Exemplar MCQs

Hint

(b) Water waves produced by a motorboat sailing on the surface of deep water are both longitudinal and transverse because the waves produce transverse as well as lateral vibrations in the particles of the medium. The water molecules at the surface move up and down; and back and forth simultaneously describing nearly circular paths as shown in Figure.

As the wave passes, water molecules at the crests move in the direction of the wave while those at the troughs move in the opposite direction.

Hint

(c) Frequency remains unchanged in both the medium.
So, \(\quad f_1=f_2 \Rightarrow \frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2} \Rightarrow \lambda_2=\left(\frac{v_2}{v_1}\right) \lambda_1\)
\(\lambda_1\) and \(\lambda_2\) are wavelengths and \(v_1\) and \(v_2\) are speeds in first and second medium respectively.
So, \(\quad \lambda_2=\left(\frac{2 v}{v}\right) \lambda_1=2 \lambda_1\)
Hence the wavelength of sound waves in the second medium is \(2 \lambda\).

Hint

(c) We know that speed of sound in air is given by
\(v=\sqrt{\frac{\gamma P}{\rho}}\), For air \(\gamma\) and \(P\) are constants.
\(v \propto \frac{1}{\sqrt{\rho}}\) where \(\rho\) is the density of air.
\(
\Rightarrow \quad \frac{v_2}{v_1}=\sqrt{\frac{\rho_2}{\rho_1}}
\)
where \(\rho_1\) is density of dry air and \(\rho_2\) is density of moist air. Due to the presence of moisture, density of air decreases.
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.

Hint

(c) Speed of sound wave in a medium \(v=\sqrt{\frac{\gamma R T}{M}}\). Here \(\gamma, R\) and \(M\) are constant.
Hence, \(v \propto \sqrt{T}\)
(where \(T\) is temperature of the medium)
It means when temperature changes, speed also changes.
As, \(v=f \lambda\), where \(f\) is frequency and \(\lambda\) is wavelength.
As frequency \((f)\) remains fixed, \(v \propto \lambda\) or \(\lambda \propto v\)
Hence wavelength \((\lambda)\) changes.

Hint

(b) A wave is a disturbance that propagates energy and momentum from one place to the other without the transport of matter. In 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.

Important point:
Characteristics of wave motion: 
1. It is a sort of disturbance which travels through a medium. 
2. Material medium is essential for the propagation of mechanical waves.
3. When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do leave their position and move with the disturbance.
4. There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1, and so on.

5. The velocity of the particle during their vibration is different at different positions.
6. The velocity of wave motion through a particular medium is constant.
7. It depends only on die nature of medium not on the frequency, wavelength, or intensity,
8. Energy is, propagated along with the wave motion without any net transport of the medium.

Hint

(c) In the case of a mechanical transverse wave propagates through a medium, the medium particles oscillate at right angles to the direction of wave motion or energy propagation. It travels in the form of crests and troughs.

When a mechanical transverse wave propagates through a medium element of the medium is subjected to shearing stress. Solids and strings have shear modulus, which is why, sustain shearing stress. Fluids have no shape of their own, they yield to shearing stress. Transverse waves can be transmitted through solids, they can be set up on the surface of liquids. But they cannot be transmitted into liquids and gases.

Hint

Key concept:

• When sound wave travels through a medium, say air, the particles of the medium disturb in the same fashion, i.e. compression and rarefaction (depression). When air particles come closer it is called compression. On the other hand, when particles go farther than their normal position it is called rarefaction. In the condition of compression, molecules of medium come closer to each other and in the condition of rarefaction, molecules of the medium go farther from each other; compared to their normal positions.
• When compression takes place in the medium, the density and pressure of the medium increase. When rarefaction takes place in the medium, the density and pressure of the medium decrease. This increase and decrease in density and pressure are temporary. Thus, compression is called the region of high density and pressure. Rarefaction is called the region of low density and pressure.

When sound wave travels through a medium:
1. Due to compression and rarefaction, density of the medium (air) changes. As density is changing, so Boyle’s law is not obeyed.
2. The Bulk modulus of the medium remains unchanged.
3. The time of compression and rarefaction is too small, i.e. we can assume an adiabatic process and hence no transfer of heat.

