Chapter 11 - Waves - 11th Science Guide Samacheer Kalvi Solutions - Chapter 11 Waves 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Waves
Textual Evaluation Solved
Multiple Choice Questions
Question 1.
A student tunes his guitar by striking a 120 Hertz with a tuning fork, and
simultaneously plays the $4^{\text {th }}$ string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the $4^{\text {th }}$ string on his guitar?
(a) 130
(b) 117
(c) 110
(d) 120
Answer:
(b) 117
Question 2.
A transverse wave moves from a medium $\mathrm{A}$ to a medium $\mathrm{B}$. In medium $\mathrm{A}$, the velocity of the transverse wave is $500 \mathrm{~ms}^{-1}$ and the wavelength is $5 \mathrm{~m}$. The frequency and the wavelength of the wave in medium $\mathrm{B}$ when its velocity is $600 \mathrm{~ms}^{-1}$, respectively are
(a) $120 \mathrm{~Hz}$ and $5 \mathrm{~m}$
(b) $100 \mathrm{~Hz}$ and $5 \mathrm{~m}$
(c) $120 \mathrm{~Hz}$ and $6 \mathrm{~m}$
(d) $100 \mathrm{~Hz}$ and $6 \mathrm{~m}$
Answer:
(d) $100 \mathrm{~Hz}$ and $6 \mathrm{~m}$

Question 3.
For a particular tube, among six harmonic frequencies below $1000 \mathrm{~Hz}$, only four harmonic frequencies are given : $300 \mathrm{~Hz}, 600 \mathrm{~Hz}, 750 \mathrm{~Hz}$ and $900 \mathrm{~Hz}$. What are the two other frequencies missing from this list?
(a) $100 \mathrm{~Hz}, 150 \mathrm{~Hz}$
(b) $150 \mathrm{~Hz}, 450 \mathrm{~Hz}$
(c) $450 \mathrm{~Hz}, 700 \mathrm{~Hz}$
(d) $700 \mathrm{~Hz}, 800 \mathrm{~Hz}$
Answer:
(b) $150 \mathrm{~Hz}, 450 \mathrm{~Hz}$
Hint:
If the tube is open at both ends so the harmonic frequencies are based on $150 \mathrm{~Hz}$. $1^{\text {st }}=150 \mathrm{~Hz} ; 2^{\text {nd }}=300 \mathrm{~Hz} ; 3^{\text {rd }}=450 \mathrm{~Hz} ; 4^{\text {th }}=600 \mathrm{~Hz} ; 5^{\text {th }}=750 \mathrm{~Hz} ; 6^{\text {th }}=900 \mathrm{~Hz}$ The above frequencies the missing frequency in the list $150 \mathrm{~Hz}, 450 \mathrm{~Hz}$

Question 4.
Which of the following options is correct?

Options for (1), (2) and (3), respectively are
(a) (B), (C) and (A)
(b) (C), (A) and (B)
(c) (A), (B) and (C)
(d) (B), (A) and (C)
Answer:
(a) (B), (C) and (A)
Question 5.
Compare the velocities of the wave forms given below, and choose the correct option.

where, $v_A, v_B, v_C$ and $v_D$ are velocities given in (A), (B), (C) and (D), respectively.
(a) $\mathrm{V}_{\mathrm{A}}>\mathrm{V}_{\mathrm{B}}>\mathrm{V}_{\mathrm{D}}>\mathrm{V}_{\mathrm{C}}$
(b) $\mathrm{V}_{\mathrm{A}}<\mathrm{V}_{\mathrm{B}}<\mathrm{V}_{\mathrm{D}}<\mathrm{V}_{\mathrm{C}}$
(c) $V_A=V_B=V_D=V_C$
(d) $V_A>V_B=V_D>V_C$
Answer:
(c) $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}$
Question 6.
A sound wave whose frequency is $5000 \mathrm{~Hz}$ travels in air and then hits the water surface. The ratio of its wavelengths in water and air is .......
(a) 4.30
(b) 0.23
(c) 5.30
(d) 1.23
Answer:

(a) 4.30
Hint.
Frequency of sound, $\mathrm{f}=5000 \mathrm{~Hz}$
Velocity of sound in air, $v_{\mathrm{a}}=343 \mathrm{~ms}^{-1}$
Velocity of sound in water, $v_{\mathrm{w}}=1480 \mathrm{~ms}^{-1}$
The ratio of wavelength $\frac{\lambda_a}{\lambda_w}=\frac{v_w}{f} \times \frac{f}{v_a}=\frac{1480}{5000} \times \frac{5000}{343}=\frac{1480}{343}$
$
=4.31
$
Question 7.
A person standing between two parallel hills fires a gun and hears the first echo after $t_1$ $\mathrm{sec}$ and the second echo after $\mathrm{t}_2 \mathrm{sec}$. The distance between the two hills is .....
(a) $\frac{v\left(t_1-t_2\right)}{2}$
(b) $\frac{v\left(t_1 t_2\right)}{2\left(t_1+t_2\right)}$
(c) $v\left(t_1+t_2\right)$
(d) $\stackrel{v\left(t_1\right.}{ }$
Answer:
(d) $\frac{v\left(t_1+t_2\right)}{2}$
Hint:
Distance between man and hill $1, d_1=v \times \frac{t_1}{2}$

