know, V = nλ

∴ 320 = 400 x λ

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

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

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

Mathematical Problems for Wave Motion

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

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

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

10 = \(\frac{1}{2}\) x 980 x t²

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

The tuning fork completes 8 vibrations in that time,

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

Mathematical Problems for Wave Motion

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

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

Velocity of sound (V) = 330 m · s^-1

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

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

= 0.375 m = 37.5 cm.

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

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

There are 5 waves within 6 consecutive crests.

So, the length of these 5 waves = 30 cm

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

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

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

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

So, the velocity of sound in the medium,

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

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

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

∴ Velocity of sound in air,

V = nλ = 512 x = 341.33 m · s^-1

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

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

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

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

Wave Motion Equation Of A Travelling Or Progressive Wave Numerical Examples

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

The general equation of a progressive wave is,

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

The equation of the given wave is

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Wave velocity, V = 400 cm · s^-1.

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

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

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

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

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

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

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

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

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

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

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

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

Solution: Given the equation,

y = 10sinπ(t-0.002x) cm….(1)

General equation of a progressive wave,

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

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

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

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

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

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

3. Acceleration of a particle,

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

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

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

Given equation,

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

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

A = y₀; V= λn

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

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

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

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

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

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

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

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

Wavelength, λ =π= 3.14 m;

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

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

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

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

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

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

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

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

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

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

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

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

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

So, phase difference = θ₂-θ₁

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

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

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

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

Solution:

The phase of the given progressive wave,

θ = \(\frac{\pi}{2}(200 t-x)\) ….(1)

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

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

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

2. For two points separated by 8 cm,

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

∴ Phase difference = θ₂ – θ₁

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

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

So, phase difference = 0°.

2. The required phase change

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

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

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

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

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

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

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

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

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

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

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

So, two points 6 rn apart have a path difference of \(\frac{\lambda}{2}\).

∴ Phase difference = \(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2}=\pi \mathrm{rad}=180^{\circ}\)

So, the two points are in the opposite phase. For example, if a crest is formed at a point, a trough would be formed 6m apart. However, the magnitudes of the amplitude would be the same.

∴ Amplitude difference = 0.

Example 14. The bulk modulus and the density of steel are 80 and 8 times, respectively, of those of water. Calculate the speed of sound in steel. Given, the speed of sound in water = 1493 m · s^-1.
Solution:

The velocity of sound in an elastic medium is

V = \(\sqrt{\frac{E}{\rho}}\)

Where E = modulus of elasticity of the medium ρ = density of the medium

∴ Velocity of sound in water, \(V_w=\sqrt{\frac{E_w}{\rho_w}}\)

Velocity of sound in steel, \(V_w=\sqrt{\frac{E_w}{\rho_w}}\)

∴ \(\frac{V_s}{V_w}=\sqrt{\frac{E_s}{\rho_s} \cdot \frac{\rho_w}{E_w}}=\sqrt{\frac{E_s}{E_w} \times \frac{\rho_w}{\rho_s}}=\sqrt{80 \times \frac{1}{8}}=\sqrt{10}=3.162\)

∴ \(V_s=3.162 V_w=3.162 \times 1493=4721 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Example 15. A wave is represented by y = \(20 \cdot \sqrt{3} \sin \left(\frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\right) \mathrm{cm}\), Calculate the displacement of a particle at the position x = \(\frac{\lambda}{6}\) on the wave at time t = \(\frac{T}{3}\)
Solution:

Given equation, \(y=20 \cdot \sqrt{3} \sin \left(\frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\right)\)

Putting x = \(\frac{\lambda}{6}\) and \(t=\frac{T}{3}\), we get

Displacement, y = \(20 \cdot \sqrt{3} \sin \left(\frac{2 \pi T}{3 T}-\frac{2 \pi \lambda}{6 \lambda}\right)=20 \cdot \sqrt{3} \sin \left(\frac{2 \pi}{3}-\frac{\pi}{3}\right)\)

= \(20 \cdot \sqrt{3} \sin \frac{\pi}{3}=20 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2}=30 \mathrm{~cm}\)

Example 16. The equations of two waves are, \(y_1=0.30 \sin (314 t-1.57 x)\),
\(y_2=0.10 \sin (314 t-1.57 x+1.57)\) Determine the phase difference and the ratio of the intensities of the two waves.
Solution:

The phase of the first wave, θ₁ = 314t – 1.57x

Phase of the second wave, θ₂ = 314t – 1.57x +1.57

∴ The phase difference between the two waves

= \(\theta_2-\theta_1=1.57 \mathrm{rad}=1.57 \times \frac{180}{\pi}=90^{\circ}\)

We know that the intensity of a wave is proportional to the square of its amplitude.

∴ Ratio of the amplitudes of the two waves = \(\frac{A_1}{A_2}=\frac{0.30}{0.10}=\frac{3}{1}\)

∴ Ratio of the intensities of the two waves = \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{1}\)

Wave Motion Velocity Of Sound In Solid And Liquid Media

Sound propagates through solids and liquids as longitudinal elastic wave tike in gaseous media. But the velocities of sound in solid and liquid are different. If E is the modulus of elasticity of the medium and ρ is its density, Newton’s formula for the velocity of sound is given by,

c = \(\sqrt{\frac{E}{D}}\)…(1)

• The density of any solid or liquid is greater than that of a gas. Bin the modulus of elasticity of a solid or a liquid is much greater than the Thar of a gas.
• So, the velocity of sound in a solid or B. Liquid is greater than that in a gas. For example, the psVjcities of sound in iron and in water are about 15 times and 4.5 times that in air, respectively.
• The velocity of sound in iron is greater than that in air—it can be easily understood from a simple experiment. If one end of a long iron pipe is struck heavily, the sound of striking is heard twice at the other end of the pipe.
• Sound is heard for the first time due to its propagation through the iron and for the second time due to its propagation through the air in the pipe. An iron pipe nearly 100 m in length should be taken to hear the two sounds distinctly.

Principle of Superposition of Waves

There is one important difference between solids and liquids as media of propagation of sound. The modulus of elasticity for the solid is, Young’s modulus (Y) while it is the bulk modulus (k) in the case of the liquid. So the equation (1)

For solid, c = \(\sqrt{\frac{Y}{\rho}}\)….(2)

and for liquid, c = \(\sqrt{\frac{k}{\rho}}\) …(3)

In case of steel, Y = 2 x 10¹¹ N · m^-2; ρ = 7850 kg · m^-3;

From equation (2) we get,

c = \(\sqrt{\frac{2 \times 10^{11}}{7850}} \approx 5048 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

In case of water, k = 2.1 x 10⁹ N · m^-2; ρ = 1000 kg · m^-3

From equation (3) we get, c = \(\sqrt{\frac{2.1 \times 10^9}{1000}}=1449 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Velocity Of Sound In Solid And Liquid Media Numerical Examples

Example 1. Two explosions are made simultaneously from a ship, one above the surface of the water and another just below it. Sound is heard in a hydrophone placed below water from another ship 5 km apart 11s earlier than the sound reaching the deck of the ship. What is the velocity of sound in water? Given that the velocity of sound in air = 348 m · s^-1.
Solution:

Distance between the two ships = 5 km = 5000 m

So time taken by the sound to come through air = \(\frac{5000}{348} \mathrm{~s}\)

According to the question, time is taken by the sound to come through water = \(\frac{5000}{348}-11=\frac{5000-3828}{348}=\frac{1172}{348} \mathrm{~s}\)

Therefore, the velocity of sound in water

= \(\frac{5000}{\frac{1172}{348}}=1485 \mathrm{~m} \cdot \mathrm{s}^{-1}\) (approx.)

Example 2. If one end of a long steel pipe is struck, the sound is heard twice at an interval of 3 s at the other end. What is the length of the pipe? (The velocities of sound in air = 350m · s^-1, in steel = 5000 m · s^-1)
Solution:

Let the length of the pipe be l m.

So, time is taken by the sound to come through the air in the pipe = \(\frac{l}{350}\)s

Again, time taken by die sound to come through steel =\(\frac{l}{5000}\)s

According to the question, \(\frac{l}{350}-\frac{l}{5000}=3 \quad \text { or, } l\left(\frac{1}{350}-\frac{1}{5000}\right)=3\)

or, \(l\left(\frac{100-7}{35000}\right)=3\)

or, \(l=\frac{35000 \times 3}{93}=1129 \mathrm{~m}=1.129 \mathrm{~km} .\)

Example 3. Sound moves through a liquid with a velocity 1340 m · s^-1. If the density of the liquid is 0.8 g · cm^-3, determine the compressibility of the liquid.
Solution:

We know, c = \(\sqrt{\frac{k}{\rho}}\); k = bulk modulus of the liquid

∴ k = c²ρ [c = 1340 m · s^-1 = 134 x 10³ cm · s^-1]

Again, compressibility = \(\frac{1}{k}=\frac{1}{c^2 \rho}\)

= \(\frac{1}{\left(134 \times 10^3\right)^2 \times 0.8}=\frac{1}{143648} \times 10^{-5}\)

= \(6.96 \times 10^{-11} \mathrm{~cm}^2 \cdot \mathrm{dyn}^{-1}\)

Wave Motion Musical Sound And Noise

Musical sound: The sound produced due to regular and periodic vibrations, that is pleasing to hear, is called musical sound. While singing, the vocal cords of a singer vibrate with regular periodic motion.