Hint

When a wave reflects from a denser medium, there is a phase change of \(\pi\). Given equation of incident wave
\(
y_i=0.6 \sin 2 \pi\left(t-\frac{x}{2}\right)
\)
Equation of reflected wave is
\(
y_r=A_r \sin 2 \pi\left(t+\frac{x}{2}+\pi\right)
\)
Amplitude of reflected wave
\(
A_r=\frac{2}{3} \times A_i=\frac{2}{3} \times 0.6=0.4 \text { units }
\)
So, \(\quad y_r=-0.4 \sin 2 \pi\left(t+\frac{x}{2}\right)\)

Hint

(b) The speed of transverse waves in any string \(v=\sqrt{\frac{T}{\mu}}\) \(\mu=\) mass per unit length
Mass \(m=2.5 \mathrm{~kg}\)
\(
\mu=\frac{m}{l}=\frac{2.5 \mathrm{~kg}}{20}=\frac{1.25}{10}=0.125 \mathrm{~kg} / \mathrm{m}
\)
Hence wave speed \(v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{0.125}}=\sqrt{1600} \mathrm{~m} / \mathrm{s}=40 \mathrm{~m} / \mathrm{s}\)
Time taken by the disturbance to travel from one end to the other end of the string, i.e. to travel a distance of \(20 \mathrm{~m}\) is
\(
t={20 \mathrm{~m}}{40 \mathrm{~m} / \mathrm{s}}=0.5 \mathrm{~s}
\)

Hint

\(
\text { When the train is approaching towards the observer. }
\)

Apparent frequency \(n_a=n_o\left(\frac{v}{v-v_s}\right)\)
It is clear that \(n_a>n_0\) :
When the train is going away from the observer.

Apparent frequency \(n_a=n_o\left(\frac{v}{v+v_s}\right)\)
It is clear that \(n_a>n_o\).
Hence, the expected curve is (c).

Hint

Given equation is
\(
y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)
\)
Option (a): Since there is +ve sign between \(\omega t\) and \(k x\), the wave travels from right to left (the positive direction of \(x\) is from left to right). Hence it is correct.
Option (b): Speed of the wave, \(v=\frac{\omega}{k}=\frac{36 \mathrm{~s}^{-1}}{0.018 \mathrm{~cm}}=2000 \mathrm{~cm} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}\) Hence it is correct.
Option (c): Frequency of the wave, \(v=\frac{\omega}{2 \pi}=\frac{36 \mathrm{~s}^{-1}}{2 \pi}=5.7 \mathrm{~Hz}\). Hence it is correct. correct.
Option (d): Least distance between two successive crests, \(\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.018 \mathrm{~cm}^{-1}}=349 \mathrm{~cm}\). Hence it is wrong.

Hint

Given equation is
\(
y(x, t)=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)
\)
Option (a): Comparing with a standard equation of stationary wave
\(
y(x, t)=2 a \sin (k x) \cos (\omega t)
\)
Clearly, the given equation belongs to stationary wave. Hence, option (a) is not correct.
Option (b): By comparing, \(\omega=120 \pi\)
\(\Rightarrow 2 \pi f=120 \pi \Rightarrow f=60 \mathrm{~Hz}\). Hence it is correct.
Option (c): \(k=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda}\)
\(\Rightarrow \quad \lambda=\) wavelength \(=3 \mathrm{~m}\)
Frequency \(=f=60 \mathrm{~Hz}\)
Speed \(=v=f \lambda=(60 \mathrm{~Hz})(3 \mathrm{~m})=180 \mathrm{~m} / \mathrm{s}\). Hence it is correct.
Option (d): Since in stationary wave, all particles of the medium execute SHM with varying amplitude nodes. Hence, option (d) is not correct.

Hint

(c, d) We define the speed of sound waves in a fluid as, \(v=\sqrt{\frac{B}{\rho}}\), Here \(B\) is the Bulk modulus and \(\rho\) is the density of the medium.
It means, \(v \propto \frac{1}{\sqrt{\rho}}\)
[ \(\therefore\) for any fluid, \(B=\) constant \(]\)
and \(\quad v \propto \sqrt{B}\)
[ \(\because\) for medium, \(\rho=\) constant \(]\)
Hence, options (c) and (d) are correct.

Hint

Key concept:
Characteristics of wave motion:
-When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do not move with the disturbance.
– Medium particles oscillate with same frequency and also the amplitude of oscillation of all the particles is equal. All the particles marked as \(1,2,3,4\) and 5 oscillate with the same frequency.
– There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1 and so on.
– The velocity of the particle during their vibration is different at different positions.
– The velocity of wave motion through a particular medium is constant. It depends only on the nature of medium not on the frequency, wavelength or intensity.

Option (a): Clearly, the particles 1,2 and 3 are having different phase.
Option (b) and (c): Particles of the wave executes SHM with same amplitude.
Option (d): The wave velocity of the mechanical wave depends only on the elastic and inertia property of medium for a progressive wave propagating in a fluid. Hence wave velocity depends upon the nature of the medium.
\(
\text { Speed }=v=\sqrt{\frac{B}{\rho}}
\)
\(
\text { Hence, } \quad v \propto \sqrt{\frac{1}{\rho}} \quad[\because B \text { is constant }]
\)
As \(\rho\) depends upon nature of the medium, hence \(v\) also depends upon the nature of the medium.

