# NCERT Solutions for Class 11 Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Chapter 15 Waves: For the preparation of class 11 exam and competitive exams like NEET and JEE Mains, solutions of NCERT class 11 chapter 15 waves are helpful. If you drop stones in still water you can see outward moving circles from the point were water surface is disturbed. Now if you put some cork piece on the disturbed surface you can observe that cork piece moves up and down but not out or away from the centre of the disturbance. This shows that the movement of water does not transfer any matter. These patterns, which move without the actual physical transfer of matter are called waves. The waves such as waves on a string, sound waves, etc, come under the mechanical waves. These waves require a medium for propagation. The CBSE NCERT solutions for class 11 physics chapter 15 waves discuss questions based on mechanical waves. Electromagnetic waves (discussed in class 12 NCERT)  do not necessarily require a medium, that is they can travel through a vacuum. Light, radio waves, X-rays, are examples of electromagnetic waves and they travel in a vacuum at a speed equal to the speed of light. The third kind of wave is Matter waves. They are associated with constituents of matter: electrons, protons, neutrons etc. For example molecules matter waves associated with electrons are employed in electron microscopes. NCERT solutions for class 11 physics chapter 15 waves help in understanding the formulas studied in the chapter. The NCERT solutions will help you to understand the concepts studied in respective chapters.

## The main topics of the chapters are listed below:

15.1 Introduction

15.2 Transverse and longitudinal waves

15.3 Displacement relation in a progressive wave

15.4 The speed of a travelling wave

15.5 The principle of superposition of waves

15.6 Reflection of waves

15.7 Beats

15.8 Doppler effect

## NCERT solutions for class 11 physics chapter 15 waves exercise:

Mass per unit length of the string is

$\\\mu =\frac{M}{l}\\ =\frac{2.50}{20}\\ =0.125\ kg\ m^{-1}$

The velocity of the transverse wave in the string will be

$\\v=\sqrt{\frac{T}{\mu }}\\ =\sqrt{\frac{200}{0.125}}\\ =\sqrt{1600} \\=40\ m\ s^{-1}$

Time taken by the disturbance to travel from one end to the other is

$\\t=\frac{l}{v}\\ =\frac{20}{40}\\ =0.5\ s$

Time taken by the stone to the pond can be calculated using the second equation of motion

s = 300 m

u = 0

a = 9.8 m s-2

$\\s=ut_{1}+\frac{1}{2}at_{1}^{2}\\ 300=4.9t_{1}^{2}\\ t_{1}=7.82\ s$

Time taken by the wave to propagate from the pond to the top of the tower is

$\\t_{2}=\frac{300}{340}\\ t_{2}=0.88\ s$

t1 + t2 = 8.7 s

The splash is heard at the top of the tower after a time of 8.7 seconds.

Mass per unit length od the wire is

$\\\mu =\frac{M}{l}\\ =\frac{2.10}{12}\\ =0.175\ kg\ m^{-1}$

The speed of a transverse wave in a wire is given by

$\\v=\frac{T}{\mu }\\ 343=\sqrt{\frac{T}{01.175}}\\ T=343^{2}\times 0.175\\ T=2.059\times 10^{4}\ N$

The tension in the wire should be $2.059\times 10^{4}\ N$ such that the speed of the transverse wave in it is equal to 343 m s-1.

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

Where $\gamma \ and \ \rho$ are the Bulk's modulus and the density respectively

As we know

$\rho =\frac{M}{V}$

where M is molecular weight of air and V is the volume of 1 mole of air

$\\v=\sqrt{\frac{\gamma P}{\rho }}\\ v=\sqrt{\frac{\gamma PV}{M}}$

From the ideal gas equation PV=nRT

since we are talking about 1 mole we take n = 1

PV=RT

The expression for the speed of sound becomes

$v=\sqrt{\frac{\gamma RT}{M}}$

$\gamma$, M and R are constant therefore at constant Temperature the speed of sound in the air do not change and it is clear that speed is independent of velocity.