• Similarly when die strings of a sitar are excited to produce a certain note, diey vibrate with regular periodic motion. The sound of clapping is usually unpleasant, but rhythmic clapping of hands may produce a pleasant sound.
• This is due to the regular vibration of the source. Similarly pattern of raindrops on tin roofs or the clickers’ sound of train wheels may be sonorous.

Noise: The sound produced by an irregular or short-lived vibration is called noise. The explosion of a cracker, the firing of a gun, the horn of a car, etc., are noises.

Tone: A musical sound of a single frequency is called a tone. When the source of sound vibrates in SHM, it produces a tone. The two prongs of a tuning fork vibrate in SHM and produce the sound of a single frequency. Thus, the sound of a tuning fork is a pure tone.

Tone note: A musical sound due to a mixture of more than one frequency is called a note. The sound produced by different musical instruments consists, in general, of more than one tone. If a tone is compared with monochromatic light such as red light, a note can be compared with compound light, i.e., white light. The tones within a note are classified on the basis of their frequencies.

• Fundamental Tone: The tone with the lowest frequency present in a note is called the fundamental tone.
• Overtones: The tones in a note, other than the fundamental tone, are called overtones.
• Harmonics: Harmonics are tones having frequencies that are integral multiples of that of the fundamental tone. The fundamental tone is also a harmonic.
• Octave: A tone whose frequency is twice that of the fundamental tone is said to be the octave of the fundamental tone.

Standing Waves and Harmonics

Suppose, a musical note consists of tones of frequencies 200 Hz, 300 Hz, 400 Hz, 500 Hz, and 600 Hz. So it can be said:

1. Each of these five different frequencies corresponds to a tone,
2. The combination of these five tones is a note,
3. The tone of 200 Hz is die fundamental tone,
4. The tones having frequencies 300 Hz, 400 Hz, 500 Hz, and 600 Hz are overtones,
5. Both of the tones of frequencies 400 Hz and 600 Hz are harmonics, as the frequencies are 2 and 3 umes that of the fundamental tone respectively. The tone of 400 Hz is called the second harmonic as its frequency is 2 times of the fundamental tone. Similarly, the tone of 600 Hz is called the third harmonic. The fundamental tone is called the first harmonic.
6. The tone of 400 Hz is the octave because this frequency is the mice die frequency of the fundamental tone.

Characteristics Of Musical Sound Or Note: There are three characteristics that differentiate musical notes. These are loudness, pitch, and quality.

1. Loudness: Loudness Is related to the sound energy reaching our ears per unit time. If the sound energy reaching our ears in 1 second goes up, we perceive a corresponding increase in the loudness.
2. Pitch: It is that characteristic that differentiates a sharp or shrill sound from a grave one. It increases with the Increase in frequency of the source.
3. Quality Or Timbre: It is that characteristic of a musical note that enables us to distinguish between a note emitted by one musical instrument from a note of the same loudness and pitch emitted by another instrument. It depends on the number of overtones present in a note and their relation with the fundamental tone.

Wave Motion Conclusion

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

• The disturbance that travels through a material medium due to a collective vibration of the particles of the medium is known as a mechanical wave.
• It is an elastic wave if the particles vibrate due to the elasticity of the medium. If the motion of the particles of a medium is simple harmonic, the corresponding wave is known as a simple harmonic wave.
• A wave, that propagates in a direction perpendicular to the direction of motion of the vibrating particles of the medium, is called a transverse wave.
• A wave, that propagates along the direction of motion of the vibrating particles of the medium, is called a longitudinal wave.
• All longitudinal waves are mechanical waves. Longitudinal waves can be transmitted through any solid, liquid, or gaseous medium. This type of wave cannot propagate without any medium.
• Transverse elastic waves cannot be produced in liquids or gases but can be created in solids.

Some physical quantities related to waves:

1. Phase: The quantity from which the motion of a wave, except the amplitude, can be known completely is called the phase of the wave.

2. Complete Wave: The wave in between two consecutive particles having the same phase at an instant is known as a complete wave.

3. Wavelength: The length of a complete wave, i.e., the distance between two consecutive particles having the same phase at an instant is called the wavelength of the wave.

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

5. Frequency: The frequency of a wave is the number of complete waves produced in unit time.

6. Amplitude: The amplitude of a wave is the maximum displacement from the mean position of a particle-producing tire wave.

7. Wave velocity: The distance traversed by a wave in unit time is called the wave velocity.

8. Wavefront: A surface normal to the direction of propagation of a wave is known as a wavefront. Particles lying on a wavefront have the same phase.

9. Ray: A normal drawn on a wavefront is called a ray.

The velocity with which sound wave propagates in a material medium depends on two properties of the medium density, and elasticity.

Characteristics of Wave Motion

• Newton assumed that the propagation of sound through a gaseous medium is an isothermal process.
• According to Laplace, the propagation of sound through a gaseous medium is an adiabatic process.
• The pressure of a gas has no effect on the velocity of sound.
• The velocity of sound in a gas is directly proportional to the square root of its absolute temperature.
• The velocity of sound in air increases by 0.61 m · s^-1 or 61 cm · s^-1 for 1°C rise in temperature.
• The velocity of sound in moist air is greater than that in dry air.
• The velocity of sound in a gas is inversely proportional to the square root of its density.
• To obtain regular reflection of sound, the reflector must be large but the surface of the reflector need not be very smooth.
• If the reflected sound is heard separately from an original sound, it is called an echo of the original sound.
• The sensation of an inarticulate sound (sound produced by gunshot, clapping, etc.) persists in our ear for about \(\frac{1}{10}\)th of a second. This is known as the persistence of hearing.
• The velocity of sound in air at 0°C is about 330 m · s^-1.

Wave Motion Useful Relations for Solving Numerical Problems

Relation between time period (T) and frequency (n): T = \(\frac{1}{n}\)

Relation between wavelength (λ) and wave number (k): \(\lambda=\frac{2 \pi}{k}\)

Relation between frequency (n) and angular frequency ω:ω = 2πn

Wave velocity (V) = frequency (n) x wavelength (λ)

Equation of a progressive wave moving along the positive and negative direction of the x-axis:

y = \(A \sin \omega\left(t \mp \frac{x}{V}\right)=A \sin (\omega t \mp k x)\)

= \(A \sin \frac{2 \pi}{\lambda}(V t \mp x)=A \sin 2 \pi\left(\frac{t}{T} \mp \frac{x}{\lambda}\right)\)

If The equation of a progressive wave can be expressed using cosine function also.

Physical quantities related to the motion of a particle in a progressive wave:

\(\begin{array}{|l|l|}
\hline \text { Displacement } & y=A \sin (\omega t-k x \pm \phi) \\
\hline \text { Velocity } & \nu=\frac{d y}{d t}=\omega A \cos (\omega t-k x \pm \phi) \\
\hline \text { Max. velocity } & \pm \omega A \\
\hline \text { Acceleration } & a=\frac{d^2 y}{d t^2}=-\omega^2 A \sin (\omega t-k x \pm \phi) \\
\hline \text { Max, acceleration } & \pm \omega^2 A \\
\hline
\end{array}\)

 

If the particle velocity is v and the wave velocity is V, then \(\nu=-V \frac{\partial y}{\partial x}\)

Velocity of Progressive waves in different media:

\(\begin{array}{|l|l|l|}
\hline \begin{array}{l}
\text { Velocity of a long- } \\
\text { tidal wave in a } \\
\text { solid medium }
\end{array} & V=\sqrt{\frac{Y}{\rho}} & \begin{array}{l}
Y=\text { Young’s modulus of } \\
\text { the medium and } \rho= \\
\text { density of the medium }
\end{array} \\
\hline \begin{array}{l}
\text { Velocity of a long- } \\
\text { tidal wave in a } \\
\text { liquid or a gaseous } \\
\text { medium }
\end{array} & V=\sqrt{\frac{E}{\rho}} & \begin{array}{l}
E=\text { bulk modulus of the } \\
\text { medium and } \rho=\text { density } \\
\text { of the medium }
\end{array} \\
\hline
\end{array}\) \(\begin{array}{|c|c|c|}
\hline \begin{array}{l}
\text { Velocity of sound } \\
\text { wave in a gaseous } \\
\text { medium }
\end{array} & V=\sqrt{\frac{\gamma p}{\rho}} & \begin{array}{l}
p=\text { pressure of the gas, } \\
\gamma=c_p / c_\nu=\text { ratio of the } \\
\text { two specific heats of the } \\
\text { gas and } \rho=\text { density of } \\
\text { the gas }
\end{array} \\
\hline \begin{array}{l}
\text { Velocity of a trans- } \\
\text { verse wave in a } \\
\text { stretched string }
\end{array} & V=\sqrt{\frac{T}{m}} & \begin{array}{l}
T=\text { tension in the string } \\
\text { and } m=\text { mass per unit } \\
\text { length of the string }
\end{array} \\
\hline
\end{array}\)

