 # NCERT Solutions For Class 11 Physics Chapter 15 : Waves

NCERT Solutions for Class 11 Physics Chapter 15 Waves has been created by our team of subject experts with an objective to help students in their Class 11 examinations. These books include the solution for all the questions given in the NCERT Physics textbooks. By practising questions from the NCERT solutions for Class 11 Physics Chapter 15, students can gain more information about the chapter and can also have a quick review before their finals.

These solutions consist of answers to extra question prepared by experts at BYJU’S along with exemplary problems, worksheets, short and long answer questions, MCQs, Tips and tricks to prepare for CBSE exams with NCERT Solutions.

This chapter comprises of comprehensive questions and solutions on a very important topic of physics such as questions on Wave dynamics etc. Questions from this chapter are repeatedly in exams and this chapter will guide you through every topics and type of waves such as tension on strings, the speed of sound in air, transverse wave, and dependence of the speed of sound in the air on factors like pressure, humidity and temperature. This chapter also has questions on concepts like the wavelength of ultrasonic sound, transverse harmonic wave and their frequency etc.

## Class 11 Physics NCERT Solutions for  Chapter 15 Waves

In this solution, we have taken a wire as a reference for signifying a wave motion and we have talked about the transverse displacement of a wire which is clamped on both sides and in which, we will be talking about finding the amplitude of a point at a certain distance in the wire. We also have questions regarding calculating the speed of sound in a wire with pistons at the end and the other end is in resonance with a tuning fork. These topics are very important during competitive exams as well because waves being one of the important topics in physics and is being repeatedly asked in every sort of exam. You can check out NCERT Solutions for class 11 Physics for chapter-wise solutions.

We can gain a lot of knowledge about waves by solving questions on the speed of sound in a steel medium. There are questions on how your guitar strings produce sounds and what kind of frequency does each string of a guitar has. We have an explanation on how a sound wave’s pressure antinode is a displacement node and vice versa also. We will be seeing how a dolphin being blind can manoeuvre through obstacles in a river and hunts preys. If a man is standing at a certain distance from an observer and he blows a horn, we can get to know the horn’s frequency depending on the man is running towards or away from the observer. A similar example can be found of a truck blowing the horn at a man in a petrol pump, finding its frequency, speed and wavelength.

Waves characteristics comprise of its frequency, amplitude, wavelength, phase, resonance, displacement of waves such as longitudinal and transverse motion in the various medium such as materials with high and low density, air, steel wires in guitars, sea waves etc.

### Subtopics of Class 11 Physics Chapter 15 Waves

1. Introduction
2. Transverse and longitudinal waves
3. Displacement relation in a progressive wave
4. The speed of a travelling wave
5. The principle of superposition of waves
6. Reflection of waves
7. Beats
8. Doppler effect.

### Important Questions of Class 11 Physics Chapter 15 Waves

Q1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Ans:

Given,
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m

Mass per unit length, μ = M /l = 5/40 = 0.125 kg / m
We know,
Velocity  of the transverse wave , v = $\sqrt{\frac{T}{\mu }}$
= $\sqrt{\frac{400}{0.125 }}$ = 56.56 m/s

Therefore, time taken by the transverse wave to reach the other side, t = l/v = 40/56.56
= 0.707 s

Q2. A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1? (g = 9.8 m s–2)

Ans:

Given,
Height of the bridge, s = 150 m
Initial velocity of the stone, u = 0
Acceleration, a = g = 9.8 m / s2
Speed of sound in air = 340 m/s

Let t be the time taken by the stone to hit the water’s surface
We know,
s = ut + ½ gt2
150 = 0 + ½ x 9.8 x t2
therefore, t2 = 300 / 9.8 = 5.53 s

Time taken by the sound to reach the bridge, t’= 150/340 = 0.44 s

Therefore from the moment the stone is released from the bridge, the sound of it splashing the water is heard after = t +t’ = 5.53 + 0.44 = 5.97s