From the equation $\\v=\sqrt{\frac{\gamma RT}{M}}\\$ it is clear that the speed of sound is linearly proportional to the square root of the temperature and therefore it will increase with the increase in temperature.

(c) increases with humidity.

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

As the humidity of air increases, the proportion of water molecules(M=18) increases and that of Oxygen(M=32) and Nitrogen(M=28) decreases thus reducing the density of air and as the speed of sound is inversely proportional to the square root of density of air, the speed will increase as the density increases and thus it increases with an increase in humidity.

$(a)\; (x-vt)^{2}$

No, the given function cannot represent a wave as x or t approach infinity the function won't be converging to a constant value and therefore the converse is not true.

$(b) log\; [(x+vt)/x_{0}]$

No, the given function cannot represent a travelling wave because as x and t become zero the given function won't be converging to a constant value and therefore the converse is not true.

$(c)1/(x+vt)$

No, the given function cannot represent a travelling wave because as x and t become zero the given function won't be converging to a constant value and therefore the converse is not true.

(a) The wavelength of the reflected sound wave which will be travelling in air is

$\\\lambda _{a}=\frac{v_{a}}{\nu }\\ \lambda _{a}=\frac{340}{10^{6}}\\ \lambda _{a}=3.4\times 10^{-4}\ m$

(b) The frequency of the transmitted sound wave would not change.

The wavelength of the transmitted sound wave which will be travelling in water is

$\\\lambda _{w}=\frac{v_{w}}{\nu }\\ \lambda _{w}=\frac{1486}{10^{6}}\\ \lambda _{w}=1.49\times 10^{-3}\ m$

The wavelength of the sound in the tissue is

$\\\lambda =\frac{V}{\nu }\\ \lambda =\frac{1.7\times 10^{3}}{4.2\times 10^{6}}\\ \lambda =4\times 10^{-4}\ m$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$

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?

The wave is travelling.

$y(x,t)=Asin(kx+\omega t+\phi )$

The wave is travelling in the negative x-direction i.e. from right to left.

$\\\omega =36\ rad\ s^{-1}\\ k=0.018\ cm^{-1}$

Speed of the wave is

$\\v=\frac{\omega }{k}\\ v=\frac{36\times 10^{-2}}{0.018} \\v=20\ m\ s^{-1}$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$

where x and y are in cm and t in s. The positive direction of x is from left to right.
(b) What are its amplitude and frequency?

Amplitude A is 3.0 cm.

Frequency is

$\\\nu =\frac{\omega }{2\pi }\\ \nu =\frac{36}{2\pi }\\ \nu =5.73\ Hz$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
(c) What is the initial phase at the origin?

The initial phase of the wave at the origin (at x = 0 and t = 0) is $\frac{\pi }{4}$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.

The difference between two consecutive crests is equal to the wavelength of the wave.

$\\\lambda =\frac{2\pi }{k}\\ \lambda =\frac{2\pi\times 10^{-2} }{0.018}\\ \lambda =3.49\ m$

$\\y(x,t)=3.0sin(36t+0.018x+\frac{\pi }{4})$

for x = 0

$\\y(t)=3.0sin(36t+\frac{\pi }{4})$

The time period of oscillation is T

$T=\frac{\pi }{18}\ s$

To make the y versus t graph we tabulate values of y(t) at different values of t as follows

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

The graph of y versus t is as follows

For other values of x, we will get a similar graph. Its time period and amplitude would remain the same, it just will be shifted by different amounts for different values of x.