Hint

Key concept:
– The points for which amplitude is minimum are called nodes in a stationary wave, nodes are equally spaced at a distance \(\lambda / 2\).
– The points for which amplitude is maximum are called antinodes. Like nodes, antinodes are also equally spaced with spacing \((\lambda / 2)\) Furthermore, nodes and antinodes are alternate with spacing \((\lambda / 4)\).
– The nodes divide the medium into segments (or loops). All the particles in a segment vibrate in the same phase but in the opposite phase with the particles in the adjacent segment. Twice in one period, all the particles pass through their mean position simultaneously with maximum velocity (A at), the direction of motion being reversed after each half cycle.
Given equation is
\(
y(x, t)=0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)
\)
This represents a stationary wave. All the points on the string between two consecutive nodes vibrate with the same frequency But are having different amplitudes of \(0.06 \sin \left(\frac{2 \pi}{3} x\right)\) and because of different amplitudes they are having different energies.
Also, all the particles between two nodes they are having the same phase of \((120 \pi t)\) at a given time.

Hint

\((a, b)\) 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.
Given, \(f_0=400 \mathrm{~Hz}, v=340 \mathrm{~m} / \mathrm{s}\) Speed of wind \(v_w=10 \mathrm{~m} / \mathrm{s}\)
Option (a): As there is no relative motion between the source and observer, hence frequency observed will be the same as natural frequency \(f_0=400 \mathrm{~Hz}\)
Option (b): 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. The speed of sound \(v=v+v_w\) \(=340+10=350 \mathrm{~m} / \mathrm{s}\)
Option (b) and (c): There will be no change in frequency because there is no relative motion between source and observer. Hence (c) and (d) are incorrect.

Hint

Option (a): In stationary wave any particle at a given position have amplitude \(2 a \sin \mathrm{kx}\).
Option (b): The time period of oscillation of all the particles is same, hence all the particles cross their mean position at the same time.
Option (c): The amplitude of all the particles are \(2 a \sin \mathrm{kx}\) which is different for different particles at different values of \(x\)
Options (d) and (e): Nodes are the points which is always at rest hence no transfer of energy across the nodes. It means the energy is a stationary wave is confined between two nodes.

Hint

Wire of sonometer is twice the length which it vibrates in its second harmonic. Thus, if the tuning fork resonates at \(\mathrm{L}\), it will resonate at \(2 \mathrm{~L}\). This can be explained as below:
The frequency of sonometer is given by
\(
f=\frac{n}{2 L} \sqrt{\frac{T}{\mu}}=\frac{n v}{2 L}(n=\text { number of loops })
\)
For a given sonometer velocity of wave will be constant. If after changing the length of wire the tuning fork still be in resonance with the wire. Then, \(\frac{n}{L}\)
\(
\begin{aligned}
&=\text { constant } \Rightarrow \frac{n_1}{L_1}=\frac{n_2}{L_2} \\
& \frac{n_1}{L_1}=\frac{n_2}{2 L_1} \Rightarrow n_2=2 n_1
\end{aligned}
\)
Hence, when the wire is doubled the number of loops also gets doubled to produce the resonance. That is it resonates in the second harmonic.

Hint

For first harmonic of open organ pipe \(L=\frac{\lambda}{2}\)
\(
\Rightarrow \quad \lambda=2 L \Rightarrow \frac{v}{f_0}=2 L \Rightarrow\left(f_0\right)_{\mathrm{open}}=\frac{v}{2 L}
\)
where \(v\) is speed of the sound wave in air.
For first harmonic of closed organ pipe \(L^{\prime}=\frac{\lambda}{4}\)
\(
\begin{aligned}
& \Rightarrow \quad \lambda=4 L^{\prime} \Rightarrow \frac{v}{f_0}=4 L^{\prime} \Rightarrow\left(f_0\right)_{\mathrm{close}}=\frac{v}{4 L^{\prime}} \\
& \Rightarrow \quad \frac{v}{2 L}=\frac{v}{4 L^{\prime}} \quad[\because \text { speed remains constant }] \\
& \Rightarrow \quad \frac{L^{\prime}}{L}=\frac{2}{4}=\frac{1}{2} \Rightarrow L^{\prime}=\frac{L}{2}
\end{aligned}
\)

Hint

When the prong of \(B\) is loaded with wax, its frequency becomes less than the original frequency. If we assume that the original frequency of \(B\) is 507, then on loading its frequency will be less than 507. The beats between \(A\) and \(B\) will be more than 5.
If we assume that the original frequency of \(B\) is 517, then on loading its frequency will be less than 517. The beats between \(A\) and \(B\) may be equal to 5. Hence the frequency of the tuning fork \(B\) when not loaded should be 517.