Distance between man and hill 2, $d_2=v \times \frac{t_2}{2}$
Distance between the two hills, $d=d_1+d_2$
$
d=\frac{v\left(t_1+t_2\right)}{2}
$
Question 8.
An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency $83 \mathrm{~Hz}$. Then the length of the air column is
(a) $1.5 \mathrm{~m}$
(b) $0.5 \mathrm{~m}$
(c) $1.0 \mathrm{~m}$
(d) $2.0 \mathrm{~m}$
Answer:
(c) $1.0 \mathrm{~m}$

Hint:
Frequency, $f=\frac{v}{\lambda}$
$83=\frac{v}{\lambda}$
For air column in a pipe closed at one end,
$
\begin{aligned}
l_1 & =\frac{\lambda}{4} \Rightarrow \lambda=4 l_1 \\
\therefore \quad 83 & =\frac{v}{4 l_1}
\end{aligned}
$
Velocity of sound in air $y=343 \mathrm{~ms}^{-1}$
$
\begin{aligned}
83 & =\frac{343}{4 l_1} \\
\therefore \quad l_1 & =\frac{343}{332}=1.033 \mathrm{~m}
\end{aligned}
$
Question 9.
The displacement $\mathrm{y}$ of a wave travelling in the $\mathrm{x}$ direction is given by $y=\left(2 \times 10^{-3}\right) \sin \left(300 t-2 x+\frac{\pi}{4}\right)$, where $\mathrm{x}$ and $\mathrm{y}$ are measured in metres and $t$ in second. The speed of the wave is

(a) $150 \mathrm{~ms}^{-1}$
(b) $300 \mathrm{~ms}^{-1}$
(c) $450 \mathrm{~ms}^{-1}$
(d) $600 \mathrm{~ms}^{-1}$
Answer:
(a) $150 \mathrm{~ms}^{-1}$
Hint:
From the standard equation of wave,
$
y=a \sin (\omega t-k x+\phi)
$
Here, $\omega=600$ and $k=2$
So, speed of wave is, $v=\frac{\omega}{k}=\frac{600}{2} ; v=150 \mathrm{~m} / \mathrm{s}$

Question 10.
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tensions, densities, lengths and diameter of the two wires are in the ratio $8: 1,1: 2, \mathrm{x}$ : $\mathrm{y}$ and $4: 1$ respectively. If the note of the higher pitch has a frequency of $360 \mathrm{~Hz}$ and the number of beats produced per second is 10 , then the value of $x: y$ is ........
(a) $36: 35$
(b) $35: 36$
(c) $1: 1$
(d) $1: 2$
Answer:
(a) $36: 35$
Question 11.
Which of the following represents a wave?
(a) $(\mathrm{x}-\mathrm{vt})^3$
(b) $x(x+v t)$
(c) $\frac{1}{(x+v t)}$
(d) $\sin (\mathrm{x}+\mathrm{vt})$
Answer:
(d) $\sin (x+v t)$
Question 12.
A man sitting on a swing which is moving to an angle of $60^{\circ}$ from the vertical is blowing a whistle which has a frequency of $2.0 \mathrm{k} \mathrm{Hz}$. The whistle is $2.0 \mathrm{~m}$ from the fixed support point of the swing. A sound detector which detects the whistle sound is kept in front of the swing. The maximum frequency the sound detector detected is ......
(a) $2.027 \mathrm{kHz}$

(b) $1.947 \mathrm{kHz}$
(c) $9.74 \mathrm{kHz}$
(d) $1.011 \mathrm{kHz}$
Answer:
(a) $2.027 \mathrm{kHz}$
Question 13.
Let. $\mathrm{y}=\frac{1}{1+x^2}$ at $\mathrm{t}=0$ s be the amplitude of the wave propagating in the positive $\mathrm{x}$ -
direction. At $\mathrm{t}=2 \mathrm{~s}$, the amplitude of the wave propagating becomes $y=\frac{1}{1+(x-2)^2}$
. Assume that the shape of the wave does not change during propagation. The velocity of the wave is .....
(a) $0.5 \mathrm{~ms}^{-1}$
(b) $1.0 \mathrm{~ms}^{-1}$
(c) $1.5 \mathrm{~ms}^{-1}$
(d) $2.0 \mathrm{~ms}^{-1}$

