Answer (b) Water waves produced by a motorboat sailing in water are both longitudinal and transverse, because the waves, produce transverse as well as lateral vibrations in the particles of the medium.

• (a) Water waves produced by a motorboat sailing in water are not “neither longitudinal nor transverse” because they involve both types of vibrations: longitudinal (parallel to the direction of wave propagation) and transverse (perpendicular to the direction of wave propagation).

• (c) Water waves produced by a motorboat sailing in water are not “only longitudinal” because they also involve transverse vibrations, where the particles of the medium move perpendicular to the direction of wave propagation.

• (d) Water waves produced by a motorboat sailing in water are not “only transverse” because they also involve longitudinal vibrations, where the particles of the medium move parallel to the direction of wave propagation.

Answer (c) Let the frequency in the first medium is $v$ and in the second medium is $v$.’ Frequency remains same in both the medium

So,

$$ v=v^{\prime} \Rightarrow \frac{v}{\lambda}=\frac{v^{\prime}}{\lambda^{\prime}} $$

$$ \Rightarrow \quad \lambda^{\prime}=(\frac{v^{\prime}}{v}) \lambda $$

$\lambda$ and $\lambda^{\prime}, v$ and $v^{\prime}$ are wavelengths and speeds in first and second medium respectively.

So,

$$ \lambda^{\prime}=(\frac{2 v}{v}) \lambda=2 \lambda $$

• Option (a) $\lambda$: This option is incorrect because the wavelength of the sound wave changes when it enters a medium with a different speed. Since the speed of sound in the second medium is twice that in the first medium, the wavelength must also change proportionally. Therefore, the wavelength cannot remain the same ($\lambda$).

• Option (b) $\frac{\lambda}{2}$: This option is incorrect because it suggests that the wavelength in the second medium is half of that in the first medium. Given that the speed of sound in the second medium is twice that in the first medium, the wavelength should increase, not decrease. Hence, the wavelength cannot be $\frac{\lambda}{2}$.

• Option (d) $4 \lambda$: This option is incorrect because it suggests that the wavelength in the second medium is four times that in the first medium. Since the speed of sound in the second medium is only twice that in the first medium, the wavelength should also be twice, not four times. Therefore, the wavelength cannot be $4 \lambda$.

Answer (c) Due to presence of moisture density of air decreases.

We know that speed of sound in air is given by $v=\sqrt{\frac{\gamma p}{\rho}}$

For air $\gamma$ and $p$ are constants.

$$ \begin{aligned} & v \propto \frac{1}{\sqrt{\rho}}, \text { where } \rho \text { is density of air. } \\ & \frac{v_2}{v_1}=\sqrt{\frac{\rho_2}{\rho_1}} \end{aligned} $$

where $\rho_1$ is density of dry air and $\rho_2$ is density of moist air.

As $\quad \rho_2<\rho_1=\frac{v_2}{v_1}>1 \Rightarrow v_2>v_1$

Hence, speed of sound wave in air increases with increase in humidity.

• (a) The speed of sound in air is not independent of temperature. In fact, it increases with an increase in temperature because the speed of sound is proportional to the square root of the absolute temperature of the air.

• (b) The speed of sound in air is not significantly affected by pressure under normal atmospheric conditions because both the pressure and density of air change proportionally with altitude, leaving the speed of sound relatively constant.

• (d) The speed of sound does not decrease with an increase in humidity. As explained, the presence of moisture decreases the density of air, which in turn increases the speed of sound.

Answer (c) Speed of sound wave in a medium $v \propto \sqrt{T}$ (where $T$ is temperature of the medium) Clearly, when temperature changes speed also changes.

As,

$$ v=v \lambda $$

where $v$ is frequency and $\lambda$ is wavelength.

Frequency $(v)$ remains fixed

$\Rightarrow \quad v \propto \lambda$ or $\lambda \propto v$

As does not change, so wavelength $(\lambda)$ changes.

• (a) Frequency of sound waves: The frequency of a sound wave is determined by the source of the sound and does not change with the temperature of the medium. The temperature affects the speed of sound, but the frequency remains constant.

• (b) Amplitude of sound waves: The amplitude of a sound wave is related to the energy and intensity of the wave, which are influenced by the source of the sound and the medium’s properties, but not directly by the temperature of the medium.

• (d) Loudness of sound waves: Loudness is a perceptual characteristic related to the amplitude and intensity of the sound wave. While temperature can affect the propagation of sound and potentially its perceived loudness due to changes in air density and absorption, it does not directly change the loudness of the sound wave itself.