Hint

Here, we have given: \(y=3 \sin \omega t+4 \cos \omega t\)
So, two components are there, \(3 \sin \omega t\) and \(4 \cos \omega t\)
where, individual amplitudes are given by
\(\mathrm{A}_1=3 \mathrm{~cms}\) and \(\mathrm{A}_2=4 \mathrm{~cms}\).
so, the resultant amplitude will be,
\(
\mathrm{A}=\sqrt{\mathrm{A}_1^2+\mathrm{A}_2^2}=\sqrt{3^2+4^2}
\)
\(
A=5 \mathrm{~cms}
\)

Hint

Given, displacement of the wave
\(
\begin{aligned}
& y=3 \sin \omega x+4 \cos \omega x \\
& \text { Let us assume, } 3=A \cos \theta \dots(i)\\
& 3=A \cos \theta \dots(ii)
\end{aligned}
\)
On dividing Eq. (ii) by Eq. (i)
\(
\tan \theta=4 / 3=>\phi=\tan ^{-1}(4 / 3)
\)
Squaring and adding equations (i) and (ii),
\(
\begin{aligned}
& A^2 \cos ^2 \theta+A^2 \sin ^2 \theta=3^2+4^2 \\
& \Rightarrow \quad A^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=25 \\
& A^2=25=>A=5 \text {. Hence, amplitude }=5 \mathrm{~cm}
\end{aligned}
\)
Frequency of vibrations, \(f=\frac{n}{2 l} \sqrt{\frac{T}{\mu}}\)
Mass per unit length
\(
\begin{aligned}
& \mu=\frac{\text { Mass }}{\text { Length }}=\frac{\pi r^2 l \rho}{l}=\pi r^2 \rho \\
& \therefore \quad f=\frac{n}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}} \Rightarrow f \propto \sqrt{\frac{1}{r^2}} \\
& \text { or } \quad f \propto \frac{1}{r} \\
&
\end{aligned}
\)
Hence, when the radius is tripled, the frequency will be \(\frac{1}{3}\) rd of the previous value.

Hint

We know that speed of sound in air \(v=\sqrt{\frac{\gamma R T}{M}} \Rightarrow v \propto \sqrt{T}\)
\(
\therefore \quad \frac{v_T}{v_0}=\sqrt{\frac{T_T}{T_0}}=\sqrt{\frac{T_T}{273}}
\)
But it is given, \(\frac{v_T}{v_0}=\frac{3}{1}\)
\(
\begin{array}{rlrl}
\Rightarrow & \frac{3}{1} & =\sqrt{\frac{T_T}{T_0}} \Rightarrow \frac{T_T}{273}=9 \\
& \therefore & T_T =273 \times 9=2457 \mathrm{~K} \\
& & =2457-273=2184^{\circ} \mathrm{C}
\end{array}
\)

Hint

If two waves of almost equal frequencies interfere, they are producing beats.
Let \(n_1>n_2\)
Beat frequency \(f_{\text {beat }}=n_1-n_2\)
\(\therefore\) Time period of beats \(T_{\text {beat }}=\frac{1}{f_{\text {beat }}}=\frac{1}{n_1-n_2}\)
This time period will be equal to the time interval between successive maxima.

Hint

Given, length of the wire \(l=12 \mathrm{~m}\)
Mass of wire \(m=2.10 \mathrm{~kg}\)
Tension in wire \(T=2.06 \times 10^4 \mathrm{~N}\)
Speed of transverse wave \(v=\sqrt{\frac{T}{\mu}}\)
where \(\mu=\) Linear mass density \(=\) Mass per unit length, \(\mu=\frac{2.10}{12} \mathrm{~kg} / \mathrm{m}\)
\(
\Rightarrow \quad v=\sqrt{\frac{2.06 \times 10^4}{\left(\frac{2.10}{12}\right)}}=343 \mathrm{~m} / \mathrm{s}
\)

Hint

Length of pipe, \(l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\) Fundamental frequency of closed organ pipe
\(
\begin{aligned}
& f_0=\frac{v}{4 L}=\frac{330}{4 \times 20 \times 10^{-2}}=412.5 \mathrm{~Hz} \\
& \frac{f_{\text {given }}}{f_0}=\frac{1237.5}{412.5}=3
\end{aligned}
\)
It means third harmonic node of the pipe is resonantly excited by the source of given frequency.