## Q.15.10 (a) For the travelling harmonic wave

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

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 $(a)\; 4\; m$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =k\Delta x$

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

$k=2\pi \times 0.008\ cm^{-1}$

Phase difference for two points separated by a distance of 4 m would be

$\\\Delta \phi =2\pi \times 0.0080\times 4\times 100\\ \Delta \phi=6.4\pi \ rad$

Q.15.10 (b) For the travelling harmonic wave

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

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

$(b)\; 0.5\; m$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =k\Delta x$

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

$k=2\pi \times 0.008\ cm^{-1}$

Phase difference for two points separated by a distance of 0.5 m would be

$\\\Delta \phi =2\pi \times 0.0080\times 0.8\times 100\\ \Delta \phi=0.8\pi \ rad$

Q.15.10 (c) For the travelling harmonic wave

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

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 $(c)\lambda /2$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$

Phase difference for two points separated by a distance of $\frac{\lambda }{2}$ would be

$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}\\ \Delta \phi =\pi\ rad$

Q.15.10 (d) For the travelling harmonic wave

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

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 $(d)3\lambda /4$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$

Phase difference for two points separated by a distance of $\frac{3\lambda }{4}$ would be

$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}\\ \Delta \phi =\frac{3\pi }{2}\ rad$

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

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$.

(a) Does the function represent a travelling wave or a stationary wave?

The given function is of the following form

$y(x,t)=2Asin\left (kx \right )cos(\omega t)$

which is the general equation representing a stationary wave and therefore the given function represents a stationary wave.

## Q.15.11 (b) The transverse displacement of a string (clamped at its both ends) is given by$y(x,t)=0.06\; \sin (\frac{2\pi }{3}x)\cos (120\; \pi t)$

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$.

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

We know that when two waves of the same amplitude, frequency and wavelength travelling in opposite directions get superimposed we get a stationary wave.

$\\y_{1}=asin(kx-\omega t)\\ \\y_{2}=asink(\omega t+kx)$

$\\y_{1}+y_{2}=asin(kx-\omega t)+asink(kx+\omega t)\\ =asin(kx)cos(\omega t)-asin(\omega t)cos(kx)+asin(\omega t)cos(kx)+asin(kx)cos(\omega t)\\ =2asink(kx)cos(\omega t)$

Comparing the given function with the above equations we get

$\\k=\frac{2\pi }{3}\\ \lambda =\frac{2\pi }{k}\\ \lambda =3\ m$

$\\\omega =120\pi \\ \nu =\frac{\omega }{2\pi }\\ \nu =60\ Hz$

$\\v=\nu \lambda \\ v=60\times 3=180\ m\ s^{-1}$

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$.

(c) Determine the tension in the string

The mass per unit length of the string is

$\\\mu =\frac{M}{l}\\ \mu =\frac{3\times 10^{-2}}{1.5}\\ \mu =0.02\ kg\ m^{-1}$

Speed of a transverse wave is given by

$\\v=\sqrt{\frac{T}{\mu }}\\ T=\mu v^{2}\\ T=0.02\times (180)^{2}\\ T=648\ N$

(i) (a) All the points vibrate with the same frequency of 60 Hz.

(b) They all have the same phase as it depends upon time.

(c) At different points, the amplitude is different and is equal to A(x) given by

$A(x)=0.06sin(\frac{2\pi }{3}x)$

(ii)

$\\A(0.375)=0.06sin(\frac{2\pi }{3}\times 0.375)\\ A(0.375)=0.06\times sin(\frac{\pi }{4})\\ A(0.375)=0.06\times \frac{1}{\sqrt{2}}\\ A(0.375)=0.042m$

$(a)\; y=2\; \cos (3x)\sin (10t)$

$(b)\; y=2\sqrt{x-vt}$

$(c)y=3\sin (5x-0.5t)+ 4\; \cos (5x-0.5t)$

$(d)\; y=\cos x\; \sin t+\cos 2x\: \sin \: 2t$

(a)  It represents a stationary wave.

(b) The given function does not represent a wave as we can see at certain values of x and t the function would not be defined.

(c) It represents a travelling wave.

(d) It represents a stationary wave. It is a superposition of two stationary waves which ultimately results in another stationary wave.