Answer (b) Propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted.

There is no movement of matter (mass) and hence momentum.

• (a) Matter: Longitudinal waves do not transmit matter through the medium; they only cause particles of the medium to oscillate back and forth around their equilibrium positions without any net movement of matter.

• (c) Energy and matter: While longitudinal waves do transmit energy, they do not result in the net movement of matter through the medium. The particles of the medium only oscillate around their equilibrium positions.

• (d) Energy, matter and momentum: Longitudinal waves transmit energy but do not cause a net movement of matter or momentum through the medium. The particles oscillate back and forth, but there is no overall transfer of matter or momentum.

Answer (c) When mechanical transverse wave propagates through a medium, the constituent of the medium oscillate perpendicular to wave motion causing change in shape. That is each, element of the medium is subjected to shearing stress. Solids and strings have shear modulus, that is why, sustain shearing stress.

Fluids have no shape of, their own, they yield to shearing stress. This is why transverse waves are possible in solids and strings but not in fluids.

• (a) Mechanical transverse waves can propagate through all mediums: This is incorrect because mechanical transverse waves require a medium that can sustain shearing stress. Solids and strings can sustain shearing stress due to their shear modulus, but fluids (liquids and gases) cannot. Therefore, mechanical transverse waves cannot propagate through all mediums, particularly not through fluids.

• (b) Longitudinal waves can propagate through solids only: This is incorrect because longitudinal waves involve the oscillation of particles in the direction of wave propagation, which results in compressions and rarefactions. These waves can propagate through solids, liquids, and gases, as all these mediums can support compressional stress.

• (d) Longitudinal waves can propagate through vacuum: This is incorrect because longitudinal waves are mechanical waves that require a medium to propagate. In a vacuum, there are no particles to oscillate and transmit the wave, so longitudinal waves cannot propagate through a vacuum.

Answer (d) (a) Due to compression and rarefactions density of the medium (air) changes. At compressed regions density is maximum and at rarefactions density is minimum

(b) As density is changing, so Boyle’s law is not obeyed

(c) Bulk modulus remains same

(d) The time of compression and rarefaction is too small i.e., we can assume adiabatic process and hence no transfer of heat

• (a) Due to compression and rarefactions density of the medium (air) changes. At compressed regions density is maximum and at rarefactions density is minimum.

• (b) As density is changing, so Boyle’s law is not obeyed.

• (c) Bulk modulus remains same.

Thinking Process

Due to reflection from a denser medium there is a phase change of $180^{\circ}$ in the reflected wave.

Answer (b) Amplitude of reflected wave

$$ A_{r}=\frac{2}{3} \times A_{i}=\frac{2}{3} \times 0.6=0.4 \text { units } $$

Given equation of incident wave

$$ y_{i}=0.6 \sin 2 \pi(t-\frac{x}{2}) $$

Equation of reflected wave is

$$ y_{r}=A_{r} \sin 2 \pi(t+\frac{x}{2}+\pi) $$

[ $\because$ At denser medium, phase changes by $\pi]$ The positive sign is due to reversal of direction of propagation

So,

$$ y_{r}=-0.4 \sin 2 \pi(t+\frac{x}{2}) \quad[\because \sin (\pi+\theta)=-\sin \theta] $$

• Option (a): The amplitude of the reflected wave is not correctly adjusted. The given amplitude is 0.6, but it should be $\frac{2}{3}$ of 0.6, which is 0.4. Additionally, the phase change of $\pi$ is not considered, which would introduce a negative sign.

• Option (c): The amplitude is correctly adjusted to 0.4, but the phase change of $\pi$ is not considered. The phase change would introduce a negative sign, which is missing in this option.

• Option (d): The amplitude is correctly adjusted to 0.4, but the direction of propagation is not reversed. The correct equation should have a positive sign in the argument of the sine function to indicate the reversal of direction.

Answer (b)

Mass $m=2.5 kg$

$$ \mu=\text { mass per unit length } $$

$$ \begin{aligned} & =\frac{m}{l}=\frac{2.5 kg}{20}=\frac{1.25}{10}=0.125 kg / m \\ \text { Speed } v & =\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{0.125}} \quad \text { [speed of transverse waves in any string] } \\ l & =v \times t \Rightarrow 20=\sqrt{\frac{200}{0.125}} \times t \\ t & =20 \times \sqrt{\frac{125}{2 \times 10^{5}}}=20 \times \sqrt{\frac{25 \times 5}{2 \times 10^{5}}} \\ & =20 \times \sqrt{25 \times \frac{1}{0.4 \times 10^{5}}} \\ & =20 \times 5 \sqrt{\frac{1}{4 \times 10^{4}}}=\frac{20 \times 5}{2 \times 10^{2}} \\ & =\frac{1}{2}=0.5 \end{aligned} $$

• Option (a) $1 s$: This option is incorrect because the calculated time for the disturbance to travel the length of the string is $0.5 s$, not $1 s$. The speed of the wave and the length of the string do not support a travel time of $1 s$.