Hint

Apparent frequency \(n_a=n_o\left(\frac{v}{v-v_s}\right)\)
Frequency of whistle \(n_0=400 \mathrm{~Hz}\)
Speed of train \(v_s=10 \mathrm{~m} / \mathrm{s}\)
Velocity of sound in air \(v=330 \mathrm{~m} / \mathrm{s}\)
Apparent frequency when source is moving
\(
n_a=n_o\left(\frac{v}{v-v_s}\right)=400\left(\frac{330}{330-10}\right)=412.5 \mathrm{~Hz}
\)

Hint

If we observe the graph there are some points on the graph which are always at rest. The points on positions \(x=10,20,30,40\) never move, always at mean position with respect to time. These are forming nodes which characterize a stationary wave.
We know the distance between two successive nodes is equal to \(\lambda / 2\)
\(
\begin{aligned}
& \Rightarrow \quad \lambda=2 \times \text { (node to node distance) } \\
& =2 \times(20-10)=20 \mathrm{~cm}
\end{aligned}
\)

Hint

Given, frequency of the wave \(f=256 \mathrm{~Hz}\)
\(
\text { Time period } T=\frac{1}{f}=\frac{1}{256} \mathrm{~s}=3.9 \times 10^{-3} \mathrm{~s}
\)
(a) In the second plot, the displacement of each particle is zero. It means all the points are crossing the mean position. At the first plot point at \(A^{\prime}\) is at amplitude position. The time taken to move from amplitude position to mean position to is equal to one-fourth of the time period.
Hence, \(t=\frac{T}{4}=\frac{1}{40}=\frac{3.9 \times 10^{-3}}{4} \mathrm{~s}=9.8 \times 10^{-4} \mathrm{~s}\)
Method 2: Wavelength \(\lambda=\frac{v}{f}=\frac{360 \mathrm{~m} / \mathrm{s}}{256 \mathrm{~Hz}}=1.406 \mathrm{~m}\)
\(
A A^{\prime}=\frac{\lambda}{4}=\frac{1.406 \mathrm{~m}}{4}=0.3516 \mathrm{~m}
\)
Time \((t)\), at which the second curve is plotted
\(
t=\frac{A A^{\prime}}{v}=\frac{0.3516 \mathrm{~m}}{360 \mathrm{~m} / \mathrm{s}}=0.000976 \mathrm{~s}=9.8 \times 10^{-4} \mathrm{~s}
\)
(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
\(
\Rightarrow \quad \lambda=\frac{v}{f}=\frac{360}{256}=1.41 \mathrm{~m}
\)

Hint

If a pipe is partially filled with water whose upper surface of the water acts as a reflecting surface of a closed organ pipe. If the length of the air column is varied until its natural frequency equals the frequency of the fork, then the column resonates and emits a loud note.
The frequency of the tuning fork, \(\mathrm{f}=512 \mathrm{~Hz}\).
For observation of the first maxima of intensity,
The frequency of the tuning fork, \(f=512 \mathrm{~Hz}\).
For observation of the first maxima of intensity,
(a) For the first maxima of intensity, the length of the air column \(l=\frac{\lambda}{4} \Rightarrow \lambda=4 l=4 \times 17 \times 10^{-2} \mathrm{~m}\)
Hence speed of sound \(v=f \lambda=512 \times\left(4 \times 17 \times \times 10^{-2}\right)\) \(=348.16 \mathrm{~m} / \mathrm{s}\)
(b) We know that \(v \propto \sqrt{T}\) where temperature \((T)\) is in kelvin.
\(
\begin{aligned}
& \frac{v_{20}}{v_0}=\sqrt{\frac{273+20}{273+0}}=\sqrt{\frac{293}{273}} \\
& \frac{v_{20}}{v_0}=\sqrt{1.073}=1.03 \\
& v_0=\frac{v_{20}}{1.03}=\frac{348.16}{1.03}=338 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
(c) The resonance will still be observed for \(17 \mathrm{~cm}\) length of air column above mercury. However, due to more complete
reflection of sound waves at mercury surface, the intensity of reflected sound increases.

Hint

Find the time taken by the wave in the solid part.
Given, As the innermost part is solid,
\(
r_1=d_1=1000 \mathrm{~km}
\)
And Outer most part is also solid,
\(
d_3=r_3-r_2
\)
\(
d_3=6400-3500=2900 \mathrm{~km}
\)
So, distance travelled by the wave in solid part, \(2\left(d_1+d_3\right)\)
And speed of the wave in soild part \(=8 \mathrm{~km} / \mathrm{s}\)
So, time taken by the wave in solid part \(=\frac{\text { distance travelled }}{\text { speed }}\)
\(
\begin{aligned}
& t_1=\frac{2\left(d_1+d_2\right)}{8} \\
& t_1=\frac{2(1000+2900)}{8}=975 \mathrm{~s}
\end{aligned}
\)
Find time taken by the wave in the liquid part.
Given, the middle part is liquid.
Liquid distance, \(=2 d_2=2\left(r_2-r_1\right)\)
\(2(3500-1000)=5000 \mathrm{~km}\)
And speed of the wave in liquid part \(=5 \mathrm{~km} / \mathrm{s}\)
So, time taken by the wave in liquid part,
\(
\begin{aligned}
& =\frac{\text { distance travelled }}{\text { speed }} \\
& t_2=\frac{2\left(d_2\right)}{5} \\
& t_2=\frac{5000}{5}=1000 \mathrm{~s}
\end{aligned}
\)
Find the total time taken by the wave
Total time taken, \(\mathrm{t}=\mathrm{t}_1+\mathrm{t}_2\)
\(
\mathrm{t}=(975+1000) \mathrm{s}
\)
\(
\mathrm{t}=32 \min 55 \mathrm{~s}
\)