Answer:
(b) $1.0 \mathrm{~ms}^{-1}$
Hint.
The general expression $\mathrm{y}$ in terms of $\mathrm{x}$
$
y=\frac{1}{1+(x-v t)^2}
$
The shape of wave does not change, also wave move in $2 \mathrm{sec}, 2 \mathrm{~m}$ in positive ' $x$ ' direction. So, wave moves $2 \mathrm{~m}$ in $2 \mathrm{sec}$.
$
\therefore \text { The velocity of the wave }=\frac{\text { displacement }}{\text { time }}=\frac{2}{2} ; v=1 \mathrm{~ms}^{-1}
$
Question 14.
A uniform rope having mass $m$ hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed $\mathrm{v}$ with height $\mathrm{h}$ from the lower end?

Answer:

Question 15 .
An organ pipe A closed at one end is allowed to vibrate in its first harmonic and another pipe $\mathrm{B}$ open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of $A$ and $B$ is ......
(a) $\frac{8}{3}$
(b) $\frac{3}{8}$
(c) $\frac{1}{6}$
(d) $\frac{1}{3}$
Answer:
(c) $\frac{1}{6}$
Hint:
(c) $\frac{1}{6}$
Short Answer Questions
Question 1.
What is meant by waves?
Answer:
The disturbance which carries energy and momentum from one point in space to another point in space without the transfer of the medium is known as a wave.
Question 2.
Write down the types of waves.
Answer:
Waves can be classified into two types:
(a) Transverse waves
(b) Longitudinal waves

Question 3.
What are transverse waves? Give one example.
Answer:
In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)
Question 4.
What are longitudinal waves? Give one example.
Answer:
In longitudinal wave motion, the constituent of the medium oscillate or vibrate about their mean positions in a direction parallel to the direction of propagation (direction of energy transfer) of waves.
Example: Sound waves travelling in air.

Question 5 .
Define wavelength.
Answer:
For transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.
Question 6.
Write down the relation between frequency, wavelength and velocity of a wave.
Answer:
Frequency, $f=\frac{1}{\text { Time period }}$, which implies that the dimension of frequency $[f]=\frac{1}{[\mathrm{~T}]}=\mathrm{T}^{-1} \Rightarrow[\lambda f]=[\lambda][f]=\mathrm{LT}^{-1}=[$ Velocity $]$
Therefore, Velocity, $\lambda f=v$
Question 7.
What is meant by interference of waves?
Answer:
Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or the same amplitude.
Question 8.
Explain the beat phenomenon.
Answer:

When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two. sources, then their difference in frequency gives the beat frequency. Number of beats per second $\mathrm{n}=\left|\mathrm{f}_1-\mathrm{f}_2\right|$ per second
Question 9.
Define intensity of sound and loudness of sound. Answer:
1. The loudness of sound is defined as "the degree of sensation of sound produced in the ear or the perception of sound by the listener".
2. The intensity of sound is defined as "the sound power transmitted per unit area taken normal to the propagation of the sound wave".

Question 10.
Explain Doppler Effect.
Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.
Question 11.
Explain red shift and blue shift in Doppler Effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth.
Question 12.
What is meant by end correction in resonance air column apparatus?
Answer:
The antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is
$
\frac{1}{4} \lambda=\mathrm{L}_1+e
$
Again taking end correction into account, we have
$
\frac{3}{4} \lambda=\mathrm{L}_2+e
$

Question 13.
Sketch the function $Y=x+a$. Explain your sketch
Answer:
When $\mathrm{a}=0, \mathrm{y}=\mathrm{x}$
when $\mathrm{a}=1 ; \mathrm{x}=1: \mathrm{y}=1+1=2$
when $\mathrm{a}=2 ; \mathrm{x}=2 ; \mathrm{y}=2+2=4$

Explanation: This implies, when increasing the value of a, the line shifts towards right side at $\mathrm{a}=0$, and line shifts towards left side at $\mathrm{a}=1,2, \ldots$..For $\mathrm{a}=\mathrm{vt}, \mathrm{y}=\mathrm{x}-\mathrm{vt}$ satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of $\mathrm{x}$ and $t$. Hence it does not represent a waves.
Question 14.
Write down the factors affecting velocity of sound in gases.
Answer:
(a) Effect of pressure
(b) Effect of temperature
(c) Effect of density
(e) Effect of wind
Question 15.
What is meant by an echo? Explain.
Answer:
Echo: An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces.
Explanation: The speed of sound in air at $20^{\circ} \mathrm{C}$ is $344 \mathrm{~m} \mathrm{~s}^{-1}$. If we shout at a wall which is at $344 \mathrm{~m}$ away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time interval between each sound is $\left(\frac{1}{10}\right)^{\text {th }}$ of a second (persistence of hearing) i.e., $0.1 \mathrm{~s}$. Then,