• Option (c) $2 s$: This option is incorrect because the calculated time for the disturbance to travel the length of the string is $0.5 s$, not $2 s$. A travel time of $2 s$ would imply a much slower wave speed, which is inconsistent with the given tension and mass per unit length.

• Option (d) data given is insufficient: This option is incorrect because the given data (mass of the string, tension, and length of the string) is sufficient to calculate the wave speed and the time it takes for the disturbance to travel the length of the string. The problem provides all necessary information to determine the correct answer.

Thinking Process

The observed frequency is apparent frequency due to Doppler shift.

Answer (c) Let the original frequency of the source is $n_0$. Let the speed of sound wave in the medium is $v$.

As observer is stationary

Apparent frequency $n_{a}=(\frac{v}{v-v_{s}}) n_0$

[when train is approaching]

$$ =(\frac{v}{v-v_{s}}) n_0=n_{a}>n_0 $$

When the train is going away from the observer

Apparent frequency $n_{a}=(\frac{v}{v+v_{s}}) n_{o}=n_{a}<n_{o}$

Hence, the expected curve is (c).

• Option (a): This option suggests a constant frequency over time, which is incorrect because the Doppler effect causes a change in frequency as the train approaches and then recedes from the observer. The frequency should increase as the train approaches and decrease as it moves away.

• Option (b): This option suggests a linear change in frequency, which is incorrect because the Doppler effect results in a sudden change in frequency at the moment the train passes the observer. The frequency should not change linearly but rather exhibit a sharp transition.

• Option (d): This option suggests a symmetrical curve with a smooth transition, which is incorrect because the Doppler effect causes an abrupt change in frequency when the train passes the observer. The frequency should show a distinct jump rather than a smooth curve.

Thinking Process

To find the characteristic parameters associated with a wave, compare the given equation of the wave with a standard equation.

Answer $(a, b, c)$

Given equation is

$$ y(x, t)=3.0 \sin (36 t+0.018 x+\frac{\pi}{4}) $$

Compare the equation with the standard form.

$$ y=a \sin (\omega t+k x+\phi) $$

(a) As the equation involves positive sign with $x$, hence the wave is travelling from right to left. Hence, option (a) is correct.

(b) Given

$\omega=36 \Rightarrow 2 \pi v=36$

$\Rightarrow$

$v=$ frequency $=\frac{36}{2 \pi}=\frac{18}{\pi}$

$k=0.018 \Rightarrow \frac{2 \pi}{\lambda}=0.018$

$$ \begin{aligned} & \Rightarrow \quad \frac{2 \pi v}{v \lambda}=0.018 \Rightarrow \frac{\omega}{v}=0.018 \quad[\because 2 \pi v=\omega \text { and } v \lambda=v] \\ & \Rightarrow \quad \frac{36}{V}=0.018=\frac{18}{1000} \\ & \Rightarrow \quad v=2000 cm / s=20 m / s \end{aligned} $$

(c) $2 \pi v=36$

$$ \Rightarrow \quad v=\frac{36}{2 \pi} Hz=\frac{18}{\pi}=5.7 Hz $$

(d) $\frac{2 \pi}{\lambda}=0.018$

$$ \begin{aligned} \Rightarrow \quad \lambda & =\frac{2 \pi}{0.018} cm \\ & =\frac{2000 \pi}{18} cm=\frac{20 \pi}{18} m=3.48 cm \end{aligned} $$

Hence, least distance between two successive crests $=\lambda=3.48 m$.

• Option (d) is incorrect: The least distance between two successive crests in the wave is given by the wavelength (\lambda). From the calculations, (\lambda = 3.48 , \text{cm}), not (2.5 , \text{cm}). Therefore, the least distance between two successive crests is not (2.5 , \text{cm}).

Answer (b, c)

Given equation is

$$ y(x, t)=0.06 \sin (\frac{2 \pi x}{3}) \cos (120 \pi t) $$

(a) Comparing with a standard equation of stationary wave

$$ y(x, t)=a \sin (k x) \cos (\omega t) $$

Clearly, the given equation belongs to stationary wave. Hence, option (a) is not correct.