Hint

R.m.s speed of molecules of a gas
\(
\begin{aligned}
& \mathrm{c}=\sqrt{\frac{3 P}{\rho}} \\
& \mathrm{c}=\sqrt{\frac{3 R T}{M}}(\mathrm{i})[\mathrm{M}=\text { Molar mass }] \\
& \because \mathrm{PV}=\mathrm{nRT} \\
& \mathrm{n}=1 \\
& \text { Or } \mathrm{P}=\frac{R T}{V}
\end{aligned}
\)
\(
\therefore \frac{p}{\rho}=\frac{R T}{M}
\)
Speed of sound wave in gas, \(v=\sqrt{\frac{r P}{\rho}}\)
\(
\mathrm{V}=\sqrt{\frac{r R T}{M}} \text { (ii) }
\)
Dividing eqn (ii) by eqn (i),
\(
\begin{aligned}
& \frac{c}{v}=\frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{r R T}{M}}} \\
& \frac{c}{v}=\sqrt{\frac{3}{r}} \text { [r = adiabatic constant for diatomic gas] } \\
& r=\frac{7}{5} \\
& \text { Thus, } \frac{c}{v}=\text { constant }
\end{aligned}
\)

Hint

No, in wave motion there is no actual transfer of matter but the transfer of energy between the points where as when the wind blows air particles moves with it.

Hint

\(
\text { It is a non-mechanical wave because this type of wave does not require a material medium to travel. }
\)

Hint

Equation of the wave is \(y=c_1 \sin \left(c_2 x+c_3 t\right)\)
When the variable of the equation is \(\left(c_2 x+c_3 t\right)\), then the wave must be moving in the negative \(x\)-axis with time \(t\).

Hint

A sine wave has a maxima and a minima and the particle displacement has phase difference of \(\pi\) radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to \(\lambda / 2\)

Hint

The minima of the sine wave function occur at 0 and after this, the next minima of the sine wave occurs at \(\pi\). The wavelength of the sine wave function is \(2 \pi\) i.e. one wave completes its full waveform in period of \(2 \pi\). then the particle closest to it having zero displacement is at a distance \(\pi\) and is equal to \(\lambda / 2\)

Hint

The direction of the displacement of a wave is perpendicular to the wave’s motion direction in transverse waves. When a transverse wave is moving in a \(y\)-direction, then the displacement of the wave will be in the \(x\)-direction, and if the wave which is travelling along the \(x\)-axis, its displacement will be towards the \(y\)-axis.
The equation for the wave travelling along the \(y\)-axis is \(x=A \sin (k y-\omega t)\)

Hint

\(
\mathrm{y}=\mathrm{A} \sin ^2(\mathrm{kx}-\omega \mathrm{t})=\frac{\mathrm{A}}{2}\{1-\cos (2 \mathrm{kx}-2 \omega \mathrm{t})\}
\)
Comparing the above equation with the standard progressive wave equation \(y=A \sin (k x-\omega t)\), we get
\(
\text { Frequency } f^{\prime}=\frac{\omega^{\prime}}{2 \pi}=\frac{2 \omega}{2 \pi}=\frac{\omega}{\pi} \text { and amplitude } \mathrm{A}^{\prime}=\mathrm{A} / 2
\)