Length of the string is l given by

$\\l=\frac{M}{\mu }\\ l=\frac{3.5\times 10^{-2}}{4\times 10^{-2}}\\ l=0.875m$

Since the wire is vibrating in the fundamental mode

$\\l=\frac{\lambda }{2}\\ \lambda =2l\\ \lambda =2\times 0.875\\ \lambda =1.75m$

Speed of the string (v) is

$\\v=\nu l\\ v=45\times 1.75\\ v=78.75\ ms^{-1}$

Tension in the string is given by

$\\T=\mu v^{2}\\ T=4\times 10^{-2}\times (78.75)^{2}\\ T=248.0625\ N$

The pipe behaves as a pipe open at one end and closed at one end. Such a pipe would produce odd harmonics i.e.

$\nu =(2n-1)\frac{v}{4l_{n}}$

Two consecutive modes of vibration are given in the question

For l1 = 25.5 cm

$\nu =(2n-1)\frac{v}{4l_{1}}$

For l= 79.3 cm

$\\\nu =(2(n+1)-1)\frac{v}{4l_{2}}\\ \nu =(2n+1)\frac{v}{4l_{2}}$

Since at both these modes the system resonates with the same frequency we have

$\\(2n-1)\frac{v}{4l_{1}} =(2n+1)\frac{v}{4l_{2}}\\ \frac{2n-1}{2n+1}=\frac{25.5}{79.3}\\ \frac{2n-1}{2n+1}\approx \frac{1}{3}\\ n=1$(our approximation is correct since the edge effects may be neglected)

$\\\nu =\frac{v}{4l_{1}}\\ v=340\times 4\times 0.255\\ v=346.8\ ms^{-1}$

When the rod is clamped at the middle at is vibrating in the fundamental mode, a node is formed at the middle of the rod and antinodes at the end. i.e.

$\\\frac{L}{2}=\frac{\lambda }{4}\\ \lambda =2L\\ \lambda =0.2m$

where L is the length of the rod.

Speed of sound in steel is

$\\v=\nu \lambda \\ v=2.53\times 10^{3}\times 0.2\\ v=5060ms^{-1}$

Let the nth harmonic mode of the pipe get resonantly excited by a 430 Hz source.

$\\\frac{(2n-1)v}{4l}=\nu \\ 2n-1=\frac{430 \times 4\times 0.2}{340}\\ n=1$

The pipe resonates with a 430 Hz source in the fundamental mode when one end is open.

Let the mth harmonic mode of the pipe get resonantly excited by a 430 Hz source when both ends are open.

$\\\lambda =\frac{2l}{m}\\ m=\frac{2l\nu }{v}\\ m=\frac{2\times 0.2\times 430}{340}\\ m=0.5$

Since m is coming out to be less than 1 the same source will not be in resonance with the pipe if both ends are open.

$\\\nu _{A}=324Hz\\ Beat\ frequency(b)=6Hz\\ \nu _{B}=\nu _{A}\pm b\\ \nu _{B}=318Hz \\or \\\nu _{B}=330Hz$

Since frequency increases with an increase in Tension, the frequency of string A must have decreased. Therefore $\nu _{B}=318Hz$.(If it were 330 Hz the beat frequency would have increased with decrease in Tension in string A)

Q.15.19 Explain why (or how) :

(a) in a sound wave, a displacement node is a pressure antinode and vice versa,

In the propagation of a sound wave the pressure increases at points where displacement decreases, Therefore maximum pressure at points of minimum displace and vice-versa i.e. a displacement node is a pressure antinode and vice versa.

Q.15.19 Explain why (or how) :

(b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

Bats emit ultrasonic waves and when these waves strike the obstacles and get reflected back to the bats they ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”.

Q.15.19 Explain why (or how) :

(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,

We can distinguish between the two notes with the same frequency as the harmonics they emit are different.