(b) By comparing,

$$ \begin{aligned} & \omega & =120 \pi \\ \Rightarrow & 2 \pi f & =120 \pi \Rightarrow f=60 Hz \end{aligned} $$

(c) $k=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda}$

$$ \Rightarrow \begin{aligned} \lambda & =\text { wavelength }=3 m \\ \text { Frequency } & =f=60 Hz \\ \text { Speed } & =v=f \lambda=(60 Hz)(3 m)=180 m / s \end{aligned} $$

(d) Since in stationary wave, all particles of the medium execute SHM with varying amplitude ’nodes.

• (a) It represents a progressive wave of frequency $60 Hz$

  The given equation ( y(x, t) = 0.06 \sin \left( \frac{2 \pi x}{3} \right) \cos (120 \pi t) ) matches the form of a stationary wave equation ( y(x, t) = a \sin (kx) \cos (\omega t) ). A progressive wave would have the form ( y(x, t) = A \sin (kx - \omega t) ) or ( y(x, t) = A \sin (kx + \omega t) ). Therefore, the given equation does not represent a progressive wave.

• (d) Amplitude of this wave is constant

  In a stationary wave, the amplitude varies along the length of the string. The amplitude is zero at the nodes and maximum at the antinodes. The given equation ( y(x, t) = 0.06 \sin \left( \frac{2 \pi x}{3} \right) \cos (120 \pi t) ) shows that the amplitude depends on the position ( x ) (specifically, it is ( 0.06 \sin \left( \frac{2 \pi x}{3} \right) )), indicating that the amplitude is not constant.

Answer (c, $d)$

Speed of sound waves in a fluid is given by $v=\sqrt{\frac{B}{\rho}}$, where $B$ is Bulk modulus and $\rho$ is density of the medium.

Clearly,

and

$$ \begin{matrix} v \propto \frac{1}{\sqrt{\rho}} & {[\therefore \text { for any fluid, } B=\text { constant }]} \\ v \propto \sqrt{B} & {[\because \text { for medium, } \rho=\text { constant }]} \end{matrix} $$

• (a) directly on density of the medium: This is incorrect because the speed of sound in a fluid is inversely proportional to the square root of the density, not directly proportional to the density. If the density increases, the speed of sound decreases.

• (b) square of Bulk modulus of the medium: This is incorrect because the speed of sound in a fluid is directly proportional to the square root of the Bulk modulus, not the square of the Bulk modulus. If the Bulk modulus increases, the speed of sound increases, but not as the square of the Bulk modulus.

Answer ( $b, c, d)$

During propagation of a plane progressive mechanical wave, like shown in the diagram, amplitude of all the particles is equal.

(i) Clearly, the particles $O, A$ and $B$ are having different phase.

(ii) Particles of the wave shown in the figure are having up and down SHM.

(iii) For a progressive wave propagating in a fluid.

Hence,

$$ \begin{aligned} \text { Speed } & =v=\sqrt{\frac{B}{\rho}} \\ v & \propto \sqrt{\frac{1}{\rho}} \end{aligned} $$

$$ [\because B \text { is constant }] $$

As $\rho$ depends upon nature of the medium, hence $v$ also depends upon the nature of the medium.

• Reason why option (a) is incorrect: During the propagation of a plane progressive mechanical wave, not all particles are vibrating in the same phase. Different particles are at different points in their cycles of motion, as indicated by the particles O, A, and B having different phases in the diagram.

Answer $(a, b, d)$

Given equation is

$$ y(x, t)=0.06 \sin (\frac{2 \pi}{3} x) \cos (120 \pi t) $$

Comparing with standard equation of stationary wave

$$ y(x, t)=a \sin (k x) \cos (\omega t) $$

It is represented by diagram.

where $N$ denotes nodes and $A$ denotes antinodes.

(a) Clearly, frequency is common for all the points.

(b) Consider all the particles between two nodes they are having same phase of $(120 \pi t)$ at a given time.

(c) and (d) But are having different amplitudes of $0.06 \sin (\frac{2 \pi}{3} x)$ and because of different amplitudes they are having different energies.

• Option (c) is incorrect because the points between two consecutive nodes have different amplitudes given by (0.06 \sin \left(\frac{2 \pi}{3} x\right)). Since energy is proportional to the square of the amplitude, different amplitudes result in different energies.

Thinking Process

When the wind is blowing in the same direction as that of sound wave then net speed of the wave is sum of speed of sound wave and speed of the wind.