 

If M is the mass of 1 mol of a gas, then c = \(\sqrt{\frac{\gamma R T}{M}}\)

where T = absolute temperature of the gas,

R = universal gas constant

If c₀ and c are the velocities of sound at 0°C and t °C respectively, then c = c₀(1+ 0.001830 t)

If the densities of two different gases at the same temperature and pressure are ρ₁ and ρ₂ and the velocities of sound in the two gases are c₁ and c₂ respectively, then \(\frac{c_1}{c_2}=\sqrt{\frac{\rho_2}{\rho_1}}\)

If the distance of a reflector from the source of sound is D, the time required to hear the echo is t, and the velocity of sound is V, then, 2D = Vt

Wave Motion Very Short Answer Type Questions

Question 1. What kind of energy is transmitted through an elastic wave?
Answer: Mechanical energy

Question 2. Give an example of an elastic wave.
Answer: Soundwave

Question 3. Is It possible for a transverse wave to propagate in a liquid?
Answer: No

Question 4. What kind of wave is an X-ray?
Answer: Electromagnetic

Question 5. If the direction of motion of a wave and the direction of vibration of the particles of a medium are parallel to each other, the wave is called a transverse wave.
Answer: False

Question 6. If the direction of motion of a wave and the direction of vibration of the particles of a medium are perpendicular to each other, the wave is called a ______ wave.
Answer: Transverse

Question 7. During propagation of sound, compressions and rarefactions occur so rapidly that the _______ of the gas cannot remain constant.
Answer: Temperature

Question 8. If λ is the wavelength of a transverse wave, what will be the distance between two consecutive crests?
Answer: λ

Question 9. If λ is the wavelength of a longitudinal wave, what will be the length of a compression?
Answer: \(\frac{\lambda}{2}\)

Question 10. If the velocity of sound is 330 m · s^-1, what will be the wavelength of the sound wave emitted from a tuning fork of frequency 220 Hz?
Answer: 1.5 m

Question 11. The time required to form a complete wave is called the time period of the wave. Is the statement true or false?
Answer: True

Question 12. What is the number of complete waves formed in unit time called?
Answer: Frequency

Question 13. The maximum displacement of a particle on the wave from its mean position is called the _______ of the wave.
Answer: Amplitude

Question 14. The surface of vibration of the particles of a medium having the same phase during the propagation of a wave is called the ________ of the wave.
Answer: Wavefront

Question 15. The frequency of a progressive wave x = velocity of the wave.
Answer: Wavelength

Question 16. The equation of a progressive wave is given by y = Acos\(\frac{2 \pi}{\lambda}\)(Vt-x). Find the frequency of the wave.
Answer: \(\frac{V}{\lambda}\)

Question 17. The equation of a progressive wave is given by y = \(4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the amplitude of the wave.
Answer: 4cm

Question 18. The equation of a progressive wave is given by y = \(4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the frequency of the wave.
Answer: 10 Hz

Question 19. The equation of a progressive wave is given by \(y=4 \sin 20 \pi\left(t-\frac{x}{100}\right) \mathrm{cm}\). Find the velocity of the wave.
Answer: 100 cm · s^-1

Question 20. Why is the sound heard in CO₂ more intense in comparison to the sound heard in the air?
Answer: Density more

Question 21. By what process does the propagation of sound through a gaseous medium take place?
Answer: Adiabatic

Question 22. What is the ratio of the velocities of sound through hydrogen and oxygen at STP?
Answer: 4: 1

Question 23. If the temperature remains constant, what will be the bulk modulus of a gas?
Answer: Equal to the pressure of the gas

Question 24. Why is the velocity of sound greater in deep water than that on its surface?
Answer: Pressure increases with the depth of water, velocity of sound increases with the increase of pressure

Question 25. Why does a violinist return his instrument on entering a warm room?
Answer: Due to an increase in temperature velocity increases

Question 26. If temperature is kept constant, the velocity of sound in a gas is independent of which of the following properties? Pressure, density, or wind flow.
Answer: Pressure

Question 27. Is the velocity of sound in dry air at constant temperature, less or greater than that in moist air?
Answer: Less

Question 28. The equation of a progressive wave is given by y = \(A \cos \omega\left(t-\frac{x}{V}\right)\). What will be the frequency and amplitude of the wave?
Answer: \(\left[\frac{\omega}{2 \pi}, A\right]\)

Question 29. What should be the minimum time interval between an original sound and its echo, so that the echo is heard separately?
Answer: \(\frac{1}{10}\)s

Question 30. The velocity of sound in air is 330 m • s^-1. What should be the minimum distance of a reflector so that a listener can hear the echo distinctly?
Answer: 16.5 m

Question 31. In a big hall, sound persists even after the original sound is stopped. What is the phenomenon called?
Answer: Reverberation

Question 32. If a sound wave enters water from the air, is the angle of refraction less or greater than the angle of incidence?
Answer: Greater

Wave Motion Assertion Reason Type Questions And Answers

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

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

Question 1.

Statement 1: A sound wave is regarded as a pressure wave.

Statement 2: Energy in this type of wave is transported due to the formation of compression and rarefaction in the medium in which pressure difference is created.

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

Question 2.

Statement 1: Any function of space and time that satisfies the following equation represents a wave \(\frac{d^2 y}{d x^2}=\frac{1}{V^2} \frac{d^2 y}{d t^2}\)

Statement 2: y s Asin ωt, y = Acosωt do not satisfy the above equations and hence do not represent waves.

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

Wave Motion Match The Columns

Question 1. A wave is transmitted from a denser to a rarer medium.

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

Question 2. Three traveling sinusoidal waves on identical strings have the same tension. The mathematical form of the waves are \(y_1=A \sin (3 x-6 t), y_2=A \sin (4 x-8 t) and y_3=A \sin (6 x-12 t)\).

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

Wave Motion Comprehension Type Questions And Answers

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

Question 1. Represents two snaps of a traveling wave on a string of mass per unit length μ = 0.25 kg · m^-1. The first snap is taken at t = 0 and the second is taken at t = 0.053 s.

1. The speed of the wave is

1. \(\frac{20}{3}\) m · s^-1
2. \(\frac{10}{3}\) m · s^-1
3. 20 m · s^-1
4. 10 m · s^-1

Answer: 2. \(\frac{10}{3}\) m · s^-1

2. The frequency of the wave is

1. \(\frac{5}{3}\) Hz
2. \(\frac{10}{3}\) Hz
3. 5 Hz
4. 10 Hz

Answer: 1. \(\frac{5}{3}\) Hz

3. The maximum speed (in m · s^-1) of the particle is

1. \(\frac{5 \pi}{13}\)
2. \(\frac{5 \pi}{13}\)
3. \(\frac{\pi}{30}\)
4. \(\frac{7 \pi}{20}\)

Answer: 3. \(\frac{\pi}{30}\)

Question 2. A sinusoidal wave is propagating in a negative x -x-direction in a string stretched along the x-axis. A particle of string at x = 2 cm is found at its mean position and it is moving in a positive y-direction at t = 1s. The amplitude of the wave, the wavelength, and the angular frequency of the wave are 0.1 m, \(\frac{\pi}{4}\) m and 4π rad • s^-1 respectively,

1. The equation of the wave is

1. y = 0.1sin[47π(t- 1) + 8(x-2)]
2. y = 0.1 sin [(t- 1) – (x-2)]
3. y = 0.1sin [4π(t- 1)-8(x-2)]
4. None of these

Answer: 1. y = 0.1sin[47π(t- 1) + 8(x-2)]

2. The speed of particle at x = 2m and t = 1 s is

1. 0.2π m · s^-1
2. 0.67π m · s^-1
3. 0.4π m · s^-1
4. 0

Answer: 3. 0.4π m · s^-1

3. The instantaneous power (in J · s^-1) transfer through x = 2 m and t = 1.125 s is

1. 10
2. \(\frac{4 \pi}{3}\)
3. \(\frac{2 \pi}{3}\)
4. Zero

Answer: 4. Zero

Question 3. Suppose a musical note consists of tones of frequencies 100 Hz, 200 Hz, 300 Hz, 400 Hz, 500 Hz, and 600 Hz.

1. What should be the frequency of the fundamental tone?

1. 200 Hz
2. 100 Hz
3. 400 Hz
4. 600 Hz

Answer: 2. 100 Hz

2. Among the frequencies the third and fourth harmonics are respectively

1. 300 Hz, 500 Hz
2. 400 Hz, 500 Hz
3. 300 Hz, 400 Hz
4. None of these

Answer: 3. 300 Hz, 400 Hz

3. Among the tones the octave is

1. 200 Hz
2. 300 Hz
3. 400 Hz
4. None

Answer: 1. 200 Hz

Wave Motion Integer Answer Type Questions

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

Question 1. A transverse wave propagating along x -axis is represented by (x, t) = 8.0sin (0.5π x – 4πt – \(\frac{\pi}{4}\)) where x is in meters and t is in seconds. Calculate the speed in m µ s^-1 of the wave.
Answer: 8

Question 2. The water waves are traveling along the surface of an ocean at a speed of 2.5 m · s^-1 and splashing up periodically against a pole. Each adjacent crest is 5 m apart. The crest splashes upon reaching the foot of the pole. How much time (in seconds) passes between each successive splashing?