Q.15.19 (d) Explain why (or how) :

(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases,

Transverse waves produce shear, gases don't have shear modulus and cannot sustain shear and therefore can only propagate longitudinal waves. Solids have both shear and bulk modulus of elasticity and can propagate both transverse and longitudinal waves.

## Q.15.19 (e) Explain why (or how): the shape of a pulse gets distorted during propagation in a dispersive medium.

As we know a pulse contains waves of different wavelengths, these waves travel at different speeds in a dispersive medium and thus the shape of the pulse gets distorted.

$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$

where $\\\nu _{o}$ is the frequency as observed by the observer, $\nu$ is the frequency of the source, v is the speed of the wave, vo is the speed of the observer and vis the speed of the source.

(i) (a) When the source is moving towards the observer and the observer is stationary.

$\\\nu _{o}=\left ( \frac{v}{v- v_{s}} \right )\nu \\ \nu _{o}=\frac{340}{340-10}\times 400\\ \nu _{o}=412Hz$

(b)

$\\\nu _{o}=\left ( \frac{v}{v+ v_{s}} \right )\nu \\ \nu _{o}=\frac{340}{340+10}\times 400\\ \nu _{o}=389Hz$

(ii) The speed of the sound does not change as it is independent of the speed of observer and source and remains equal to 340 ms-1.

Speed of the wind v= 10 m s-1

Speed of sound in still air va = 340 m s-1

Effective speed with which the wave reaches the observer = v = vw + va = 10 + 340= 350 m s-1

There is no relative motion between the observer and the source and therefore the frequency heard by the observer would not change.

The wavelength of the sound as heard by the observer is

$\\\lambda =\frac{v}{\nu }\\ \lambda =\frac{350}{400}\\ \lambda =0.875m$

The above situation is not identical to the case when the air is still and the observer runs towards the yard as then there will be relative motion between the observer and the source and the frequency observed by the observer would change.

$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$

The frequency as heard by the observer is

$\\\nu _{o}=\frac{340+10}{340}\times 400\\ \nu _{o}=411.76Hz$

$\\\lambda=\frac{340}{400}\\ \lambda=0.85m$

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$
(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?

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$

The displacement of oscillation of a point at x = 1 cm and t = 1 s is

$\\y(1,1)=7.5sin(0.0050\times 1+12\times 1+\frac{\pi }{4})\\ y(1,1)=7.5sin(12.79)\\ y(1,1)=7.5sin(\frac{12.266\times180^{o} }{\pi })\\ y(1,1)=7.5sin(733.18^{o})\\ y(1,1)=1.71cm$

The general expression for the velocity of oscillation is

$\\v_{y}(x,t)=\frac{\mathrm{dy(x,t)} }{\mathrm{d} t}\\=\frac{d}{dt}\left [ sin\left ( 7.5sin(0.0050x+12t+\frac{\pi }{4} \right ) \right ]\\ =90cos(0.0050x+12t+\frac{\pi }{4})$

$\\v_{y}(1,1)=90cos(0.0050\times 1+121+\frac{\pi }{4})\\ =90cos(12.79)\\ =90cos(733.18^{o})\\ =87.63\ cm\ s^{-1}$

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$

k=0.005 cm-1

$\omega =12rad/s$

The velocity of propagation of the wave is

$\\v=\nu \lambda \\ v=\frac{\omega }{2\pi }\times \frac{2\pi }{k}\\ v=\frac{\omega }{k}\\ v=\frac{12}{0.005\times 100}\\ v=24ms^{-1}$

The velocity of oscillation of point at x = 1 cm and t = 1 cm is not equal to the propagation of the wave.

NCERT solutions for class 11 physics chapter 15 waves additional exercise:

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$

(b) Locate the points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

The wavelength of the given wave is

$\\\lambda =\frac{2\pi }{k}\\ \lambda =\frac{2\pi }{0.005 cm^{-1}}\\ \lambda =1256cm\\ \lambda =12.56m$

The points with the same displacements and velocity at the same instant of time are separated by distances $n\lambda$.