Hint

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

Hint

\(f\) The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

Hint

The velocity of a transverse wave in a stretched string \({v}=\sqrt{\frac{T}{\mu}}\) where \(T\) is the tension of the string and \(\mu\) is mass per unit length of the string. \(\mu=\pi r^2 \times \rho\)
where \(r\) is the radius or the string and \(\rho\) is the density of the material of the string.
\(
\therefore {v}=\frac{1}{\mathrm{r}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}
\)
Since \(T, \rho\) are constant
\(
\begin{aligned}
& \therefore {v} \propto \frac{1}{\mathrm{r}} \\
& \frac{{v}_{{A}}}{{v}_{{B}}}=\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}=\left(\frac{\mathrm{r}_{\mathrm{B}}}{2 \mathrm{r}_{\mathrm{B}}}\right)=\frac{1}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& T_{A B}=T \\
& T_{C D}=2 T
\end{aligned}
\)
where
\(T_{A B}\) is the tension in the string \(A B\)
\(T_{C D}\) is the tension in the string \(C D\)
The eelation between tension and the wave speed is given by \(v=\sqrt{\frac{T}{\mu}}\)
\(v \propto \sqrt{T}\)
where
\(v\) is the wave speed of the transverse wave
\(\mu\) is the mass per unit length of the string
\(
\frac{v_1}{v_2}=\sqrt{\frac{T}{2 T}}=\frac{1}{\sqrt{2}}
\)

Hint

The sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

Hint

As \(v=\sqrt{\frac{f}{\mu}}\)
A wave pulse travels faster in a thinner string. The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

Hint

Given, \(y_1=A \sin (k x-\omega t)\)
\(
\begin{aligned}
& \mathrm{y}_2=\mathrm{A} \cos (\mathrm{kx}-\omega \mathrm{t}) \\
& \text { or } \mathrm{y}_2=\mathrm{A} \sin \left(\mathrm{kx}-\omega \mathrm{t}+\frac{\pi}{2}\right)
\end{aligned}
\)
Phase difference of two waves \(=\frac{\pi}{2}\)
\(\because\) Resultant amplitude
\(
\begin{aligned}
& \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{AA} \cos \phi} \\
& =\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos \frac{\pi}{2}} \\
& =\sqrt{2 \mathrm{~A}^2} \quad\left(\because \cos \frac{\pi}{2}=0\right)
\end{aligned}
\)

Hint

the information is insufficient to find the relation between \(t_1\) and \(t_2\).
\(
v=\sqrt{\frac{\eta}{\rho}}
\)
But because the length of wires \(A\) and \(B\) is not known, the relation between \(A\) and \(B\) cannot be determined.

Hint

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

Hint

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

Hint

We know resultant amplitude is given by
\(A_{\text {net }}=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}\) For maximum resultant amplitude \(A_{\max }=A_1+A_2\) For minimum resultant amplitude \(A_{\min }=A_1-A_2\)
So, the difference between \(\mathrm{A}_{\max }\) and \(\mathrm{A}_{\min }\) is
\(
A_{\max }-A_{\min }=A_1+A_2-A_1+A_2=2 A_2
\)

Hint

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle \((\phi)\). So, the resultant amplitude lies between the maximum resultant amplitude \(\left(A_{\max }\right)\) and the minimum resultant amplitude \(\left(A_{\min }\right)\).
\(
\begin{aligned}
& A_{\max }=A+A=2 A \\
& A_{\min }=A-A=0
\end{aligned}
\)

Hint

We know the resultant amplitude is given by
\(
\begin{aligned}
& R_{n e t}=\sqrt{A^2+A^2+2 A^2 \cos 120^{\circ}}\left(\phi=120^{\circ}\right) \\
& =\sqrt{2 A^2-A^2}\left[\because \cos 120^{\circ}=\frac{-1}{2}\right] \\
& =A
\end{aligned}
\)

Hint

The relation between fundamental frequency and the length of the string is given by
\(
f=\frac{1}{2 l} \sqrt{\frac{F}{\mu}}
\)
where
\(l\) is the length of the string
\(F\) is the tension
\(\mu\) linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
\(
f \propto \frac{1}{l}
\)

Hint

This is not the case of free oscillations. This is forced oscillation where natural frequency is \(240 \mathrm{~Hz}\) and the driver frequency is \(480 \mathrm{~Hz}\). It vibrates with driver frequency i.e., \(480 \mathrm{~Hz}\), which corresponds to the second mode of vibration of the wire.

Hint

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

Hint

When the length is doubled keeping other things constant, the fundamental frequency must be halved as \(f\) \(\propto \frac{1}{l}\) But the tuning fork is still vibrating with 416 \(\mathrm{Hz}\), which will be the second harmonic of the string. The string will vibrate with this frequency only

Hint

According to the relation of the fundamental frequency of a string
\(
f=\frac{1}{2 l} \sqrt{\frac{F}{\mu}}
\)
where \(l\) is the length of the string
\(F\) is the tension
\(\mu\) is the linear mass density of the string
We know that \(f_1=416 \mathrm{~Hz}, l_1=l\) and \(l_2=2 l\).
Also, \(m_1=4 \mathrm{~kg}\) and \(m_2=\) ?
\(
\begin{aligned}
& f_1=\frac{1}{2 l_1} \sqrt{\frac{m_1 g}{\mu}} \ldots (1)\\
& f_2=\frac{1}{2 l_2} \sqrt{\frac{m_2 g}{\mu}} \ldots (2)
\end{aligned}
\)
So, in order to maintain the same fundamental mode
\(
f_1=f_2
\)
squaring both sides of equations (1) and (2) and then equating
\(
\begin{aligned}
& \frac{1}{4 l^2} \frac{4 g}{\mu}=\frac{1}{16 l^2} \frac{m_2 g}{\mu} \\
& \Rightarrow m_2=16 \mathrm{~kg}
\end{aligned}
\)

Hint

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

Hint

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the \(Z\)-axis, the particles will move in the \(X-Y\) plane.