Q3. Steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1

Ans:

Given,
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.0 kg
Velocity of the transverse wave, v = 343 m/s

Mass per unit length, μ = M /l = 2/24 = 0.083 kg / m
We know,
Velocity of the transverse wave , v = $\sqrt{\frac{T}{\mu }}$
Therefore, T = v2 μ
= 3432 x 0.083 = 9764.87 N

Q4. Using the formula v = $\sqrt{\frac{\gamma P}{\rho }}$  explain why the speed of sound in air
( a ) does not depend upon pressure.
( b ) increases with temperature and humidity.
( c ) increases with humidity.

Ans:

Given,
v = $\sqrt{\frac{\gamma P}{\rho }}$

We know,
PV = nRT              ( for n moles of ideal gas )
=> PV = RT (m/M )
Where, m is the total mass and M is molecular mass of the gas.
therefore, P  = m (RT/M)
=> P = $\frac{\rho RT}{M}$
=> $\frac{P}{\rho} = \frac{ RT}{M}$

( a ) For a gas at a constant temperature,$\frac{P}{\rho}$
= constant
Thus, as P increases ρ and vice versa. This means that P/ρ ratio always remains constant meaning

v =$\sqrt{\frac{\gamma P}{\rho }}$
= constant i.e., velocity of sound does not depend upon the pressure of the gas.

( b ) Since, $\frac{P}{\rho} = \frac{ RT}{M}$
v = $\sqrt{\frac{\gamma P}{\rho }}$   = v = $\sqrt{\frac{\gamma RT}{M }}$

We can see that v $\propto \sqrt{T}$ i.e., speed of sound increases with temperature.

( c ) When humidity increases, effective density of the air decrease. This means $v \propto \frac{1}{\sqrt{\rho }}$, thus velocity increases.

Q5.We know that the function y = f (x, t) represents a wave travelling in one direction, where x and t must appear in the combination x + vt or x– vt or i.e. y = f (x ± vt). Is the converse true?
Can the following functions for y possibly represent a travelling wave:
( i ) (x – vt) 2
( ii ) log [ ( x + vt)/ x0 ]
( iii ) 1 / (x + vt )

Ans:

No, the converse is not true, because it is necessary for a wave function representing a travelling wave to have a finite value for all values of x and t.
As none of the above functions satisfy the above condition, thus, none represent a travelling wave.

Q6.  A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s –1 and in water 1486 m s–1

Ans:

Given,
Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz
Speed of sound in air, vA = 340 m/s
We know,

( i ) The wavelength ( λR ) of the reflected sound is :
λR = vA /v
= 340/106 = 3.4 x 10-4 m

( ii ) Speed of sound in water, vW = 1486 m/s
Therefore, the wavelength (λT )of the transmitted sound is :
λT = 1486 / 106
= 1.49 x 10-3 m

Q7. An ultrasonic scanner operating at 4.2 MHz is used to locate tumours in tissues. If the speed of sound is 2 km /s in a certain tissue, calculate the wavelength of sound in this tissue?

Ans:

Given,
Speed of sound in the tissue, vT = 2 km/s = 2 × 103 m/s
Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz

Therefore, the wavelength of sound:
λ = vT / v
= (2 × 103)/ (4.2 × 106)
= 4.76 x 10-4 m

Q8. A transverse harmonic wave on a wire is expressed as:
y( x, t ) =3 sin ( 36t +0.018x +
π/4 )
( i ) Is it a stationary wave or a travelling?
( ii ) If it is a travelling wave, give the speed and direction of its propagation.
( iii ) Find its frequency and amplitude.
( iv ) Give the initial phase at the origin.
( v ) Calculate the smallest distance between two adjacent crests in the wave?
[X and y are in cm and t in seconds. Assume the left to right direction as the positive direction of x]

Ans:

Given,
y(x, t) =3 sin (36t +0.018x + π/4)                 . . . . . . . . .  . . ( 1 )