Answer: 2

Wave Motion Equation Of A Travelling Or Progressive Wave

WBBSE Class 11 Displacement Relation of Progressive Waves Notes

Progressive Waves

A wave, that propagates through a medium in a fixed direction, is called a progressive wave in one dimension. Suppose, due to the propagation of the wave, the panicles in the medium are subjected to forced simple harmonic motions. Obviously, the wave reaches two different panicles at two different times.

• As a result, there arises a time lag, and consequently a phase lag, between the vibrations of two different particles along the direction of wave motion.
• This time lag or phase lag depends on the velocity of the wave. If the wave travels from left to right, each particle starts vibrating later than its predecessor on the left. So, the phase of a panicle lags behind that of any other panicle on its left.

Let the positive direction of the x-axis be chosen along the direction of propagation of a one-dimensional progressive wave. O is the origin.

Progressive Wave Equation

The particles of the medium vibrate simply harmonically. The displacement of the particle at O at any instant t from its mean position is given by, y = Asinωt

where A is the amplitude of vibration and ω is the angular frequency of the SHM executed by the particle.

If n is the frequency of the particle, ω = 2πn

y = Asin2πnt…(1)

The wave is traveling along the positive x -x-direction.

So, time taken by the wave to reach P at a distance x on the right-hand side of O is \(\frac{x}{V}\) i.e., with respect to time, the point P always lags behind the point O by \(\frac{x}{V}\). If t and t’ are the times at the points O and P, respectively, t’ = t – \(\frac{x}{V}\).

So, displacement of the particle at P is given by,

y = \(A \sin \omega t^{\prime}=A \sin \omega\left(t-\frac{x}{V}\right)=A \sin 2 \pi n\left(t-\frac{x}{V}\right)\)….(2)

This is the equation of a progressive wave traveling in the positive direction of x-axis.

If the wave propagates in the opposite direction, i.e., in the negative x-direction, we put -x in place of +x. So in that case, the equation of the progressive wave will be,

y = \(A \sin \omega\left(t+\frac{x}{V}\right)=A \sin 2 \pi n\left(t+\frac{x}{V}\right)\)…(3)

From equations (2) and (3) it is seen that the displacement of a vibrating panicle on the path of a progressive wave changes

1. With time and
2. With distance.

Each of these changes, with time or with distance, is periodic. Clearly, in each of the cases, the displacement graph of the particle (y-t graph or y-x graph) is a sine graph.

Any of the harmonic functions in sine and cosine forms may be used to express a simple harmonic motion. So we can express equations (1), (2), and (3) by cosine functions.

In equations (2) and (3), y is the displacement of a particle with respect to its mean position.

1. In the case of a transverse wave, y is perpendicular to the x-axis.
2. In the case of a longitudinal wave, y is parallel to the x-axis.

A Few Alternative Forms Of The Progressive Wave Equation: In equation (2), the angular function is, \(\theta=\omega\left(t-\frac{x}{V}\right).\)

By expressing θ in different ways, the equation of a progressive wave can be expressed in a few alternative forms

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Mathematical Representation of Progressive Waves

Frequency, \(n=\frac{\omega}{2 \pi} ;\) time period, \(T=\frac{1}{n}=\frac{2 \pi}{\omega}\) wavelength, \(\lambda=\frac{V}{n}\);

Wave number, \(k=\frac{2 \pi}{\lambda}=\frac{2 \pi n}{V}=\frac{\omega}{V}\)

1. \(\theta=\omega\left(t-\frac{x}{V}\right)=\omega t-\frac{\omega}{V} x=\omega t-k x\)
2. \(\theta=\omega\left(t-\frac{x}{V}\right)=\frac{\omega}{V}(V t-x)=k(V t-x)=\frac{2 \pi}{\lambda}(V t-x)\)
3. \(\theta=\omega t-k x=2 \pi\left(\frac{\omega}{2 \pi} t-\frac{k}{2 \pi} x\right)=2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\)

Using these values of θ, the equation of a progressive wave can be written as

\(\left.\begin{array}{rl}
y & =A \sin \omega\left(t-\frac{x}{V}\right) \\
& =A \sin (\omega t-k x) \\
& =A \sin \frac{2 \pi}{\lambda}(V t-x) \\
& =A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)
\end{array}\right\}\)…(4)

Any of these alternative forms may be used as per convenience. Remember that if we put -x in place of x, we shall get the equation of a progressive wave moving in the negative direction of the x-axis.

Phase: The concept of phase has been discussed in the chapter Simple Harmonic Motion.

The equation of a progressive wave is written as y = Asinθ

where \(\theta=\frac{2 \pi}{\lambda}(V t-x)\)…(5)

If a progressive wave is not damped, the amplitude of vibration A remains constant. Except for amplitude, all other information about the wave is obtained from the angle θ. This angle θ is called the phase angle or phase. If this phase angle is known,

1. From the values of 2 and V we get,

frequency, \(n=\frac{V}{\lambda}\); time period T = \(\frac{1}{n}=\frac{\lambda}{V}\), etc.

2. For a particle in position x, the displacement at an instant t can be determined. Obviously, the phase 6 of the wave depends on x and t. At any instant, the phase changes with distance x. Again, at any point, the phase changes with time t.

Phase Difference: The phase difference of two particles at any two positions along the progressive wave at a particular instance of time is actually the difference of the phase angles at the two positions at that instant.

If x₁ and x₂ are the positions of the two particles along the direction of propagation of the wave, the path difference on the wave between the two particles is x₂-x₁.

From equation (5), at any instant t, the phase of the two particles are, respectively,

⇒ \(\theta_1=\frac{2 \pi}{\lambda}\left(V t-x_1\right) \text { and } \theta_2=\frac{2 \pi}{\lambda}\left(V t-x_2\right)\)

∴ \(\theta_1-\theta_2=\frac{2 \pi}{\lambda}\left(x_2-x_1\right)\)

i.e., phase difference = \(\frac{2 \pi}{\lambda}\) x path difference….(6)

From equation (6) we get,

1. If the path difference of two particles is 0, λ, 2 λ,…, the phase difference becomes 0, 2π, 4π,… In this case, the particles are in the same phase.

2. If the path difference of two particles \(\frac{\lambda}{2}, \frac{3 \lambda}{2}, \frac{5 \lambda}{2}, \ldots\) the phase difference becomes π, 3π, 5π,… In this case, the particles are in opposite phases.

Cosine form of the equation of a progressive wave: The equation of a progressive wave can be written using a cosine function instead of the sine function. Then,

y = \(A \cos (\omega t-k x)=A \sin \left(\omega t-k x+\frac{\pi}{2}\right)\)….(7)

Comparing equations (4) and (7), it is evident that the phase difference between these two progressive waves is \(\frac{\pi}{2}\) or 90°.

Initial Condition: In equation (4), if x = 0 and t = 0, y = 0, i.e., at the beginning, the displacement of the particle at the origin is zero. Similarly, in equation (7), if x = 0 and t = 0, y = A, i.e., at the beginning, the displacement of the particle at the origin is maximum. Obviously, due to different initial conditions, the phase difference between the two progressive waves given by equations (4) and (7) is 90°.

Generally, the displacement of a particle at t = 0 and x = 0 may have any value between +A and -A. So, the phase of a progressive wave may be different from those in equations (4) or (7). Denoting the initial phase by ø, the general form of a progressive wave can be written as:

\(\left.\begin{array}{rl}
y & =A \sin (\omega t-k x \pm \phi) \\
& =A \sin \left[\omega\left(t-\frac{x}{V}\right) \pm \phi\right] \\
& =A \sin \left[\frac{2 \pi}{\lambda}(V t-x) \pm \phi\right] \\
& =A \sin \left[2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right) \pm \phi\right]
\end{array}\right\}\)…..(8)

According to equation (8), at t = 0 and x = 0, the phase angle of the particle is ±ø. This is called the epoch of the progressive wave at the origin.

Partial Derivatives: The equations in this section show that y is a function of two independent variables—x and r. So y has two independent derivatives one with respect to x and the other with respect to t. They are the partial derivatives:

1. \(\frac{\partial y}{\partial x}=\frac{d y}{d x} .\), when t is considered to be a constant = rate of change of y with respect to x when t is a constant.