$
\begin{aligned}
& \text { Velocity }=\frac{\text { Distance travelled }}{\text { Time taken }}=\frac{2 d}{t} \\
& 2 \mathrm{~d}=344 \times 0.1=34.4 \mathrm{~m} ; \mathrm{d}=17.2 \mathrm{~m}
\end{aligned}
$
The minimum distance from a sound reflecting wall to hear an echo at $20^{\circ} \mathrm{C}$ is 17.2 meter.
Long Answer Questions
Question 1.
Discuss how ripples are formed in still water.
Answer:
Suppose we drop a stone in a trough of still water, we can see a disturbance produced at the place where the stone strikes the water surface. We find that this disturbance spreads out (diverges out) in the form of concentric circles of ever increasing radii (ripples) and strike the boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the water molecules on the surface. Actually the particles of the water (medium) themselves do not move outward with the disturbance. This can be observed by keeping a paper strip on the water surface. The strip moves up and down when the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their - mean positions.
Question 2.
Briefly explain the difference between travelling waves and standing waves.

Question 3.
Show that the velocity of a travelling wave produced in a string is $\mathrm{v}=\sqrt{\frac{T}{\mu}}$
Answer:
Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string. When a jerk is given at one end (left end) of the rope, the wave pulses move towards right end with a velocity $\mathrm{v}$. This means that the pulses move with a velocity $\mathrm{v}$ with respect to an observer who is at rest frame. Suppose an observer also moves with same velocity $\mathrm{v}$ in the direction of motion of the wave pulse, then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity $\mathrm{v}$. Consider an elemental segment in the string. Let A and B be two points on the string at an instant of time. Let $\mathrm{dl}$ and $\mathrm{dm}$ be the length and mass of the elemental string, respectively. By definition, linear mass density, $\mu$ is
$
\begin{array}{r}
\mu=\frac{d m}{d l} \\
d m=\mu d l
\end{array}
$
The elemental string $\mathrm{AB}$ has a curvature which looks like an arc of a circle with centre at $\mathrm{O}$, radius $\mathrm{R}$ and the arc subtending an angle $\theta$ at the origin $\mathrm{O}$. The angle $\theta$ can be written in terms of arc length and radius as $\theta=\frac{d l}{R}$. The centripetal acceleration supplied

by the tension in the string is
$
a_{c p}=\frac{v^2}{\mathrm{R}}
$
Then, centripetal force can be obtained when mass of the string $(\mathrm{dm})$ is included in equation (3)
$
\mathrm{F}_{c p}=\frac{(d m) v^2}{\mathrm{R}}
$
The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get

$\frac{(d m) v^2}{\mathrm{R}}=\frac{\mu v^2 d l}{\mathrm{R}}$

The tension $\mathrm{T}$ acts along the tangent of the elemental segment of the string at $\mathrm{A}$ and $\mathrm{B}$. Since the arc length is very small, variation in the tension force can be ignored. We can resolve $\mathrm{T}$ into horizontal component $\mathrm{T} \cos \left(\frac{\theta}{2}\right)$ and vertical component $\mathrm{T} \sin \left(\frac{\theta}{2}\right)$ The
horizontal component at A and B are equal in magnitude but opposite in direction; therefore, they cancel each other. Since the elemental arc length $A B$ is taken to be very small, the vertical components at A and B appears to acts Vertical towards the centre of the arc and hence, they add up. The net radial force $F_r$ is
$
\mathrm{F}_r=2 \mathrm{~T} \sin \left(\frac{\theta}{2}\right)
$
Since the amplitude of the wave is very small when it is compared with the length of the spring, the sine of small angle is approximated as $\sin \left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$. Hence equation (6) can be written as
$
\mathrm{F}_r=2 \mathrm{~T} \times \frac{\theta}{2}=\mathrm{T} \theta
$

But $\theta=\frac{d l}{R}$, , therefore substituting in equation (7), we get
$
\mathrm{F}_r=\mathrm{T} \frac{d l}{\mathrm{R}}
$
Applying Newton's second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (5)
and equation (8), we have $\mathrm{T} \frac{d l}{\mathrm{R}}=\mu v^2 \frac{d l}{\mathrm{R}}$
$
v=\sqrt{\frac{\mathrm{T}}{\mu}} \text { measured in } \mathrm{ms}^{-1}
$
Question 4.
Describe Newton's formula for velocity of sound waves in air and also discuss the Laplace's correction.
Answer:
Newton's formula for speed of sound waves in air: Sir Isaac Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. That is, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such that the temperature of the medium remains constant. Therefore, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle's law,