Hint

Longitudinal waves possess all the characteristics of a wave. It has unique wavelength and velocity, and also transfer energy from one place to another.
In a Longitudinal wave, vibration of the particles of the medium is along the direction of propagation of wave.

For polarisation to occur, vibration of the particles of the medium should be perpendicular to the direction of propagation.
Hence, a Longitudinal wave cannot be polarised.
As sound waves are a good example of longitudinal waves, Sound waves does not exhibit polarisation.

Hint

In solids, the molecules can vibrate in any direction. So one can set both longitudinal and transverse waves in a solid.

Hint

\(
\text { Because particles in a gas are far apart, only longitudinal wave can travel through it. }
\)

Hint

Equation of the traveling wave
\(y=a \sin (\omega t-k x+\phi) \dots(i)\).
Phase of particle (1)
\(\phi_1=\omega t-k x_1+\phi \ldots\) (ii)
\(\Rightarrow\) Phase of particle (2)
\(\phi_2=\omega t-k x_2+\phi \ldots\) (iii)
Now, it is given that, \(\phi_2-\phi_1=\pi\)
Using (i).
\(
\begin{aligned}
& y_1=a \sin \phi_1 \\
& y_2=a \sin \left(\phi_1+\pi\right) \\
& \Rightarrow y_2=-a \sin \phi_1=-y_1 \\
& \Rightarrow\left|y_2\right|=\left|y_1\right|
\end{aligned}
\)
Also, the velocity of the particle is
\(
\begin{aligned}
& v_p=\left.\frac{\partial y}{\partial t}\right|_x \\
& \Rightarrow v=a \omega \cos (\omega t-k x+\phi) \\
& \therefore v_1=a \omega \cos \phi_1 \\
& \text { and } v_2=a \omega \cos \left(\phi_1+\pi\right) \\
& \Rightarrow v_2=-a \omega \cos \phi_1=-v_1 \\
& \therefore A \text { and } B \text { move in opposite directions. }
\end{aligned}
\)
\(A\) and \(B\) have a phase difference of \(\pi\). So, when a sine wave passes through the region, they move in opposite directions and have equal displacement.

Hint

\(
y=(0 \cdot 001 m m) \sin \left[\left(50 s^{-1}\right) t+\left(2 \cdot 0 m^{-1}\right) x\right]
\)
Equating the above equation with the general equation, we get:
\(
\begin{aligned}
y & =A \sin (\omega t-k x) \\
\omega & =\frac{2 \pi}{T}=2 \pi f \\
k & =\frac{2 \pi}{\lambda}
\end{aligned}
\)
Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, \(k\) is the wave number and \(\lambda\) is the wavelength.
\(
A=0.001 \mathrm{~mm}
\)
Now,
\(
\begin{aligned}
& 50=2 \pi f \\
& \Rightarrow f=\frac{25}{\pi} H z
\end{aligned}
\)

Hint

The frequency (f) of a standing wave, fixed at one end and free at the other end is:
\(
\mathrm{f}=\left(\frac{2 n+1}{2}\right) \frac{v}{2 L}
\)
Since \(v=f \lambda\)
\(
\begin{aligned}
& \Rightarrow \mathrm{f}=\left(n+\frac{1}{2}\right) \frac{v \lambda}{2 L} \\
& \Rightarrow \mathrm{L}=\left(\frac{2 n+1}{4}\right) \lambda \\
& \Rightarrow \mathrm{L}=\frac{\lambda}{4}, \frac{3 \lambda}{4}, \ldots
\end{aligned}
\)
Therefore the length of the string is an odd integral multiple of \(\frac{\lambda}{4}\).

Hint

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

Hint

In a stationary wave, all the particles between consecutive nodes vibrate in phase. Option (d) is correct. The particles on the different sides of a node do not vibrate in phase but they have a phase difference of \(\pi\). So the alternate parts between the consecutive nodes vibrate in phase. Thus the options (a) and (b) are not true but (c) is true.

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

\(
\text { Thus, particles in alternate antinodes vibrate in the same phase. }
\)