2. Similarly, \(\frac{\partial y}{\partial t}\) = rate of change of y with respect to t, when x is a constant.

Particle velocity and acceleration in a progressive wave: Displacement of a particle in a progressive wave, y = \(A \sin (\omega t-k x \pm \phi)\)……(9)

So, the velocity of the particle,

v = \(\frac{\partial y}{\partial t}=\omega A \cos (\omega t-k x \pm \phi)\)

or, \(\nu =v_0 \cos (\omega t-k x \pm \phi)\)…..(10)

The maximum value of the velocity of the particle = \(\pm \omega A= \pm v_0\).

This v₀ is called the velocity amplitude of the particle in a progressive wave.

From equation (9), \(\sin (\omega t-k x \pm \phi)=\frac{y}{A}\)

So, \(\cos (\omega t-k x \pm \phi)=\sqrt{1-\frac{y^2}{A^2}}=\frac{1}{A} \sqrt{A^2-y^2}\)

i.e., \(v=\omega A \cdot \frac{1}{A} \sqrt{A^2-y^2}\)

or, v = \(\omega \sqrt{A^2-y^2}\)

Differences Between Progressive And Non Progressive Waves

The phase difference between a sine function and a cosine function is 90°. So, the phase difference between displacement y and velocity v is also 90°.

From equation (11), it is also seen that if y = 0, v = ωA = v₀ and again if y = A, v = 0,

i.e., if displacement is zero, velocity is maximum and if displacement is maximum, velocity is zero.

Acceleration of the particle in a progressive wave,

a = \(\frac{\partial^2 y}{\partial t^2}=-\omega^2 A \sin (\omega t-k x \pm \phi)\)

or, a = \(-\omega^2 y\)….(12)

From this equation it is evident that the displacement of the particle y and its acceleration a are in opposite phases, i.e., the phase difference between them is 180°.

Accordingly, the phase difference between velocity and acceleration is 90°.

The maximum acceleration of the particle = \(\pm \omega^2 A= \pm a_0\)

This a₀ may be called the amplitude of acceleration of the particle in a progressive wave.

Relation between particle velocity and wave velocity in a progressive wave:

Displacement of the particle in a progressive wave, y = \(A \sin (\omega t-k x \pm \phi)\)

So, particle velocity, \(\nu=\frac{\partial y}{\partial t}=\omega A \cos (\omega t-k x \pm \phi)\)

Again, \(\frac{\partial y}{\partial x}=-k A \cos (\omega t-k x \pm \phi)\)

Wave Function and Displacement Relation

So, \(\frac{\partial}{\partial y}=-\frac{\omega}{k x}=-V\)

(As V is the wave velocity, \(k=\frac{\omega}{V}\) )

∴ v = -V \(\frac{\partial y}{\partial x}\)….(13)

This is the relation between particle velocity v and wave velocity V.

In case of a transverse wave, the direction of particle velocity is always perpendicular to that of wave velocity. In the case of a longitudinal wave, the direction of particle velocity is along or opposite to the direction of wave velocity.

Differential Equation Of A Progressive Wave: If y = \(A \sin (\omega t-k x \pm \phi)\)

⇒ \(\frac{\partial^2 y}{\partial t^2}=-\omega^2 A \sin (\omega t-k x \pm \phi)\)

and \(\frac{\partial^2 y}{\partial x^2}=-k^2 A \sin (\omega t-k x \pm \phi)\)

So, \(\frac{\frac{\partial^2 y}{\partial x^2}}{\frac{\partial^2 y}{\partial t^2}}=\frac{k^2}{\omega^2}=\frac{1}{V^2}\) (because wave velocity, \(V=\frac{\omega}{k}\))

or, \(\frac{\partial^2 y}{\partial x^2}=\frac{1}{V^2} \frac{\partial^2 y}{\partial t^2}\)…(14)

This equation is the differential equation of a progressive wave. Conversely, if any one-dimensional disturbance satisfies this equation, it is obviously a progressive wave.

Difference Between Particle Velocity And Wave Velocity:

1. The particles of a medium in a progressive wave do not change their positions due to their velocities. Every particle only vibrates on both sides of its mean position But the progressive wave advances through the medium with its wave velocity.
2. The velocities of the particles of the medium change continuously. It is maximum at the mean positions and zero at their extreme positions. However wave velocity remains constant for a particular medium. It depends only on the properties of the medium.
3. A progressive wave possesses energy due to the motion of the particles of the medium. This energy propagates with the wave through the medium with the velocity of the wave.
4. In a transverse wave, the direction of particle velocity is perpendicular to that of wave velocity. On the other hand, for a longitudinal wave, these two velocities are parallel.

The velocity of progressive waves in different media:

1. Velocity of a longitudinal wave in a solid medium, V = \(\sqrt{\frac{Y}{\rho}}\); Y = Young’s modulus of the medium, and ρ = density of the medium.

2. Velocity of a longitudinal wave in a liquid or gaseous medium, V = \(\sqrt{\frac{E}{\rho}}\); E = Bulk modulus of the medium, and ρ = density of the medium.

3. Velocity of sound wave in a gaseous medium, V = \(\sqrt{\frac{\gamma p}{\rho}}\); p = pressure of the gas, and γ = ratio of the two specific heats \(\left(\frac{c_p}{c_w}\right)\) of the gas.

Real-Life Examples of Displacement in Waves

4. Velocity of a transverse wave in a stretched string, V = \(\sqrt{\frac{T}{m}}\); T = tension in the string, and m = mass per unit length of the string.

5. Velocity of a longitudinal wave in a stretched string, \(V=\sqrt{\frac{Y}{\rho}}\); Y = Young’s modulus of the material of the string, and p ~ density of the material of the string.

6. Velocity of an electromagnetic wave, V = \(\frac{1}{\sqrt{\mu \epsilon}} ; \mu\) = permeability and ∈ = permittivity of the medium.

Characteristics Of Progressive Waves:

1. Progressive wave continuously propagates through a medium, and if not damped, it can propagate to infinity.
2. The progressive wave moves with a definite velocity. The wave velocity depends on the elastic properties and on the density of the medium. Energy is transferred with the wave through the medium with the velocity of the wave.
3. Each particle of the medium vibrates about its mean position with identical frequency and amplitude. The direction of movement of the particles may be perpendicular (transverse wave) or parallel (longitudinal wave) with respect to the direction of wave motion.
4. The velocity with which the phase of a vibrating particle of the medium is transferred to the next particle is called the wave velocity. For the same reason, it is also called the phase velocity. The phase difference between two vibrating particles is proportional to the distance of separation of the two particles along the line of wave propagation.
5. The progressive wave carries energy from one point to another without displacing the particles of the medium. Energy is transferred perpendicular to the direction of the wavefronts, i.e., along the direction of the rays.
6. Pressure and density in the medium, through which the progressive wave advances, follow the sinusoidal form of variation, like that of displacement, velocity, and acceleration.
7. A progressive wave has double periodicity. One is time periodicity determined by the time period (T) of the wave and the other is space periodicity determined by the wavelength (λ).

Different Properties Of Progressive Waves: The distinctive properties of progressive waves are explained below

Absorption: While moving through any medium, damping of the progressive wave takes place. It means that the energy of the wave gradually decreases with the increase of distance because a part of the energy of the wave is absorbed by the medium as the wave propagates.

Reflection: When a progressive wave traveling in a homogeneous medium is incident on an interface with another medium, a part of the incident wave comes back to the first medium. This phenomenon is known as the reflection of a wave. The amount of energy of the incident wave that will be reflected depends on the nature of the interface of the two media.

Refraction: When a progressive wave traveling through a homogeneous medium is incident on an interface with another medium, a part of the incident wave is transmitted into the second medium. This phenomenon is known as the refraction of a wave. In refraction, the direction of wave motion generally changes.

Interference: Let us consider two progressive waves having the same wavelength and velocity. The phase difference of the two waves is always a constant. When these two waves superpose, the amplitude of the resultant wave increases at some places and decreases at some other places of the medium. This successive increase and decrease of the amplitude of the resultant wave is called interference of waves.

Diffraction: When a progressive wave passes through the edge of an opening or of an obstacle, the direction of the wave may change. This is called the diffraction of a wave.

Scattering: In the course of propagation, when a progressive wave falls on a material particle, the particle is subject to a forced vibration. So this particle also acts as a secondary source of wave, i.e., weaves propagate in all directions from the vibrating particle. This phenomenon is called the scattering of a wave.

Polarisation: During propagation of a transverse wave through a medium, each particle of the medium vibrates on a plane perpendicular to the direction of motion of the wave.

• The plane is called the normal plane. If the vibrations of the particles on such planes arc somehow are restricted to a particular direction, then this phenomenon is called the polarisation of a wave. Obviously, polarization does not take place in case of longitudinal waves.
• It is to be noted that, polarisation does not take place in the case of sound waves. This shows that sound waves are longitudinal. On the other hand, light waves can be polarised by suitable arrangements. So light waves are transverse.

In optics, these properties of waves are discussed in detail. In this chapter, the phenomena of reflection and refraction of sound waves are mainly discussed.

Wave Motion Multiple Choice Questions And Answers

Question 1. Which one of the following phenomena differentiates transverse waves from longitudinal waves?