Mathematically
$\mathrm{PV}=$ constant
Differentiating equation (1), we get
$
\mathrm{P} d \mathrm{~V}+\mathrm{V} d \mathrm{P}=0 \text { (or) } \mathrm{P}=-\mathrm{V} \frac{d \mathrm{P}}{d \mathrm{~V}}=\mathrm{B}_{\mathrm{T}}
$
where, $\mathrm{B}_{\mathrm{T}}$ is an isothermal bulk modulus of air. Substituting equation (2) in equation the speed of sound in air is
$
v_{\mathrm{T}}=\sqrt{\frac{\mathrm{B}_{\mathrm{T}}}{\rho}}=\sqrt{\frac{\mathrm{P}}{\rho}}
$
Since $P$ is the pressure of air whose value at NTP (Normal Temperature and Pressure) is $76 \mathrm{~cm}$ of mercury, we have
$
\begin{aligned}
& P=\left(0.76 \times 13.6 \times 10^3 \times 9.8\right) \mathrm{N} \mathrm{m}^{-2} \\
& \rho=1.293 \mathrm{~kg} \mathrm{~m}^{-3}
\end{aligned}
$
Here $\rho$ is density of air, then the speed of sound in air at Normal Temperature and Pressure (NTP) is

$
\begin{aligned}
v_{\mathrm{T}} & =\sqrt{\frac{\left(0.76 \times 13.6 \times 10^3 \times 9.8\right.}{1.293}} \\
& =279.80 \mathrm{~ms}^{-1} \approx 280 \mathrm{~ms}^{-1} \text { (theoretical value) }
\end{aligned}
$
But the speed of sound in air at $0^{\circ} \mathrm{C}$ is experimentally observed as $332 \mathrm{~m} \mathrm{~s}^{-1}$ which is close upto $16 \%$ more than theoretical value (Percentage error is
$
\frac{(332-280)}{332} \times 100 \%=15.6 \%
$

This error is not small.
Laplace's correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson's law (not Boyle's law as Newton assumed), which is

$
\mathrm{PV}^\gamma=\text { constant }
$
where, $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_v}$ which is the ratio between specific heat at constant pressure a at constant volume. Differentiating equation (4) on both the sides, we get
$
\text { . } \mathrm{V}^\gamma \mathrm{dP}+\mathrm{P}\left(\gamma \mathrm{V}^{\gamma-1} d \mathrm{~V}\right)=0 \text { (or) } \gamma \mathrm{P}=-\mathrm{V} \frac{d \mathrm{P}}{d \mathrm{~V}}=\mathrm{B}_{\mathrm{A}}
$
where, $B_A$ is the adiabatic bulk modulus of air. Now, substituting equation $v=\sqrt{\frac{\mathrm{B}}{\rho}}$, the speed of sound in air is $v_{\mathrm{A}}=\sqrt{\frac{\mathrm{B}_{\mathrm{A}}}{\rho}}=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\gamma} v_{\mathrm{T}}$

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take $\gamma=$ 1.47.
Hence, speed of sound in air is $v_{\mathrm{A}}=(\sqrt{1.4})\left(280 \mathrm{~m} \mathrm{~s}^{-1}\right)=331.30 \mathrm{~m} \mathrm{~s}^{-1}$, whis closer to experimental data.
Question 5.
Write short notes on reflection of sound waves from plane and curved surfaces.
Reflection of sound through the plane surface
Answer:
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. Suppose a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts (say, compressions moving like a spherical wave front).
Therefore, the reflected wave front on the plane surface is also spherical, such that its centre of curvature (which lies on the other side of plane surface) can be treated as the image of the sound source (virtual or imaginary loud speaker) which can be assumed to be at a position behind the plane surface.

Reflection of sound through the curved surface: The behaviour of sound is different
when - it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) which is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones.