Answer ( $a, b$ )

Given,

Speed of wind

$$ \begin{aligned} & v_0=400 Hz, v=340 m / s \\ & v_{w}=10 m / s \end{aligned} $$

(a) As both source and observer are stationary, hence frequency observed will be same as natural frequency $v_0=400 Hz$

(b) The speed of sound $v=v+v_{w}$

$$ =340+10=350 m / s $$

(c) and (d) There will be no effect on frequency,because there is no relative motion between source and observer hence (c),(d) are incorrect.

• (c) The frequency of sound as heard by the observer standing on the platform will increase: This is incorrect because there is no relative motion between the source (train) and the observer (standing on the platform). The wind affects the speed of sound but not the frequency when both the source and observer are stationary.

• (d) The frequency of sound as heard by the observer standing on the platform will decrease: This is incorrect for the same reason as (c). There is no relative motion between the source and the observer, so the frequency remains unchanged.

Answer $(a, b, d, e)$

Consider the equation of a stationary wave $y=a \sin (k x) \cos \omega t$

(a) clearly every particle at $x$ will have amplitude $=a \sin k x=$ fixed

(b) for mean position $y=0$

$$ \begin{matrix} \Rightarrow & & \cos \omega t & =0 \\ \Rightarrow & \omega t & =(2 n-1) \frac{\pi}{2} \end{matrix} $$

Hence, for a fixed value of $n$, all particles are having same value of

$$ \text { timet }=(2 n-1) \frac{\pi}{2 \omega} \quad[\because \omega=\text { constant }] $$

(c) amplitude of all the particles are a $\sin (k x)$ which is different for different particles at different values of $x$

(d) the energy is a stationary wave is confined between two nodes

(e) particles at different nodes are always at rest.

• (c) The amplitude of all the particles is (a \sin (kx)), which is different for different particles at different values of (x). Therefore, not all particles are oscillating with the same amplitude.

Answer Wire of twice the length vibrates in its second harmonic. Thus, if the tuning fork resonates at $L$, it will resonate at $2 L$. This can be explained as below

The sonometer frequency is given by

$$ v=\frac{n}{2 L} \sqrt{\frac{T}{m}} \quad(n=\text { number of loops }) $$

Now, as it vibrates with length $L$, we assume $v=v_1$

$$ \begin{aligned} n & =n_1 \\ \therefore \quad v_1 & =\frac{n_1}{2 L} \sqrt{\frac{T}{m}} \end{aligned} $$

When length is doubled, then

$$ v_2=\frac{n_2}{2 \times 2 L} \sqrt{\frac{T}{m}} $$

Dividing Eq. (i) by Eq. (ii), we get

$$ \frac{v_1}{v_2}=\frac{n_1}{n_2} \times 2 $$

To keep the resonance

$$ \begin{aligned} & \frac{v_1}{v_2}=1=\frac{n_1}{n_2} \times 2 \\ \Rightarrow \quad & n_2=2 n_1 \end{aligned} $$

Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.

Thinking Process

We should not confuse between pressure wave and displacement wave. By considering any type of wave outcome will be same.

Answer Consider the situation shown in the diagram

As the organ pipe is open at both ends, hence for first harmonic

$$ \begin{gathered} l=\frac{\lambda}{2} \\ \Rightarrow \quad \lambda=2 l \Rightarrow \frac{c}{v}=2 l \Rightarrow v=\frac{c}{2 l} \end{gathered} $$

where $c$ is speed of the sound wave in air.

For pipe closed at one end

$$ v^{\prime}=\frac{C}{4 L^{\prime}} $$

c for first harmonic

Hence,

$$ \begin{matrix} \Rightarrow & \frac{C}{2 L} & =\frac{C}{4 L^{\prime}} \\ \Rightarrow & \frac{L^{\prime}}{L} & =\frac{2}{4}=\frac{1}{2} \Rightarrow L^{\prime}=\frac{L}{2} \end{matrix} $$

Answer Frequency of tuning fork $A$,

$$ v_{A}=512 Hz $$

Probable frequency of tuning fork $B$,

$$ v_{B}=v_{A} \pm 5=512 \pm 5=517 \text { or } 507 Hz $$

when $B$ is loaded, its frequency reduces.

If it is $517 Hz$, it might reduced to $507 Hz$ given again a beat of $5 Hz$.

If it is $507 Hz$, reduction will always increase the beat frequency, hence $v_{B}=517 Hz$

Note For production of beats frequencies of the two tuning forks must be nearly equal i.e., slight difference in frequencies.