1. Reflection
2. Refraction
3. Polarisation
4. Interference

Answer: 3. Polarisation

Question 2. A sound wave is a

1. Transverse elastic wave
2. Transverse electromagnetic wave
3. Longitudinal elastic wave
4. Longitudinal electromagnetic wave

Answer: 3. Longitudinal elastic wave

Question 3. A light wave is a

1. Transverse elastic wave
2. Transverse electromagnetic wave
3. Longitudinal elastic wave
4. Longitudinal electromagnetic wave

Answer: 2. Transverse electromagnetic wave

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. Polarisation of the sound wave does not occur, because the sound wave is

1. Transverse wave
2. Longitudinal wave
3. Progressive wave
4. Elastic wave

Answer: 2. Longitudinal wave

Question 5. If a child stands up while swinging on a swing, then the period of the swing will

1. Increase
2. Decrease
3. Remain same
4. Increase if the child is long and decrease if the child is short

Answer: 2. Decrease

Question 6. A wave is advancing along the surface of water with a velocity of 80 cm · s^-1. If the distance between two consecutive crests is 20 cm, then what will be the wavelength?

1. 80 cm
2. 20 cm
3. 4 cm
4. \(\frac{1}{4}\) cm

Answer: 2. 20 cm

WBBSE Class 11 Wave Motion MCQs

Question 7. When a wave coming from a medium enters another medium, the property of the wave which remains unchanged is

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

Answer: 2. Frequency

Question 8. A wave is advancing along the surface of water with a velocity of 80 cm • s^-1. If the distance between two consecutive crests is 20 cm, then what will be the frequency of the wave?

1. 80Hz
2. 20Hz
3. 4 Hz
4. \(\frac{1}{4}\) Hz

Answer: 3. 4 Hz

Question 9. When a force F₁ acts on a particle, the frequency is 6Hz, and when a force F₂ acts, the frequency is 8 Hz. Now if both the forces act simultaneously in the same direction, then their frequency’ becomes

1. 20Hz
2. 14Hz
3. 10Hz
4. 2Hz

Answer: 3. 10Hz

Question 10. If the wavelength in a medium is reduced by 50%, keeping its velocity constant, the percentage change in its frequency is

1. 50%
2. 25%
3. 100%
4. 200%

Answer: 3. 100%

Question 11. The equation of a progressive wave is given by y = \(5 \sin \frac{\pi}{2}(100 t-x) \mathrm{cm}\). The wavelength is

1. 2 cm
2. 4 cm
3. 50 cm
4. 100 cm

Answer: 2. 4 cm

Question 12. The equation of a progressive wave is given by y = \(5 \sin \frac{\pi}{2}(100 t-x) \mathrm{cm}\). The frequency of the wave is

1. 25 Hz
2. 50 Hz
3. 100 Hz
4. 50π Hz

Answer: 1. 25 Hz

Question 13. The equation of a wave is given by y = \(10^{-4} \sin (60 t+2 x)\). Here x is expressed in metres and r in seconds. This represents a wave

1. Which is moving in the negative direction of the x-axis with a velocity of 30 m • s^-1
2. Whose wavelength is  πm
3. Whose frequency’ is \(\frac{30}{\pi}\) hz
4. All the above options are correct

Answer: 4. All the above options are correct

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Question 14. The equation of a tramverse wave is given by Y = \(Y_0 \sin 2 \pi\left(f t-\frac{x}{\lambda}\right)\). If the wave velocity, then

1. \(\lambda=\frac{\pi Y_0}{4}\)
2. \(\lambda=\frac{\pi Y_0}{2}\)
3. \(\lambda=\pi Y_0\)
4. \(\lambda=2 \pi Y_0\)

Answer: 2. \(\lambda=\frac{\pi Y_0}{2}\)

Key Terms in Wave Motion Explained

Question 15. Which one of the following physical quantities moves with a wave at the same velocity?

1. Amplitude
2. Mass
3. Momentum
4. Energy

Answer: 4. Energy

Question 16. When temperature increases, the frequency of a turning fork,

1. Increases
2. Decreases
3. Remains same
4. Increases or decreases depending on the material

Answer: 2. Decreases

Question 17. The pressure, density, and ratio of two specific heats of a gas are p, ρ, and γ respectively. The velocity of sound in the gas is

1. \(\frac{\sqrt{p}}{\rho}\)
2. \(\gamma \frac{\sqrt{p}}{\rho}\)
3. \(\sqrt{\frac{\gamma p}{\rho}}\)
4. \(\sqrt{\frac{3 \gamma p}{\rho}}\)

Answer: 3. \(\sqrt{\frac{\gamma p}{\rho}}\)

Question 18. If the absolute temperature of a gas is T and the velocity’ of sound in the gas is V, then

1. V ∝ T
2. V ∝ √T
3. \(V \propto \frac{1}{T}\)
4. \(V \propto \frac{1}{\sqrt{T}}\)

Answer: 2. V ∝ √T

Question 19. The velocity of sound in moist air is greater than that in dry air, because

1. Moist air is comparatively light
2. Moist air is comparatively heavy
3. The pressure of moist air is low
4. The pressure of moist air is high

Answer: 1. Moist air is comparatively light

Question 20. Laplace concluded that the temperature of a gaseous medium cannot remain constant during the propagation of sound through it, because

1. The pressure of the gas remains constant
2. The volume of the gas remains constant
3. The particles of the medium vibrate with a large amplitude
4. The compressions and the rarefactions occur very rapidly

Answer: 4. The compressions and the rarefactions occur very rapidly

Question 21. According to Laplace’s correction, the propagation of sound through a gaseous medium is

1. An isochoric process
2. An isobaric process
3. An isothermal process
4. An adiabatic process

Answer: 4. An isochoric process

Common Wave Motion MCQs for Class 11

Question 22. In an adiabatic process, the bulk modulus of a gas is

1. Equal to its density
2. Greater than its density
3. Equal to its pressure
4. Greater than its pressure

Answer: 4. Greater than its pressure

Question 23. The velocity of sound in air at STP is 330 m· s^-1. If the atmospheric pressure becomes 75 cm Hg, then the velocity of sound at the same temperature will be

1. 330 m · s^-1
2. Less than 330 m · s^-1
3. Greater than 330 m · s^-1
4. Given data is insufficient

Answer: 1. 330 m · s^-1

Question 24. For 1°C rise in temperature, the velocity of sound in air increases by

1. About 2 m · s^-1
2. About 2 ft · s^-1
3. About 61 m · s^-1
4. About 61 ft · s^-1

Answer: 2. About 2 ft · s^-1

Question 25. The ratio of the velocities of sound through hydrogen and oxygen at STP is

1. 1:16
2. 1:4
3. 4:1
4. 16:1

Answer: 3. 4:1

Question 26. At what temperature the velocity of sound in a gas will increase by 10% of its value at 27 °C?

1. 29.7°C
2. 32.7°C
3. 57°C
4. 90°C

Answer: 4. 90°C

Question 27. TWo monatomic ideal gases of molecular masses m₁ and m₂ respectively are enclosed in separate containers and kept at the same temperature. The ratio of the speed of sound of 1st and 2nd gas is given by

1. \(\sqrt{\frac{m_1}{m_2}}\)
2. \(\sqrt{\frac{m_2}{m_1}}\)
3. \(\frac{m_1}{m_2}\)
4. \(\frac{m_2}{m_1}\)

Answer: 2. \(\sqrt{\frac{m_2}{m_1}}\)

Question 28. Velocity of sound is maximum in

1. Water
2. Air
3. Vacuum
4. Steel

Answer: 4. Steel

Types of Waves MCQs

Question 29. The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300K is

1. \(\sqrt{\frac{2}{7}}\)
2. \(\sqrt{\frac{1}{7}}\)
3. \(\frac{\sqrt{3}}{5}\)
4. \(\frac{\sqrt{6}}{5}\)

Answer: 3. \(\frac{\sqrt{3}}{5}\)

Question 30. During a speech, the vibration of the diaphragm of a microphone is

1. Natural vibration
2. Damped vibration
3. Forced vibration
4. Resonant vibration

Answer: 3. Forced vibration

Question 31. Phon is the unit of

1. Pitch
2. Quality
3. Timbre
4. Loudness

Answer: 4. Loudness

Question 32. A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

1. 1000
2. 10000
3. 10
4. 100

Answer: 4. 100

Question 33. Which one of the following phenomena does not take place in the case of sound waves?

1. Reflection
2. Refraction
3. Polarisation
4. Interference

Answer: 3. Polarisation

Question 34. A balloon full of carbon dioxide and another full of water behaves as lenses for the refraction of sound. What will be the nature of these lenses?

1. Both are convergent
2. Both are divergent
3. The first one is divergent and the second one is convergent
4. The first one is convergent and the second one is divergent

Answer: 4. The first one is convergent and the second one is divergent

Question 35. If a man standing at any place between two hills fires a gun, how many echoes will he hear?