( i ) We know, the equation of a progressive wave travelling from right to left is:
y (x, t) = a sin (ωt + kx + Φ)         . . . . . . . . . . . . . . . . . . .  ( 2 )

Comparing equation ( 1 ) to equation ( 2 ), we see that it represents a wave travelling from right to left and also we get:
a = 3 cm,  ω = 36 rad/s , k = 0.018 cm and ϕ = π/4

( ii )Therefore the speed of propagation , v = ω/k = 36/ 0.018 = 20 m/s

( iii ) Amplitude of the wave, a = 3 cm
Frequency of the wave v = ω / 2π
= 36 /2π = 5.7 hz
( iv ) Initial phase at the origin = π/4
( v ) the smallest  distance between two adjacent crests in the wave, λ = 2π/ k = 2π / 0.018
= 349 cm
Q9. For the wave in the above question (Q8), plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. ( i ) Give the shapes of these plots.
( ii ) With respect to which aspects (amplitude, frequency or phase) does the oscillatory motion in a travelling wave differ from one point to another?

Ans:

Given,
y(x, t) =3 sin (36t +0.018x + π/4)                . . . . . . . . .  . . ( 1 )

For x= 0, the equation becomes :
y( 0, t ) =3 sin ( 36t +0 + π/4 )       . . . . . . . . .  . . ( 2 )
Also,
ω = 2 π/t = 36 rad/s
=> t = π/18 secs.

Plotting the displacement (y) vs. (t) graphs using different values of t listed below:

 t 0 T/ 8 2T/ 8 3T/ 8 4T/ 8 5T/ 8 6T/ 8 7T/ 8 T y $\frac{3}{\sqrt{2}}$ 3 $\frac{3}{\sqrt{2}}$ 0 $\frac{-3}{\sqrt{2}}$ -3 $\frac{-3}{\sqrt{2}}$ 0 $\frac{3}{\sqrt{2}}$

Similarly graphs are obtained for x = 0, x = 2 cm, and x = 4 cm. The oscillatory motion in the travelling wave is different from each other only in terms of phase. Amplitude and frequency are invariant for any change in x.
The y-t plots of the three waves are shown in the given figure: Q10. A travelling harmonic wave is given as: y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35).
What is the  phase difference between the oscillatory motion of two points separated by a distance of:
( i ) 8 m,
( ii ) 1 m,
( iii ) λ /2,
( iv ) 6λ/4
[ X and y are in cm and t is in secs ].

Ans:

Given,
Equation for a travelling harmonic wave :
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
= 2.0 cos (20πt – 0.016πx + 0.70 π)
Where,
Propagation constant, k = 0.0160 π
Amplitude, a = 2 cm
Angular frequency, ω= 20 π rad/s

We know,
Phase difference Φ = kx = 2π/ λ

( i ) For x = 8 m = 800 cm
Φ = 0.016 π × 800 = 12.8 π rad

( ii )For x =  1 m = 100 cm
Φ = 0.016 π × 100 = 1.6 π rad

( iii ) For x =λ /2
Φ = (2π/ λ)  x  ( λ/2)

( iv ) For x =λ6 /4
Φ = (2π/ λ)  x  ( 6λ/4)

Q11. The transverse displacement of a wire (clamped at both its ends) is described as :
y (x, t) = $0.06sin(\frac{2\pi}{3}x)cos(120\pi t)$
The mass of the wire is 6 x 10-2 kg and its length is 3m.
Provide answers to the following questions:
( I ) Is the function describing a stationary wave or a travelling wave?
( ii ) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave.
( iii ) Calculate the wire’s tension.
[X and y are in meters and t in secs]

Ans:

We know,
The standard equation of a stationary wave is described as:
y (x, t) = 2a sin kx cos ωt

Our given equation y (x, t) =$0.06sin(\frac{2\pi}{3}x)cos(120\pi t)$  is similar to the general equation .
( i ) Thus, the given function describes a stationary wave.

( ii ) We know, a wave travelling in the positive x-direction can be represented as :
y1 = a sin( ωt – kx )
Also,
A wave travelling in the negative x-direction is represented as :
y2 = a sin( ωt + kx )

Super- positioning  these two waves gives us :
y = y1 + y2
= a sin( ωt – kx ) – a sin( ωt + kx )
= asin(ωt)cos(kx) – asin(kx)cos(ωt) – asin(ωt)cos(kx) – asin(kx)cos(ωt)
= – 2asin(kx)cos(ωt)
= $-2asin(\frac{2\pi }{\lambda }x)cos(2\pi vt)$ . . . . . . . . . . . . . . ( 1 )

The transverse displacement of the wires is described as :
$0.06sin(\frac{2\pi}{3}x)cos(120\pi t)$             . . . . . . . . . . . . . . . . .( 2 )

Comparing equations ( 1 ) and ( 2 ) , we get :
2π/ λ = 2π/ 3
Therefore, wavelength  λ = 3m

Also, 2πv/λ = 120π
Therefore, speed v = 180 m/s

And, Frequency = v/λ = 180/3
= 60 Hz

( iii )Given,
Velocity of the transverse wave, v = 180 m / s
The string’s mass, m = 6 × 10-2 kg
String length, l = 3 m
Mass per unit length of the string, μ = m/l = (6 x 10-2 )/3
= 2 x 10-2 kg/m
Let the tension in the wire be T
Therefore, T = v2 μ
= 1802 x 2 x 10-2
= 648 N.

Q12. Considering the wave described in the above question ( Q11 ) answer the following questions;
( a ) Are all the points in the wire oscillating at the same values of   (i) frequency, (ii) phase, (iii) amplitude? Justify your answers.
( b ) Calculate the amplitude of a point 0.4 m away from one end?

Ans:

( a ) As the wire is clamped at both its ends, the ends behave as nodes and the whole wire vibrates in one segment. Thus,
( i ) Except at the ends which have zero frequency, all the particles in the wire oscillate with the same frequency.
( ii ) All the particles in the wire lie in one segment, thus they all have the same phase. Except for the nodes.
( iii ) Amplitude, however, is different for different points.

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

For x = 0.4m and t =0

Amplitude = displacement =$0.06sin(\frac{2\pi}{3}x)cos0$
= $0.06sin(\frac{2\pi}{3}\times 0.4)1$
= 0.044 m

Q13. Present below are functions of x and t to describe the displacement (longitudinal or transverse) of an elastic wave. Identify the ones describing  ( a ) a stationary wave, ( b ) a travelling wave and            ( c ) neither of the two :
( i ) y = 3 sin( 5x – 0.5t ) + 4cos( 5x – 0.5t )
( ii ) y = cosxsint + cos2xsin2t.
( iii ) y = 2 cos (3x) sin (10t)
( iv ) y = $2\sqrt{x – vt}$

Ans:

( i ) This equation describes a travelling wave as the harmonic terms ωt and kx are in the combination of kx – ωt.
( ii ) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately in the equation. In fact, this equation describes the superposition of two stationary waves.
( iii ) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately.
( iv ) This equation does not contain any harmonic term. Thus, it is neither a travelling wave nor a stationary wave.

Q14. A string clamped at both its ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 Hz. The linear mass density of the string is 4.0 × 10-2 kg/ m and its mass is 2 × 10 -2 kg. Calculate:
( i ) the velocity of a transverse wave on the string,
( ii ) the tension in the string.

Ans:

Given,
Mass of the string, m = 2 x 10-2 kg
Linear density of the string = 4 x 10-2 kg
Frequency, vF = 45 Hz

We know, length of the wire = m/µ
= (2 x 10-2)/ (4 x10-2) = 0.5 m

We know, λ = 2l/n
Where, n = number of nodes in the wire.

For fundamental node, n =1
=> λ = 2l
= 2 x 0.5 = 1m

( i ) Therefore speed of the transverse wave, v = λ vF
= 1 x 45 = 45 m/s
( ii ) Tension in the string = µ v2
= 4 x10-2 x 45 = 81 N

Q15. A 1m long pipe with a movable piston at one end and an opening at the other will be in resonance with a tuning fork vibrating at 340 Hz, if the length of the pipe is 79.3 cm or 25.5 cm. Calculate the speed of sound in air. Neglect the edge effects.

Ans:

Given,
Frequency of the turning fork, νF = 340 Hz
Length of the pipe, l1 = 0.255 m

As the given pipe is has a piston at one end, it will behave as a pipe with one end closed and the other end open, as depicted in the figure below: This kind of system creates odd harmonics. We know, fundamental note in a closed pipe is written as:
l1 = λ / 4
0.255 x 4 = λ = 1.02m
Therefore speed of sound, v = λ vF
= 340 × 1.02 = 346.8 m/s
Q16. A steel bar of length 200 cm is nailed at its midpoint. The fundamental frequency of the longitudinal vibrations of the rod is 2.53 kHz. At what speed will the sound be able to travel through steel?

Ans:

Given,
Length , l = 200 cm = 2 m
Fundamental frequency of vibration, νF = 2.53 kHz = 2.53 × 103 Hz

The bar is then plucked at its midpoint, forming an antinode (A) at its centre, and nodes (N) at its two edges, as depicted in the figure below : The distance between two successive nodes is   λ / 2
=> l =  λ / 2
Or,   λ = 2 x 2 = 4m

Thus, sound travels through steel at a speed of v = νλ
v = 4 x 2.53 x 103
=10.12 km / s

Q17. One end of A 20 cm long tube is closed. Find the harmonic mode of the tube that will be resonantly excited by a source of frequency 430 Hz. lf both the ends are open, can the same source still produce resonance in the tube?    (Sound travels in air at 340 m /s).

Ans:

Given,
Length of the pipe, l = 20 cm = 0.2 m
Frequency of the source = nth
the normal mode of frequency, νN = 430 Hz
Speed of sound, v = 340 ms -1

We know, that in a closed pipe the nth normal mode of frequency vN  = ( 2n -1 )v/4l
where n is an integer = 0, 1, 2, 3, 4, . . . . .

430 = ( 2n – 1 )( 340/4 x0.2)
2n = 2.01
n ≈ 1
Thus, the given source resonantly excites the first mode of vibration frequency

Now, for a pipe open at both the ends, the nth  mode of vibration frequency:
VR = nv/2l
n = VR 2l/v
n = (2 x 0.2 x 430)/340 = 0.5
As the mode of vibration ( n ) has to be an integer, this source is not in resonance with the tube.

Q18. Guitar strings X and Y striking the note ‘Ga’ are a little out of tune and give beats at 6 Hz. When the string X is slightly loosened and the beat frequency becomes 3 Hz. Given that the original frequency of X is 324 Hz, find the frequency of Y.

Ans:

Given,
Frequency of  X, fX = 324 Hz
Frequency of  Y = fY
Beat’s frequency, n = 6 Hz
Also,
n = $\left | f_{X}+f_{Y} \right |$
6 = $324 \pm f_{Y}$
=> fY = 330 Hz or 318 Hz
As frequency drops with a decrease in tension in the string, thus fY cannot be 330 Hz
=> fY = 318 Hz

Q19. Explain how:
( i ) A sound wave’s pressure antinode is a displacement node and vice versa.
( ii ) The Ganges river dolphin despite being blind, can manoeuvre and swim around obstacles and hunt down preys.
( iii ) A guitar note and violin note are being played at the same frequency, however, we can still make out which instrument is producing which note
( iv ) Both transverse and longitudinal wave can propagate through solids, but only longitudinal waves can move through gases.
( v ) In a dispersive medium, the shape of a pulse propagating through it gets distorted.

Ans:

( i ) An antinode is a point where pressure is the minimum and the amplitude of vibration is the maximum. On the other hand, a node is a point where pressure is the maximum and the amplitude of vibration is the minimum.
( ii )  The Ganges river dolphin sends out click noises which return back as vibration informing the dolphin about the location and distances of objects in front of it. Thus, allowing it to manoeuvre and hunt down preys with minimum vision.
( iii ) The guitar and the violin produce overtones of different strengths. Thus, one can differentiate between the notes coming from a guitar and a violin even if they are vibrating at the same frequencies.
( iv ) Both solids and fluids have a bulk modulus of elasticity. Thus, they both allow longitudinal waves to propagate through them. However, unlike solids, gases do not have shear modulus. Thus, transverse waves cannot pass through gases.
( v ) A pulse is a combination of waves of un-similar wavelengths. These waves move at different velocities in a dispersive medium. This causes the distortion in its shape.

Q20. A man standing at a certain distance from an observer blows a horn of frequency 200 Hz in still air. ( a ) Find the horn’s frequency for the observer when the man ( i ) runs towards him at 20 m/s  ( ii ) runs away from him at 20 m /s.
( b ) Find the speed of sound in both the cases.
[Speed of sound in still air is 340 m/ s]

Ans:

Given,
Frequency of the horn, νH = 200 Hz
Velocity of the man, vT = 20 m/ s
Velocity of sound, v = 340 m/ s

( a ) We know,
(i) The apparent frequency of the horn as the man approaches the observer is:
v’ = vH [ v/(v – vT) ]
= 200 [ 340 /(340 – 20) ]
= 212.5 Hz
(ii) The apparent frequency of the horn as the man runs away from  the observer is:
v’’ = vH [ v/(v + vT) ]
= 200 [ 340 /(340 + 20) ]
= 188.88 Hz
( b ) The speed of sound is 340 m/s in both the cases. The apparent change in frequency is a result of the relative motions of the observer and the source.

Q21. A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air.  The Wind then starts blowing towards the petrol pump at 20 m /s. Calculate the wavelength, speed, and frequency of the horn’s sound for a man standing at the petrol pump. Is this situation completely identical to a situation when the observer moves towards the truck at 20 m /sand the air is still?

Ans:

For the standing observer:
Frequency, νH = 200 Hz
Velocity of sound, v = 340 m/s
Speed of the wind, vW = 20 m/s

The observer will hear the horn at 200 Hz itself because there is no relative motion between the observer and the truck.

Given that the wind blows in the observer’s direction at 20 m/s.
Effective velocity of the sound, vE = 340 + 20 = 360 m/s

The wavelength ( λ ) of the sound :
λ = vE /vH  = 360/200
λ = 1.8 m

For the observer running towards the train :
Speed of the observer, vo = 20 m/s

We know,
The apparent  frequency of the sound as the observers move towards the truck is :
v’ =   vH [(v + vo)/v ]
=  200[ (20 + 340 )/ 340 ] =  211.764 Hz

As the air is still the effective velocity of sound is still 340 m/s.
As the truck is stationary the wavelength remains 1.8 m.

Thus, the two cases are not completely identical.

 Also Access NCERT Exemplar for Class 11 Physics Chapter 15 CBSE Notes for Class 11 Physics Chapter 15

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These NCERT Solutions for Class 11 Physics, Chapter 15, Waves provides complete information regarding the topic along with the definitions and examples. These NCERT solutions are available in downloadable PDF format. Students can download BYJU’S App to avail effective and Interactive lessons of CBSE Class 11 subjects. Related Links NCERT Solution Of Class 7 Maths NCERT Class 8 Maths Answers Of Maths NCERT Class 9 Answers Of Maths NCERT Class 10 CBSE Class 12 NCERT Maths Solutions NCERT Class 7 Science Answers Of Science NCERT Class 8 NCERT Class 9 Science CBSE NCERT Solutions For Class 10 Science NCERT Class 12 Biology NCERT Class 12 Physics NCERT Class 11 Chemistry