Answer Given, displacement of an elastic wave $y=3 \sin \omega t+4 \cos \omega t$ Assume,

$$ 3=a \cos \phi $$

$$ 4=a \sin \phi $$

On dividing Eq. (ii) by Eq. (i)

Also, $\quad a^{2} \cos ^{2} \phi+a^{2} \sin ^{2} \phi=3^{2}+4^{2}$

$\Rightarrow \quad a^{2}(\cos ^{2} \phi+\sin ^{2} \phi)=25$

Hence,

where

$$ a^{2} \cdot 1=25 \Rightarrow a=5 $$

Hence, amplitude $=5 cm$

Thinking Process

In a sitar wire, the vibration is assumed to be similar as a wire fixed at both ends.

Answer Frequency of vibrations produced by a stretched wire

$$ \begin{aligned} & v=\frac{n}{2 l} \sqrt{\frac{T}{\mu}} \\ & \text { Mass per unit length } \mu=\frac{\text { Mass }}{\text { Length }}=\frac{\pi r^{2} l \rho}{l}=\pi r^{2} \rho \quad[\because M=v p=A l \rho=\pi r^{2} l \rho] \\ & \therefore \quad v=\frac{n}{2 l} \sqrt{\frac{T}{\pi r^{2} \rho}} \Rightarrow v \propto \sqrt{\frac{1}{r^{2}}} \\ & v \propto \frac{1}{r} \\ & \text { Hence, when radius is tripled, } v \text { will be } \frac{1}{3} rd \text { of previous value. } \end{aligned} $$

Answer We know that speed of sound in air $v \propto \sqrt{T}$

$ \therefore \frac{v_{T}}{v_0} =\sqrt{\frac{T_{T}}{T_0}}=\sqrt{\frac{T_{T}}{273}}$

$\text { But } \frac{v_{T}}{v_0} =\frac{3}{1} $

$\therefore \frac{3}{1} =\sqrt{\frac{T_{T}}{T_0}} \Rightarrow \frac{T_{T}}{273}=9 $

$\therefore \quad \text { [where } T \text { is in kelvin] }$

$ T_{T} =273 \times 9=2457 K$

$ =2457-273=2184^{\circ} C$

Thinking Process

When two waves of almost equal frequencies interfere, they are producing beats.

Answer Let,

Beat frequency

$$ \begin{aligned} & n_1>n_2 \\ & v_{b}=n_1-n_2 \end{aligned} $$

Time period of beats $=T_{b}=\frac{1}{v_{b}}=\frac{1}{n_1-n_2}$

Answer Given, length of the wire

Mass of wire

Tension

Speed of transverse wave

$$ \begin{aligned} l & =12 m \\ m & =2.10 kg \\ T & =2.06 \times 10^{4} N \\ v & =\sqrt{\frac{T}{\mu}} \quad \quad \quad[\text { where } \mu=\text { mass per unit length] } \\ & =\sqrt{\frac{2.06 \times 10^{4}}{(\frac{2.10}{12})}}=\sqrt{\frac{2.06 \times 12 \times 10^{4}}{2.10}}=343 m / s \end{aligned} $$

Answer Length of pipe

$$ \begin{aligned} l & =20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m} \\ v_{\text {funda }} & =\frac{V}{4 L}=\frac{330}{4 \times 20 \times 10^{-2}} \quad \quad \quad \text{(for closed pipe)} \\ v_{\text {funda }} & =\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\ \frac{v_{\text {given }}}{v_{\text {funda }}} & =\frac{1237.5}{412.5}=3 \end{aligned} $$

Hence, 3rd harmonic node of the pipe is resonantly excited by the source of given frequency.

Answer As the source (train) is moving towards the observer (platform) hence apparent frequency observed is more than the natural frequency.

Frequency of whistle

Speed of train

Velocity of sound in air

$$ \begin{aligned} v & =400 Hz \\ v_{t} & =10 m / s \\ v & =330 m / s \end{aligned} $$

Apparent frequency when source is moving $v_{app}=(\frac{v}{v-v_{t}}) v$

$$ \begin{aligned} & =(\frac{330}{330-10}) 400 \\ \Rightarrow \quad v_{\text {app }} & =\frac{330}{320} \times 400=412.5 Hz \end{aligned} $$

Answer We have to observe the displacement and position of different points, then accordingly nature of two wave is decided.

Points on positions $x=10,20,30,40$ never move, always at mean position with respect to time. These are forming nodes which characterise a stationary wave.