1. 1
2. 2
3. 4
4. More than 4

Answer: 4. More than 4

Question 36. A sound is made in sea water from the bottom of a ship and its echo is heard after a time t. If v is the velocity of sound in sea water, the depth of the sea at that place is

1. vt
2. \(\frac{v t}{2}\)
3. \(\frac{v t}{10}\)
4. \(\frac{v t}{20}\)

Answer: 2. \(\frac{v t}{2}\)

Wave Properties and Characteristics MCQs

Question 37. An echo coming from a hill is heard 20 s after the original sound. If the velocity of sound in air is 330 m • s^-1, the distance of the hill will be

1. 3.3 km
2. 6.6 km
3. 1.65 km
4. 13.2 km

Answer: 1. 3.3 km

Question 38. If the velocity of sound in air and in water are 330 m · s^-1 and 1400 m · s^-1 respectively, what will be the critical angle for refraction of sound going to water from air?

1. About 29°
2. About 13.6°
3. About 76.4°
4. No critical angle in this case

Answer: 2. About 13.6°

Question 39. A source of sound of frequency 600 Hz is placed in water. The speed of sound in water and air is 1500 m/s and 300 m/s respectively. The frequency of sound recorded by an observer who is standing in air is

1. 660 Hz
2. 600 Hz
3. 540 Hz
4. None of the above

Answer: 2. 600 Hz

In this type of question, more than one option is correct.

Question 40. A wave equation that gives the displacement along the y-direction is given by y = 10^-4 sin(60t+2x) where x and y are in meters and t is time in seconds. This represents a wave

1. Traveling with a velocity of 30 m · s^-1 in the negative x -direction
2. Of wavelength πm
3. Of frequency \(\frac{30}{\pi}\) hz
4. Of amplitude 10^-4 m traveling along the negative x -direction

Answer: All options are correct

Question 41. For a transverse wave on a string, the string displacement is described by, y = f(x- at) where f represents a function and A is a negative constant. Then which of the following is/are correct statement(s)?

1. The shape of the string at time t = 0 is given by f(x)
2. The shape of the waveform doesn’t change as it moves along the string.
3. Waveform moves in the positive x -x-direction
4. The speed of the waveform is a.

Answer:

1. The shape of the string at time t = 0 is given by f(x)
2. The shape of the waveform doesn’t change as it moves along the string.

Question 42. Mark the correct statement(s) concerning waves

1. A wave can have both transverse and longitudinal components
2. A wave doesn’t result in the bulk flow of the materials of its medium
3. A wave is a traveling disturbance
4. A wave can be there even in the absence of an elastic medium

Answer: All options are correct

Real-Life Examples of Wave Motion Applications

Question 43. Mark the correct statements.

1. Note emitted from any musical instrument is generally a combination of tones
2. A tone can be compared with compound light
3. Fundamental tone can be treated as the 1st harmonic
4. Harmonics are only the odd multiples of the fundamental tone

Answer:

1. Note emitted from any musical instrument is generally a combination of tones

3. Fundamental tone can be treated as the 1st harmonic

Question 44. Suppose a musical instrument emits sound having frequencies 250 Hz, 300 Hz, 350 Hz, 400 Hz, 450 Hz, and 500 Hz respectively.

1. Tones having frequencies 250 Hz, 300 Hz, 350 Hz, 400 Hz, 450 Hz, and 500 Hz are overtones.
2. The tone of 250 Hz is the fundamental tone
3. The combination of these six tones forms a note
4. The tone of 500 Hz is the octave

Answer:

2. The tone of 250 Hz is the fundamental tone

3. Combination of these six tones forms a note

4. The tone of 500 Hz is the octave

Wave Motion Question and Answers

Question 1. In the case of a transverse wave, do the particles of the medium vibrate in the u plane parallel to the plane of propagation of the wave?
Answer:

Let us assume that the wave is propagating along the z-axis. Then the plane of propagation of the wave, or the wavefronts, will be perpendicular to z -the axis, i.e., parallel to the xy-plane.

Again the direction of motion of tire vibrating particles of the medium in a transverse wave is perpendicular to z -the axis, i.e., parallel to the xy-plane. So, the particles of the medium vibrate in a plane parallel to the plane of propagation of the wave.

Question 2. The equation of a progressive wave In a stretched string Is given by, y = Asin(kx – ωt). What is the maximum particle velocity?
Answer:

Particle velocity, v = \(\frac{\partial y}{\partial t}=\frac{\partial}{\partial t}\{A \sin (k x-\omega t)\}\)

= \(-\omega A \cos (k x-\omega t)\)

Since the maximum value of a cosine function is ±1, the magnitude of maximum particle velocity is ωA.

Wave Properties and Characteristics Q&A

Question 3. Regular reflection of sound takes place from a large rough reflector, but not regular reflection of light. On the other hand, regular reflection of light takes place from a small smooth reflector, but not regular reflection of sound. What Is the reason for this?
Answer:

To obtain effective reflection of a wave from a reflector, its size should be greater than the wavelength. The wavelength of audible sound in air varies from 1.5 cm to 16 in.

• On the other hand, the wavelength of visible light varies from 4 x 10^-7 m to 8 x 10^-7 m, i.e., the wavelength of sound is much greater than that of light. So light can be reflected from a very small reflector, but not sound.
• Again for regular reflection, the size of the notches of the surface of a reflector should be smaller than the wavelength of the wave. The wavelength of light is very small.
• So the size of the notches of the reflector should be very small, i.e., the reflector should be smooth. On the other hand, the wavelength of sound is large. So regular reflection of sound takes place even from a rough reflector.

Question 4. A sound wave travels from air to water. The angle of incidence at the surface of separation between air and water is i₁ and the angle of refraction is l₂. Snell’s law is applicable in this case. Determine which one of i₁ and i₂ is greater.
Answer:

According to Snell’s law, \(\frac{\sin i_1}{\sin i_2}\) = constant.

The value of this constant is equal to the ratio of the velocities of the wave in the two media.

i.e., \(\frac{\sin i_1}{\sin i_2}=\frac{V_1}{V_2}\)

Here, the first medium is air and the second medium is water. The velocity of sound in water is greater than that in air (i.e., V₂>V₁).

So, \(\sin i_2>\sin i_1 \quad \text { or, } \quad i_2>i_1\)

∴ i₁ is smaller than i₂.

WBBSE Class 11 Wave Motion Q&A

Question 5. A balloon full of carbon dioxide and another frill of water behaves as lenses for the refraction of sound. What type of lens will they resemble?
Answer:

The velocity of sound in carbon dioxide is less than that in air, i.e., when sound enters carbon dioxide from air, its velocity decreases. So in the case of refraction of sound, carbon dioxide is a denser medium relative to air.

• Hence, a balloon full of carbon dioxide, when placed in the air, behaves like a convex lens. In the case of a convex lens of glass, light rays incident on it become convergent. Similarly, sound rays passing through a balloon full of carbon dioxide become convergent.
• On the other hand, when sound enters water from the air, its velocity increases, because the velocity of sound in air is 330 m · s^-1, but that in water is about 1500 m · s^-1. So, in case of refraction of sound, water is a rarer medium than air.
• Hence, a balloon full of water behaves in an opposite manner. So in this case, since the sound wave moves to a rarer medium from a denser medium, the waves will diverge. Therefore, the balloon containing water will behave like a concave lens.

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Question 6. In a progressive wave, the maximum velocity and maximum acceleration of a particle in the medium are v₀ and a₀, respectively. Find the amplitude and the frequency of the waves.
Answer:

Let the equation of a progressive wave be y = \(A \sin \omega\left(t-\frac{x}{V}\right)\)

So, particle velocity v = \(\frac{\partial y}{\partial t}=\omega A \cos \omega\left(t-\frac{x}{V}\right)\),

i.e., v₀ = ωA

Again, acceleration, a = \(\frac{\partial v}{\partial t}=-\omega^2 A \sin \omega\left(t-\frac{x}{V}\right),\)

i.e., a₀ = ω²A

So, amplitude, A = \(\frac{(\omega A)^2}{\omega^2 A}=\frac{v_0^2}{a_0}\)

And frequency, n = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \cdot \frac{\omega^2 A}{\omega A}=\frac{a_0}{2 \pi v_0}\).

Question 7. A plane wave of sound traveling in the air is incident upon a plane’s water surface. The angle of incidence is 60. Assuming Snell’s law to be valid for sound waves, it will be refracted into the water away from the normal. State whether the statement is true or false and explain.
Answer:

The velocity of sound in water is about 4 times that in air. So for sound waves, air is a denser medium and water is a rarer medium. In this case, the sound wave is incident on a rarer medium (water) from the denser medium (air).

• For sound waves, the critical angle in air with respect to water is about \(\sin ^{-1}\left(\frac{1}{4}\right)=14.5^{\circ}\).
• So if the angle of incidence is 60°, sound waves will not be refracted in water, rather there will be a total internal reflection from the surface of the water. So the statement is false.

Question 8. Prove that the equation \(y=4 \sin \frac{2 \pi}{15}(60 t-x) \mathrm{cm}\) represents a progressive wave.
Answer:

Here the velocity of the wave is 60 cm · s^-1.