We know that any surface (smooth or rough) can absorb sound. For example, the sound produced in a big hall or auditorium or theatre is absorbed by the walls, ceilings, floor, seats etc. To avoid such losses, a curved sound board (concave board) is kept in front of the speaker, so that the board reflects the sound waves of the speaker towards the audience. This method will minimize the spreading of sound waves in all possible direction in that hall and also enhances the uniform distribution of sound throughout the hall. That is why a person sitting at any position in that hall can hear the sound without any disturbance.
Question 6.
Briefly explain the concept of superposition principle.
Answer:
Superposition Principle: When a jerk is given to a stretched string which is tied at one end, a wave pulse is produced and the pulse travels along the string. Suppose two persons holding the stretched string on either side give a jerk simultaneously, then these two wave pulses move towards each other, meet at some point and move away from each other with their original identity. Their behaviour is very different only at the crossing/meeting points; this behaviour depends on whether the two pulses have the same or different shape.
When the pulses have the same shape, at the crossing, the total displacement is the algebraic sum of their individual displacements and hence its net amplitude is higher than the amplitudes of the individual pulses. Whereas, if the two pulses have same amplitude but shapes are $180^{\circ}$ out of phase at the crossing point, the net amplitude vanishes at that point and the pulses will recover their identities after crossing.

Only waves can possess such a peculiar property and it is called superposition of waves. This means that the principle of superposition explains the net behaviour of the waves when they overlap. Generalizing to any number of waves i.e., if two are more waves in a medium move simultaneously, when they overlap, their total displacement is the vector sum of the individual displacements. We know that the waves satisfy the wave equation which is a linear second order homogeneous partial differential equation in both space coordinates and time. Hence, their linear combination (often called as linear superposition of waves) will also satisfy the same differential equation. To understand mathematically, let us consider two functions which characterize the displacement of the waves, for example, $\mathrm{y}_1=\mathrm{A}_1 \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_2=\mathrm{A}_2 \cos (\mathrm{kx}-\omega \mathrm{t})$
Since, both $y_1$ and $y_2$ satisfy the wave equation (solutions of wave equation) then their algebraic sum $\mathrm{y}=\mathrm{y}_1+\mathrm{y}_2$
also satisfies the wave equation. This means, the displacements are additive. Suppose we multiply $\mathrm{y}_1$ and $\mathrm{y}_2$ with some constant then their amplitude is scaled by that constant Further, if $\mathrm{C}_1$ and $\mathrm{C}_2$ are used to multiply the displacements $\mathrm{y}_1$ and $\mathrm{y}_2$, respectively, then, their net displacement $\mathrm{y}$ is $\mathrm{C}=\mathrm{C}_1 \mathrm{y}_1+\mathrm{C}_2 \mathrm{y}_2$
This can be generalized to any number of waves. In the case of $n$ such waves in more than one dimension the displacements are written using vector notation. Here, the net displacement $\vec{y}$ is
$
\vec{y}=\sum_{i=1}^n \mathrm{C}_i \vec{y}_i
$
The principle of superposition can explain the following :

(a) Space (or spatial) Interference (also known as Interference)
(b) Time (or Temporal) Interference (also known as Beats)
(c) Concept of stationary waves
Waves that obey principle of superposition are called linear waves (amplitude is much smaller than their wavelengths). In general, if the amplitude of the wave is not small then they are called non-linear waves. These violate the linear superposition principle, e.g., laser. In this chapter, we will focus our attention only on linear waves.

Question 7.
Explain how the interference of waves is formed.
Answer:
Consider two harmonic waves having identical frequencies, constant phase difference $\mathrm{cp}$ and same wave form (can be treated as coherent source), hut having amplitudes $\mathrm{A}_1$ and $\mathrm{A}_2$, then
$
\begin{aligned}
& y_1=\mathrm{A}_1 \sin (k x-\omega t) \\
& y_2=\mathrm{A}_2 \sin (k x-\omega t+\phi)
\end{aligned}
$
Suppose they move simultaneously in a particular direction, then interference occurs (i.e., overlap of these two waves). Mathematically $\mathrm{y}=\mathrm{y}_1+\mathrm{y}_2 \ldots(3)$
Therefore, substituting equation (1) and equation (3) in equation (3), we get
$
y=\mathrm{A}_1 \sin (k x-\omega t)+\mathrm{A}_2 \sin (k x-\omega t+\text { Times New F }
$
Using trigonometric identity $\sin (\alpha+\beta)=(\sin \alpha \cos \beta+\cos \alpha \sin \beta)$, we get
$
\begin{aligned}
& y=A_1 \sin (k x-\omega t)+A_2[\sin (k x-\omega t) \cos \phi+\cos (k x \\
& y=\sin (k x-\omega t)\left(A_1+A_2 \cos \phi\right)+A_2 \sin \phi \cos (k x-
\end{aligned}
$
Let us re-define $A \cos \theta=\left(A_1+A_2 \cos \phi\right)$
and $\quad \mathrm{A} \sin \theta=\mathrm{A}_2 \sin \phi$
then equation (4) can be rewritten as $y=\mathrm{A} \sin (k x-\omega t) \cos \theta+\mathrm{A} \cos (k x-\omega$
$
\begin{aligned}
& y=\mathrm{A}(\sin (k x-\omega t) \cos \theta+\sin \theta \cos (k x-\omega t)) \\
& y=\mathrm{A} \sin (k x-\omega t+\theta)
\end{aligned}
$
By squaring and adding equation (5) and equation (6), we get
$
A_2=A_1^2+A_2^2+2 A_1 A_2 \cos \phi
$
Since, intensity is square of the amplitude $\left(I=A_2\right)$, we have
$
\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi
$
This means the resultant intensity at any point depends on the phase difference at that point.