$\because$ Distance between two successive nodes $=\frac{\lambda}{2}$

$\Rightarrow$

$$ \begin{aligned} \lambda & =2 \times(\text { node to node distance }) \\ & =2 \times(20-10) \\ & =2 \times 10=20 cm \end{aligned} $$

Answer Given, frequency of the wave $v=256 Hz$

Time period

$$ T=\frac{1}{v}=\frac{1}{256} S=3.9 \times 10^{-3} S $$

(a) Time taken to pass through mean position is

$$ t=\frac{T}{4}=\frac{1}{40}=\frac{3.9 \times 10^{-3}}{4} s=9.8 \times 10^{-4} s $$

(b) Nodes are $A, B, C, D, E$ (i.e., zero displacement) Antinodes are $A^{\prime}, C^{\prime}$ (i.e., maximum displacement)

(c) It is clear from the diagram $A^{\prime}$ and $C^{\prime}$ are consecutive antinodes, hence separation $=$ wavelength $(\lambda)$

$$ =\frac{v}{v}=\frac{360}{256}=1.41 m $$

$$ [\therefore v=v \lambda] $$

Thinking Process

The pipe partially filled with water, acts as closed organ pipe. According to this, we will find associated frequencies.

Answer Consider the diagram frequency of tuning fork $v=512 Hz$.

For observation of first maxima of intensity

(a) $L=\frac{\lambda}{4} \Rightarrow \lambda=4 L$

[for closed pipe]

$$ \begin{aligned} v & =v \lambda=512 \times 4 \times 17 \times 10^{-2} \\ & =348.16 m / s \end{aligned} $$

(b) We know that $v \propto \sqrt{T}$

where temperature $(T)$ is in kelvin.

$$ \begin{aligned} \frac{v_{20}}{v_0} & =\sqrt{\frac{273+20}{273+0}}=\sqrt{\frac{293}{273}} \\ \frac{v_{20}}{v_0} & =\sqrt{1.073}=1.03 \\ v_0 & =\frac{v_{20}}{1.03}=\frac{348.16}{1.03}=338 m / s \end{aligned} $$

(c) Resonance will be observed at $17 cm$ length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.

Answer Let, there are $n$ number of loops in the string.

Length corresponding each loop is $\frac{\lambda}{2}$.

Now, we can write

$ L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n} $

$\Rightarrow \frac{v}{v} =\frac{2 L}{n} \Rightarrow[\because v=v \lambda] $

$\Rightarrow v =\frac{n}{2 L} v=\frac{n}{2 L} \sqrt{\frac{T}{\mu}} \quad[\because \text { velocity of transverse waves }=\sqrt{T / \mu}] $

$\Rightarrow v \propto n \quad[\because \text { length and speed are constants }] $ $\text { So, } $

Answer Speed of wave in solid $=8 km / s$

Speed of wave in liquid $=5 km / s$

Required time $=[\frac{1000-0}{8}+\frac{3500-1000}{5}+\frac{6400-3500}{8}] \times 2 \quad[\because$ diameter $=$ radius $\times 2]$

$$ \begin{matrix} =[\frac{1000}{8}+\frac{2500}{5}+\frac{2900}{8}] \times 2 & {[\text { time }=\frac{\text { distance }}{\text { speed }}]} \\ =[125+500+362.5] \times 2=1975 & \end{matrix} $$

As we are considering at diametrically opposite point, hence there is a multiplication of 2.

Answer We know that rms speed of molecules of a gas

$$ c=\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}} $$

where $M=$ molar mass of the gas.

Speed of sound wave in gas $v=\sqrt{\frac{\gamma \rho}{\rho}}=\sqrt{\frac{\gamma R T}{M}}$

On dividing Eq. (i) by Eq. (ii), we get

$$ \frac{c}{V}=\sqrt{\frac{3 R T}{M} \times \frac{M}{\gamma R T}} \Rightarrow \frac{c}{V}=\sqrt{\frac{3}{\gamma}} $$

where $\gamma=$ adiabatic constant for diatomic gas

Hence,

$$ \gamma=\frac{7}{5} $$

$$ [\text { since } \gamma=\frac{C_{p}}{C_{V}}] $$

Thinking Process

To predict the nature of wave we have to compare with standard equations.

Answer (a) The equation $y=100 \cos (100 \pi t+0.5 x)$ is representing a travelling wave along $x$-direction.

(b) The equation $y=5 \cos (4 x) \sin (20 t)$ represents a stationary wave, because it contains sin, cos terms i.e., combination of two progressive waves

(c) As the equation $y=10 \cos [(252-250) \pi t] \cdot \cos [(252+250) \pi t]$ involving sum and difference of two near by frequencies 252 and 250 have this equation represents beats formation.