If y₁ is the displacement of a particle at the point (x + 60) at time (t+1), then

⇒ \(y_1=4 \sin \frac{2 \pi}{15}[60(t+1)-(x+60)]\)

= \(4 \sin \frac{2 \pi}{15}(60 t-x)=y\)

So, the motion of a particle at a distance x cm at time t is identical to the motion of the particle at a distance (x + 60) cm at time (t + 1). In this case, the disturbance in the medium moves 60 cm in 1 s. So, the given equation is an equation of a progressive wave.

Question 9. If the wavelength of an X-ray is 3Å, what is the frequency of the wave? [1A = 10^-8 cm]
Answer:

Velocity of X-ray = velocity of light

= c = 3 x 10¹⁰ cm · s^-1

Wavelength, λ = 3Å = 3 x 10^-8 cm

∴ Frequency, n = \(\frac{c}{\lambda}=\frac{3 \times 10^{10}}{3 \times 10^{-8}}=10^{18} \mathrm{~Hz} .\)

Question 10. Which property of waves proves that sound wave is longitudinal?
Answer:

The polarisation property of wave proves it. It is not possible for any longitudinal wave to exhibit the phenomenon of polarisation, as in longitudinal waves, the vibrating particles always move along the direction of propagation of the wave. Since sound waves do not exhibit the phenomenon of polarisation, they are longitudinal.

Question 11. Which property of waves shows that light wave is transverse?
Answer:

The polarisation property of wave proves it. Any transverse wave exhibits the phenomenon of polarisation, as the vibrating particles always move perpendicular to the direction of propagation of the wave. Since light waves exhibit the phenomenon of polarisation, they are transverse.

Question 12. If the amplitude of a progressive wave at a distance r from a point source is A, what will be the amplitude at a distance 2r?
Answer:

Intensity of a wave ∝ (amplitude)²

Again, the intensity is inversely proportional to the square of the distance from a point source,

i.e., \(\text { intensity } \propto \frac{1}{(\text { distance })^2}\)

So, \(\text { amplitude } \propto \frac{1}{\text { distance }}\)

Hence, the amplitude will be half, i.e., \(\frac{A}{2}\) at a distance 2r,

Question 13. The equation of a wave la given by y = \(A \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)\), Determine the amplitude, the frequency, and the wavelength of the wave.
Answer:

y = \(A \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)=\frac{A}{2} \cdot 2 \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)\)

= \(\frac{A}{2}\left[1+\cos \left(4 \pi n t-4 \pi \frac{x}{\lambda}\right)\right] \)

= \(\frac{A}{2}+\frac{A}{2} \cos \left(4 \pi n t-4 \pi \frac{x}{\lambda}\right)\)

Here, the second term, i.e., the term containing the cosine function represents the wave. Comparing this term with the general equation y = A’ cos (ωt- kx), we have,

Amplitude, A’ = \(\frac{A}{2}\)

∴ \(\omega=4 \pi n, \text { i.e., frequency }=\frac{\omega}{2 \pi}=2 n,\)

k = \(\frac{4 \pi}{\lambda}, \text { i.e., wavelength }=\frac{2 \pi}{k}=\frac{2 \pi}{\frac{4 \pi}{\lambda}}=\frac{\lambda}{2}\).

Step-by-Step Solutions to Wave Motion Problems

Question 14. The equation of a progressive wave Is given by y = \(A \sin 2 \pi\left(p t-\frac{x}{5}\right)\). What Is the ratio of the maximum particle velocity of the medium to the wave velocity?
Answer:

y = \(A \sin 2 \pi\left(p t-\frac{x}{5}\right)=A \sin \left(2 \pi p t-\frac{2 \pi}{5} x\right)\)

Comparing the given equation with the general equation y = Asin(ωt- kx) we have,

∴ \(\omega=2 \pi p \text { and } k=\frac{2 \pi}{5}\)

∴ Wave velocity, V = \(\frac{\omega}{k}=\frac{2 \pi p}{\frac{2 \pi}{5}}=5 p\)

Again, particle velocity,

v = \(\frac{\partial y}{\partial t}=2 \pi p A \cos \left(2 \pi p t-\frac{2 \pi}{5} t\right)\)

Maximum particle velocity, v₀ = 2πpA

So, the ratio of the maximum panicle velocity to the wave velocity = \(\frac{v_0}{V}=\frac{2 \pi p A}{5 p}=\frac{2}{5} \pi \lambda\)

Question 15. Even if the ratio \(\frac{p}{\rho}\) for helium and oxygen are kept equal, why la the velocity of sound not equal hi the two gases?
Answer:

The velocity of sound, c = \(\sqrt{\frac{\gamma p}{\rho}}\), Though the value of the ratio \(\frac{p}{\rho}\) for the two gases are equal, the velocity of sound c differs due to the different values of γ. Oxygen Is a diatomic gas and helium Is a monatomic gas. The values of γ for these two gases are \(\frac{7}{5}\) and \(\frac{5}{3}\) respectively. So, the velocity of sound In them is not equal.

Question 16. Why Is the velocity of sound In the air higher In the rainy season than in winter?
Answer:

Both the temperature and the humidity of the air are higher in the rainy season than In winter. We know, the velocity of sound in air is given by,

c = \(\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma R T}{M}}\)

Since c ∝ √T, the velocity of sound increases with the increase in temperature of the air.

Again with the increase in humidity, the amount of water vapour in air also increases. The density of water vapor Is less than the density of air. So the density ρ Is less in the rainy season and the velocity of sound increases.

Hence, the velocity of sound in air is higher in the rainy season In comparison to that in winter.

Question 17. At what temperature the velocity of sound In air Is twice its value at 0°C?
Answer:

The velocity of sound (c) in the air is proportional to the square root of its absolute temp (T), i.e., c α √T.

So the velocity of sound is doubled if the absolute temperature becomes four times. Since the initial temperature is 0°C or 273 K,

The required final temperature = 4x 273 K = 1092 K

= (1092 – 273)°C = 819°C.

Question 18. The velocity of sound is generally greater in solids than in gases why?
Answer:

The velocity of sound in a solid is given by, \(V_s=\sqrt{\frac{Y}{\rho}} ;\)

Y = Young’s modulus of the solid, ρ = density of the solid

The velocity of sound in a gas is given by, \(V_g=\sqrt{\frac{k}{\rho^{\prime}}}\)

k = bulk modulus of the gas, ρ’ = density of the gas.

Although the density of a solid medium is more than that of a gaseous medium, the modulus of elasticity of a solid (Y) is many times greater than that of a gas (k).

Hence, \(\frac{Y}{\rho}>\frac{k}{\rho^{\prime}}\)

Therefore, V_s > V_g, i.e., the velocity of sound is greater in solids than in gases.

Question 19. How will the velocity of sound be affected at high altitudes?
Answer:

The temperature decreases with increasing altitude. Furthermore, the humidity of air is generally less in hill areas. So, the velocity of sound will decrease due to a decrease in both temperature and humidity.

Real-Life Examples of Wave Motion Applications

Question 20. The speed of sound in water and in air are 1500 m • s^-1 and 300 m • s^-1 respectively. A source of sound of frequency 600 Hz is placed inside water. If the emitted sound enters air from the source, what will be its frequency?
Answer:

The frequency of the emitted sound will remain the same, i.e., 600 Hz, because the frequency of a wave does not change due to refraction.

Question 21. Determine the relationship between the rms velocity. f of the molecule of a gas and the velocity of sound in the gas.
Answer:

From the kinetic theory of gases, we have,

rms velocity of the gas molecules = \(\sqrt{\frac{3 p}{\rho}}\)

Again from the calculations of Newton and Laplace we have, the velocity of sound in a gaseous medium = \(\sqrt{\frac{\gamma p}{\rho}}\)

[γ = ratio of the two specific heats of the gas]

So, the ratio of the rms velocity and the velocity of sound in a gaseous medium = \(\sqrt{\frac{3}{\gamma}}\)

Question 22. If the velocity of sound in oxygen at STP is v, what will be the velocity of sound in helium under the same conditions?
Answer:

Velocity of sound, c = \(\sqrt{\frac{\gamma p}{\rho}}\),

i.e., if the pressure remains the same, for two different gases, \(\frac{c_1}{c_2}=\sqrt{\frac{\gamma_1}{\gamma_2}} \cdot \sqrt{\frac{\rho_2}{\rho_1}}\)

The ratio of the densities of oxygen and helium = \(\frac{16}{2}\) = 8

Again, for oxygen \(\gamma_1=\frac{7}{5}\) (oxygen is diatomic)

And for helium \(\gamma_2=\frac{5}{3}\) (helium is monatomic)

So, \(\frac{\gamma_1}{\gamma_2}=\frac{7 / 5}{5 / 3}=\frac{21}{25}\)

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{21}{25}} \times \sqrt{\frac{1}{8}}\)

As \(c_1=v, c_2=v \sqrt{\frac{25 \times 8}{21}}=3.1 v\).