(a) For constructive interference:
When crests of one wave overlap with crests of another wave, their amplitudes will add up and we get constructive interference. The resultant wave has a larger amplitude than the individual waves as shown in figure (a).

The constructive interference at a point occurs if there is maximum intensity at that point, which means that $\cos \varphi=+1 \Rightarrow \varphi=0,2 \pi, 4 \pi, \ldots=2 n \pi$, where $n=0,1,2, \ldots$
This is the phase difference in which two waves overlap to give constructive interference. Therefore, for this resultant wave,
$
I_{\text {maximum }}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=\left(\mathrm{A}_1+\mathrm{A}_2\right)^2
$
Hence, the resultant amplitude
$
\mathrm{A}=\mathrm{A}_1+\mathrm{A}_2
$

(b) For destructive interference: When the trough of one wave overlaps with the crest of another wave, their amplitudes "cancel" each other and we get destructive interference as shown in figure (b). The resultant amplitude is nearly zero. The destructive interference occurs if there is minimum intensity at that point, which means $\cos \varphi=-1$ $\Rightarrow \varphi=\pi, 3 \pi, 5 \pi, \ldots=(2 \mathrm{n}-1) \mathrm{K}$, where $\mathrm{n}=0,1,2, \ldots$ i.e. This is the phase difference in which two waves overlap to give destructive interference. Therefore,
$
\mathrm{I}_{\text {minimum }}=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2=\left(\mathrm{A}_1-\mathrm{A}_2\right)^2
$
Hence, the resultant amplitude $\mathrm{A}=\left|\mathrm{A}_1-\mathrm{A}_2\right|$
Question 8.
Describe the formation of beats.
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second $\mathrm{n}=\left|\mathrm{f}_1-\mathrm{f}_2\right|$

$x_m=\left(\frac{2 m+1}{2}\right) \frac{\lambda}{2}$, where, $m=0,1,2 \ldots$
For $m=0$ we have maximum at $x_0=\frac{\lambda}{2}$
For $m=1$ we have maximum at $x_1=\frac{3 \lambda}{4}$
For $m=2$ we have maximum at $x_2=\frac{5 \lambda}{4}$ and so on.
The distance between two successive antinodes can be computed by
$
x_m-x_{m-1}=\left(\frac{2 m+1}{2}\right) \frac{\lambda}{2}-\left(\frac{(2 m+1)+1}{2}\right) \frac{\lambda}{2}=\frac{\lambda}{2}
$
Similarly, the minimum of the amplitude A' also occurs at some points in the space, and these points can be determined by setting $\sin (\mathrm{kx})=0 \Rightarrow \mathrm{kx}=0, \pi, 2 \pi, 3 \pi, \ldots=\mathrm{n} \pi$
where $\mathrm{n}$ takes integer or integral values. Note that the elements at these points do not vibrate (not move), and the points are called nodes. The $\mathrm{n}^{\text {th }}$ nodal positions is given by, $x_n=n \frac{\lambda}{2}$ where, $n=0,1,2, \ldots$
For $n=0$ we have minimum at $\quad x_0=0$
For $n=1$ we have minimum at $\quad x_1=\frac{\lambda}{2}$
For $\mathrm{n}=2$ we have maximum at $x_2=\lambda$ and so on.
The distance between any two successive nodes can be calculated as
$
x_n-x_{n-1}=n \frac{\lambda}{2}-(n-1) \frac{\lambda}{2}=\frac{\lambda}{2}
$
Characteristics of stationary waves:
1. Stationary waves are characterised by the confinement of a wave disturbance between two rigid boundaries. This means, the wave does not move forward or between two rigid boundaries. This means, the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Therefore, they are called "stationary waves or standing waves".
2. Certain points in the region in which the wave exists have maximum amplitude, called as anti-nodes and at certain points the amplitude is minimum or zero, called as nodes.
3. The distance between two consecutive nodes (or) anti-nodes is $\frac{\lambda}{2}$
4. The distance between a node and its neighbouring anti-node is $\frac{\lambda}{4}$
5. The transfer of energy along the standing wave is zero.
Question 10.
Discuss the law of transverse vibrations in stretched strings.
Answer:
Laws of transverse vibrations in stretched strings: There are three laws of transverse vibrations of stretched strings which are given as follows:
(i) The law of length: For a given wire with tension $\mathrm{T}$ (which is fixed) and mass per unit length $\mu$ (fixed) the frequency varies inversely with the vibrating length. Therefore, $f \propto \frac{1}{l} \Rightarrow f=\frac{\mathrm{C}}{l}$
$\Rightarrow 1 \times \mathrm{f}=\mathrm{C}$, where $\mathrm{C}$ is a constant
(ii) The law of tension: For a given vibrating length 1 (fixed) and mass per unit length $\mathrm{p}$ , (fixed) the frequency varies directly with the square root of the tension $\mathrm{T}$,
$
f \propto \sqrt{\mathrm{T}}
$
$\Rightarrow \quad f=\mathrm{A} \sqrt{\mathrm{T}}$, where $\mathrm{A}$ is a constant