The points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s would be at a distance of

$\pm \lambda ,\pm 2\lambda ,\pm 3\lambda ...$ from x = 1cm.

$\lambda =12.56m$

Therefore all points at distances $\pm 12.56m,\pm 25.12m,\pm 37.68m$ from the point x=1cm would have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

(a) The pulse does not have a definite frequency or wavelength however the wave has definite speed given the medium is non-dispersive.

(b) The frequency of the note produced by the whistle is not 0.05 Hz. It only implies the frequency of repetition of the pip of the whistle is 0.05 Hz,

$y(x,t)=Asin(\omega t\pm kx+\phi )$

A=0.05 m

Tension in the string is T=mg

$\\T=90\times 9.8\\ T=882N$

The speed of the wave in the string is v

$\\v=\sqrt{\frac{T}{\mu }}\\ v=\sqrt{\frac{882}{8\times 10^{-3}}}\\ v=332ms^{-1}$

Angular frequency of the wave is

$\\\omega =2\pi \nu \\ \omega =2\pi \times 256\\ \omega =1608.5rad/s$

$\\k=\frac{2\pi }{\lambda }\\ k=\frac{2\pi\nu }{v}\\ k=4.84m^{-1}$

Since at t=0, the left end (fork end) of the string x=0 has zero transverse displacement (y=0) and is moving along the positive y-direction, the initial phase is zero. $\left (\phi =0\ rad \right )$

Taking the left to the right direction as positive we have

$y(x,t)=0.05sin(1608.5t-4.84x)$

Here t is in seconds and x and y are in metres.

Frequency of SONAR$(\nu )$ =40 kHz

Speed of enemy submarine vo=360 km h-1 = 100 m s-1

$\\\nu _{o}=(\frac{v+v_{o}}{v})\nu \\ =\frac{1450+100}{1450}\times 40\times 10^{3}\\ =42.76\ kHz$

This is the frequency which would be observed and reflected by the enemy submarine but won't appear the same to the SONAR(source) as again there is relative motion between the source(enemy submarine) and the observer(SONAR)

The frequency which would be received by the SONAR is

$\\\nu'_{o} =\left ( \frac{v}{v-v_{s}} \right )\nu _{o}\\ =\frac{1450}{1450-100}\times 42.76\times 10^{3}\\ =45.92\ kHz$

Let us assume the earthquake occurs at a distance s.

$\\\Delta t=4\times 60=240s\\ \Delta t=\frac{s}{v_{s}}-\frac{s}{v_{p}} \\\Delta t=\frac{s}{4}-\frac{s}{8}\\ 240=\frac{s}{8}\\ s=1960km$

The origin of the earthquake is at a distance of 1960 km.

Apparent frequency striking the wall and getting reflected is

The frequency emitted by the bats is $\nu =40kHz$

Speed of sound is v

Speed of bat is 0.03v

$\\\nu'=(\frac{v}{v-v_{s}})\nu \\ =\frac{v}{v-0.03v}\times 40kHz\\ =41.237kHz$

Frequency of sound as heard by the bat

$\\\nu''=(\frac{v+v_{o}}{v})\nu'\\ =\frac{v+0.03v}{v}\times \nu' \\=1.03\times 41.237kHz \\ =42.474kHz$

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## Subject wise solutions for NCERT class 11

 NCERT solutions for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry

## Importance of NCERT solutions for class 11 physics chapter 15 waves:

• In  NCERT class 11 physics chapter 15 waves, you will study mathematical and physical aspects of the longitudinal, transverse and physical wave.
• The solutions of NCERT class 11 physics chapter 15 waves help students in self-preparation of the chapter.  In both board exam and competitive exams like Jee Main and NEET beats and the Doppler effects are important topics.
• There are questions based on these topics in the  CBSE NCERT solutions for class 11 physics chapter 15 waves.