(d) As the equation $y=4 \sin (5 x-t / 2)+3 \cos (5 x-t / 2)$ involves negative sign with $x$, have if represents a travelling wave along $x$-direction.

Note We must not confuse with sign connected with $x$ and direction of propagation of wave. It is just reversed, positive sign with $x$ shown propagation of the wave in negative $x$-direction and vice-versa.

Answer Standard equation of a progressive wave is given by

$$ y=a \sin (\omega t-k x+\phi) $$

This is travelling along positive $x$-direction.

Given equation is $\quad y=5 \sin (100 \pi t-0.4 \pi x)$

Comparing with the standard equation

(a) Amplitude $=5 m$

(b) $k=\frac{2 \pi}{\lambda}=0.4 \pi$

$\therefore \quad$ Wavelength $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.4 \pi}=\frac{20}{4}=5 m$

(c) $\omega=100 \pi$

$$ \begin{aligned} \omega & =2 \pi \nu=100 \pi \\ \therefore \quad \text { Frequency } v & =\frac{100 \pi}{2 \pi}=50 Hz \end{aligned} $$

(d) Wave velocity $v=\frac{\omega}{k}$, where $k$ is wave number and $k=\frac{2 \pi}{\lambda}$.

(e)

$$ \begin{aligned} & =\frac{100 \pi}{0.4 \pi}=\frac{1000}{4} \\ & =250 m / s \\ y & =5 \sin (100 \pi t-0.4 \pi x) \\ \frac{d y}{d t} & =\text { particle velocity } \end{aligned} $$

From Eq. (i),

$$ \frac{d y}{d t}=5(100 \pi) \cos [100 \pi t-0.4 \pi x] $$

For particle velocity amplitude $(\frac{d y}{d t})_{\max }$

Which will be for ${\cos [100 \pi t-0.4 \pi x]}_{\max }=1$

$\therefore$ Particle velocity amplitude

$$ \begin{aligned} & =(\frac{d y}{d t})_{\max }=5(100 \pi) \times 1 \\ & =500 \pi m / s \end{aligned} $$

Answer Given, wave functions are

$$ \begin{aligned} y & =2 \cos 2 \pi(10 t-0.0080 x+3.5) \\ & =2 \cos (20 \pi t-0.016 \pi x+7 \pi) \end{aligned} $$

Now, standard equation of a travelling wave can be written as

$$ y=a \cos (\omega t-k x+\phi) $$

On comparing with above equation, we get

$$ \begin{aligned} & a=2 cm \\ & \omega=20 \pi rad / s \\ & k=0.016 \pi \end{aligned} $$

Path difference $=4 cm$

(a) Phase difference $\Delta \phi=\frac{2 \pi}{\lambda} \times$ Path difference

$$ \begin{aligned} \Delta \phi & =0.016 \pi \times 4 \times 100 \\ & =6.4 \pi rad \end{aligned} $$

(b) $\Delta \phi=\frac{2 \pi}{\lambda} \times(0.5 \times 100)$

$[\because$ Path difference $=0.5 m]$

$$ \begin{aligned} & =0.016 \pi \times 0.5 \times 100 \\ & =0.8 \pi rad \end{aligned} $$

(c) $\Delta \phi=\frac{2 \pi}{\lambda} \times(\frac{\lambda}{2})=\pi rad$

(d) $\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}=\frac{3 \pi}{2} rad$

(e) $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{20 \pi}=\frac{1}{10} s$

$$ \therefore \quad \begin{aligned} \text { At } x & =100 cm \\ t & =T \\ \phi_1 & =20 \pi T-0.016 \pi(100)+7 \pi \\ & =20 \pi(\frac{1}{10})-1.6 \pi+7 \pi=2 \pi-1.6 \pi+7 \pi \end{aligned} $$

Again, at $x=100 cm, t=5 s$

$$ \begin{aligned} \phi_2 & =20 \pi(5)-0.016 \pi(100)+7 \pi \\ & =100 \pi-(0.016 \times 100) \pi+7 \pi \\ & =100 \pi-1.6 \pi+7 \pi \end{aligned} $$

$\therefore$ From Eqs. (i) and (ii), we get

$\Delta \phi=$ phase difference $=\phi_2-\phi_1$

$$ \begin{aligned} & =(100 \pi-1.6 \pi+7 \pi)-(2 \pi-1.6 \pi+7 \pi) \\ & =100 \pi-2 \pi=98 \pi rad \end{aligned} $$