(iii) The law of mass: For a given vibrating length 1 (fixed) and tension $\mathrm{T}$ (fixed) the frequency varies inversely with the square root of the mass per unit length $\mu$, $\begin{aligned} f & \propto \frac{1}{\sqrt{\mu}} \\ \Rightarrow \quad f & =\frac{\mathrm{B}}{\sqrt{\mu}}, \text { where } \mathrm{B} \text { is a constant }\end{aligned}$
Question 11.
Explain the concepts of fundamental frequency, harmonics and overtones in detail. Fundamental frequency and overtones in detail.
Answer:
Fundamental frequency and overtones: Let us now keep the rigid boundaries at $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}$ and produce a standing waves by wiggling the string (as in plucking strings in a guitar). Standing waves with a specific wavelength are produced. Since, the amplitude must vanish at the boundaries, therefore, the displacement at the boundary must satisfy
the following conditions
$
x(x=0, t)=0 \text { and } y(x=L, t)=0
$
Since, the nodes formed at a distance $\frac{\lambda_n}{2}$ apart, we have $n\left(\frac{\lambda_n}{2}\right)=\mathrm{L}$, where $\mathrm{n}$ is an integer, $\mathrm{L}$ is the length between the two boundaries and $\lambda_{\mathrm{n}}$ is the specific wavelength that satisfy the specified boundary conditions. Hence,
$
\lambda_n=\left(\frac{2 \mathrm{~L}}{n}\right)
$
Therefore, not all wavelengths are allowed. The (allowed) wavelengths should fit with the specified boundary conditions, i.e., for $\mathrm{n}=1$, the first mode of vibration has specific wavelength $\lambda_1=2 L$. Similarly for $n=2$, the second mode of vibration has specific wavelength

$
\lambda_2=\left(\frac{2 \mathrm{~L}}{2}\right)=\mathrm{L}
$
For $\mathrm{n}=3$, the third mode of vibration has specific wavelength
$
\lambda_3=\left(\frac{2 L}{3}\right) \text { and so on. }
$
The frequency of each mode of vibration (called natural frequency) can be calculated.
We have,
$
f_n=\frac{v}{\lambda_n}=n\left(\frac{v}{2 \mathrm{~L}}\right)
$
The lowest natural frequency is called the fundamental frequency.
$
f_1=\frac{v}{\lambda_1}=\left(\frac{v}{2 \mathrm{~L}}\right)
$
The second natural frequency is called the first over tone.
$
f_2=2\left(\frac{v}{2 \mathrm{~L}}\right)=\frac{1}{\mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}
$
The third natural frequency is called the second over tone.
$
f_3=3\left(\frac{v}{2 \mathrm{~L}}\right)=3\left(\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}\right) \text { and so on. }
$
Therefore, the $\mathrm{n}^{\text {th }}$ natural frequency can be computed as integral (or integer ) multiple of fundamental frequency, i.e., $\mathrm{f}_{\mathrm{n}}=\mathrm{nf}_1$ where $\mathrm{n}$ is an integer ...(5)
If natural frequencies are written as integral multiple of fundamental frequencies, then the frequencies are called harmonics. Thus, the first harmonic is $\mathrm{f}_1=\mathrm{f}_1$ (the fundamental frequency is called first harmonic), the second harmonic is $f_2=2 \mathrm{f}_1$, the third harmonic is $\mathrm{f}_3=3 \mathrm{f}_1$ etc.
Question 12.
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using sonometer.
Answer:
Stationary waves in sonometer: Sono means sound related, and sonometer implies sound-related measurements.
It is a device for demonstrating the relationship between the frequency of the sound produced in the transverse standing wave in a string, and the tension, length and mass per unit length of the string. Therefore, using this device, we can determine the following quantities: