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,

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

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.

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.

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.

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.

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

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,

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

Answer (b)

Mass $m=2.5 kg$

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

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

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).

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 $=\lambda=3.48 m$.

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.

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

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

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.

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.

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.

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.

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 $=\phi_2-\phi_1$

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