NCERT Solutions for Class 11 Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Here, M = 2.50 kg, T = 200 N, length of the string, 1 = 20 m
Therefore, mass per unit length of the string,

Question 2.
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)
Answer:
Here, h = 300 m, g = 9.8 m s ^-2, v = 340 m s^-1 if h = time taken by stone to strike the surface of water in the pond, then from

Question 3.
A 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 speed of a transverse wave on the wire equals to the speed of sound in dry air at 20 °C = 343 m s^_1.
Answer:
Here, l = 12.0 m, M = 2.10 kg, T = ?, ν = 343 m s^-1
Mass per unit length ,μ

Question 4.
Use the formula  \(v=\sqrt { \cfrac { \Upsilon \rho }{ \rho } } \) to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Answer:
(a) According to Boyle’s law, we know that
PV = constant at a constant temperature … (i)
Now volume of gas, \(v=\cfrac { M }{ \rho }\)
∴ From (i) and (ii), we get
\(v=\cfrac {PM }{ \rho }\) = constant.
Since mass of a gas remains constant.
\(v=\cfrac { p }{ \rho }\) = constant.
g is always constant for a given gas or air
∴  from \(v=\sqrt { \cfrac { \Upsilon \rho }{ \rho } }\) equation ,we conclude that velocity of sound in air always remaun constant if its temperature is constant.

where M = ρV = molecular weight of the  gas
As ϒ,R and M are Constants, So α√T
i.e. velocity of sound in a gas is directly proportional to the square root of its I      temperature, hence we conclude that the velocity of sound in air increases with increase in temperature.
(c) Effect of humidity : The presence of water vapours in  air changes  the  density of air, thus the velocity of sound changes  with  humidity of air.
Let ρ_m = density of moist air.
ρ_d = density of dry air.
_υm = velocity of sound in moist air.
_υd = velocity of sound in dry air.

Also we know that density of water vapours is less than the density of dry air i.e. dry air is heavier than water vapours as the molecular mass of water is less than that of N₂ (28) and O₂ (32), so p_m <  ρ_d.

i.e. velocity of sound in air increases with humidity, i.e. velocity of sound in moist air is greater than velocity of sound in dry air. That is why sound travels faster on rainy day than on a dry day.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination (x – vt) to (x + vt), i.e. y = f (x ± vt). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
(a) (x-vt)²
(b) log [(x + vt)/x₀]
(c) 1/(x + vt)
Answer:
No, the converse is not true.
The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, the wave function must have a finite value.
Out of the given functions for y, none of these satisfies this condition, so these functions do not represent a travelling wave.

Question 6.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If this 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 ms-¹.
Answer:
Here, υ = 1000 kHz = 1000 x 10³ Hz = 10⁶ Hz;
speed of sound in air,υ_a = 340 m s^_1;
speed of sound in water, v_w = 1,486 m s^_1
(a) The reflected sound: After reflection, the ultrasonic sound continues to travel in air. If λ_a is wavelength in air, then

(b) The transmitted sound : The transmitted ultrasonic sound travels in water. If λ_w is wavelength of ultrasonic sound in water, then

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s¹? The operating frequency of the scanner is 4.2 MHz.
Answer:
Here, υ= 1.7 km s^_1 = 1.7 x 10³ m s^_1;
υ= 4.2 MHz = 4.2 x 10⁶ Hz

Question 8.
A transverse harmonic wave on a string is described by
y(x, t) = 3.0 Sin (36 t + 0.018 x + π/4)
where x and y are in cm and t in s. The positive direction of x is from left to right.
(a)  Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Answer:
The equation of the form

represents a harmonic wave of amplitude A, wavelength l and travelling from right to left with a
velocity v.
Now, the given equation for the transverse harmonic wave is

(a) Since the equation (i) and (ii) are of the same form, the given equation also represents a travelling wave propagating from right to left. Further, the coefficient of t gives the speed of the wave. Therefore, v = 2000 cm s^_1 = 20 m s^-1
(b) Obviously, amplitude, A = 3.0 cm

(c) Initial phase at the origin, φ=\(\cfrac { \Pi }{ 4 }\) rad
(d) Least distance between two successive crests in the wave is equal to wavelength.

Question 9.
For the wave described in question no. 8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:

For different values of t, we calculate y using equation (i). These values are tabulated below : On plotting y versus t graph, we obtain a sinusoidal curve as shown in above figure.

Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m
(b) 0.5 m
(c) λ/2
(d) 3λ/4
Answer:
Given equation of a travelling harmonic wave is
y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35) …….(i)
The standard equation of a travelling harmonic wave

Comparing equation (i) and (ii), we get

Question 11.
The transverse displacement of a string (clamped at its two ends) is given by
y(x,f) = 0.06sin\((\cfrac { 2\Pi }{ 3 } x)\) cos(120 πt)
Where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 x 10 ² kg. Answer the following:
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
Answer:
Here, the equation for transverse displacement is given by
y(x,f) = 0.06sin\( \cfrac { 2\Pi }{ 3 } x \) cos 120 πt …………(i)
(a) The displacement, which involves harmonic functions of x and f separately, represents a stationary wave and the displacement, which is harmonic function of the form (υt ± x), represents a travelling wave. Hence, the equation given above represents a stationary wave.
(b) When a wave pulse

Question 12.
(i) For the wave on a string described in question no. 11, do all the points on the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
(i) All the points on the string
(a) have the same frequency of oscillation (except at the nodes, where the frequency is zero).
(b) have the same phase between two adjacent nodes,
(c) but not the same amplitude.
(ii) Now, the amplitude of the given wave is
A = 0.06 sin \( \cfrac { 2\Pi }{ 3 } x \)
At x = 0.375 m, the amplitude is given by
A(at x = 0.375 m) = 0.06 sin \( \cfrac { 2\Pi }{ 3 } x \) x 0.375
= 0 06 sin \(\frac { \Pi }{ 4 }\) = 0.06 x 0.707 = 0.042 m 4

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent
(1) a travelling wave,
(2) a stationary wave or
(3) none at all:

(a) y=2cos(3x)sin(10t)
(b) y = 2-√x-vt
(c) y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
(d) y = cos x sin t + cos 2x sin 2t

Answer:
(a) This equation has two harmonic functions of each x and t separately, so it represents stationary wave.
(b) This function does not represent any wave as it contains no harmonic function.
(c) It represents progressive/travelling harmonic wave as the arguments of cosine and sin functions are same.
(d) This equation is the sum of two functions cos x sin t and cos 2x sin 2t each represen­ting a stationary wave. Therefore it represents superposition of two stationary waves

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear mass density is 4.0 x 10 ² kg m^-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Answer:
Here, υ = 45 Hz, M = 3.5 x 10^-2 kg;
mass/length =μ = 4.0 x  10^-2  kg m^-1

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore, the pipe is in resonance with the fundamental note and the third harmonic (79.3 cm is about 3 times 25.5 cm]
In the fundamental mode, \(\frac { \Pi }{ 4 }\) = l₁= 25.5 cm
λ = 4 x 25.5 = 102 cm = 1.02 m.
Speed of sound in air,
υ = υλ= 340 x 1.02 = 346.8 m s^-1.

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is speed of sound in steel?
Answer:

A rod clamped in the middle has antinodes (A) at its ends and node (N) at the point of clamping. In fundamental mode, thus the length of the rod is
l=\(\frac { \Pi }{ 2 }\) or 2l
Where l = length of rod and also λ = wavelength of the wave
Here, l = 100 cm
υ = 2.53 kHz = 2.53 x 10 ³ Hz
∴ λ= 2 x 100 = 200 cm
If υ be the speed of sound in steel, then
υ = υ λ = 2.53 x 10³ x 200
= 506 x 10³ cm s^_1
= 5.06 x 10³ ms^-1
υ= 5.06 km s^_1.

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^_1
Answer:
Here, L = 20 cm = 0.2 m, υ„ = 430 Hz, υ= 340 m s^_1
The frequency of n^th normal mode of vibration of closed pipe is


Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slight out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Answer:
We know that  υ ∝ √T
where u = frequency, T = tension
The decrease in the tension of a string decreases its frequency.
So let us assume that original frequency υ_A of A is more than the frequency υ_B of B.
Thus υ_A – υ_B = ± 6 Hz (given)
and υ_A = 324 Hz
∴  324 – υ_B  = ± 6
or υ_B = 324 ± 6 = 318 Hz or 330 Hz.
On reducing tension of A, Δυ= 3 Hz
If υ_B  = 330 Hz and on decreasing tension in A, will be reduced i.e. no. of beats will increase, but this is not so because no. of beats becomes 3.
∴ must be 318 Hz because on reducing the tension in string A, its frequency may be reduced to 321 Hz, thus giving 3 beats with υ_B = 318 Hz.

Question 19.
Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa,
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”,
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) In a sound wave, a node is a point where the amplitude of oscillation i.e. displacement is zero as here a compression and a rarefaction meet and the pressure is maximum, so it is called pressure antinode. While an antinode is a point where the amplitude of oscillation is maximum i.e. displacement is maximum but pressure is minimum. So this point is called pressure node.
Hence displacement node coincides with pressure antinode and displacement antinode with pressure node.
(b) Bats emit ultrasonic waves of large frequencies (small wavelength) when they fly. These ultrasonic waves are received by them after reflection from the obstacle. Their ears are so sensitive and trained that they not only get the information of the distance of the obstacle but also that of the nature of the reflecting surface.
(c) The quality of the sound produced by an instrument depends upon the number of overtones. Since the number of overtones is different in the cases of sounds produced by violin and sitar therefore we can distinguish through them.
(d) Solids possess both the volume elasticity and the shear elasticity. Therefore they can support both longitudinal and transverse waves.
On the other hand, gases have only the volume elasticity and no shear elasticity, so only longitudinal waves can propagate in gases.
(e) A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions e. with different velocities. So the shape of the pulse gets distorted i.e. a plane wave front in a non- dispersive medium does not remain a plane wave front in a dispersive medium.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
(i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1.
(b) recedes from the platform with a speed of 10 m s^-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms¹.
Answer:
Here, frequency of source of sound,υ = 400 Hz; v = 340 m s^_1 = speed of sound
Speed of source = υ_s = 10m s^_1
(i) (a) When the train approaches the platform, the apparent frequency as heard by the observer on the platform will be

(b) When the train recedes from the platform, the apparent frequency as heard by the observer is given by according to the formula :

(ii) The speed of sound in each case remains same i.e.340 ms^-1.

Question 21.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Here, v = 340 m s^-1; υ = 400 Hz
(a) Speed of the wind, v_w = 10 m s^-1
As the direction of blow of wind (yard to station) is the same as the direction of sound, therefore, for the observer standing on the platform; velocity of sound,
υ’ =υ + υ_w = 340 + 10 = 350 m s^-1
As there is no relative motion between source of the sound and the observer, the frequency of sound will remain unchanged.
Thus, frequency of sound = 400 Hz
Wavelength of sound,

(b) Speed of the observer, υ₀ = 10 m s^-1 (towards yard). When the observer moves towards the source of sound, the apparent frequency,

The wavelength of sound waves is not affected due to the motion of the observer and hence the wavelength of the sound waves will remain unchanged.
Speed of sound relative to the observer
= 340 + 10 = 350 m s^-1
Therefore, situations (a) and (b) are not equivalent.

Question 22.
A travelling harmonic wave on a string is described by y (x, t) = 7.5 sin (0.0050 + 12t + π/4)
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacement and velocity as the x = 1 cm point at f = 2 s, 5 s and 11s.
Answer:



Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation?
If the pulse rate is 1 after every 20 s, that is the whistle is blown for a split of second after every 20 s, is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
(a) A narrow sound pulse such as a short pip by a whistle does not have a definite wavelength or frequency. However, being a sound wave, it has a definite speed.
(b) If a short pip is produced after every 20 s, then frequency of the note produced by the whistle cannot be called 1/20 or 0.05 Hz. We may call 0.05 Hz as the frequency of repetition of the short pip.

Question 24.
One end of a long string of linear mass density 8 x 10 ³ kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, μ = 8.0 x 10^-3 kg m^-1, u = 256 Hz,
T= 90 kg = 90 x 9.8 = 882 N
Amplitude of wave, A =0 cm = 0.05 m.
As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by

As, the wave is propagating along positive x direction the equation of the wave is
y(x-t) =A sin(ωt-kx)
= 0.05 sin (1.61 x 10³ t- 4.84 x)
Here x, y, are in meter and t is in second.

Question 25.
A SONAR system fixed in a submarine operator at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Here, μ = 40.0 kHz ;
Speed of sound wave in water, υ = 1, 450 m s^_1
The frequency of the waves from SONAR system will undergo a change in the following two steps:
(1) Frequency of waves from SONAR system as received by the enemy submarine moving towards the system :

(2) Frequency of waves from the enemy’s submarine as received by SONAR system : The enemy submarine will reflect the waves of frequency υ’ = 42.756 kHz and will thus act as a source of waves moving with a speed; υ_s = 100 m s^_1 toward the SONAR system (listener). If υ” is the apparent frequency as received by SONAR system, then

Question 26.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about km s^-1, and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer:
Let υ₁,υ₂ be the velocities of S waves and P waves, and t₁ ,t₂ be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, then
l = υ₁t₁ = υ₂t₂       -(i)
Now, υ₁= 4 km s^_1 and υ₂ = 8 km s^_1
∴  4t₁ =8 t₂ or t₁ = 2t₂               …(ii)
Also, t₁ – t₂ = 4 min = 240 s.
Using (ii), 2t₂ – t₂ = 240 s ;t₂ = 240 s
t₁= 2 x t₂ = 2 x 240 = 480 s
Now, from (i) l = υ₁t₁ = 4 x 480 = 1920 km.
Hence earthquake occurs 1920 km away from the seismograph.

Question 27.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of the sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, υ = 40 kHz,
speed of sound in air,υ = 340 m s^_1
∴  v’ = speed of bat
= 0.03 x υ = 0.03 x 340 = 10.20 m s^-1

The bat moving towards wall acts as a moving source and for the sound waves reflected from the wall, it acts as a moving observer. Thus the source and observer are approaching each other with same speed. i.e. υ_s = υ₀ = υ’ = 10.2 m s^-1
Thus apparent frequency of the reflected sound waves heard by the bat is given by

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NCERT Solutions for Class 11 Physics