NCERT Solutions for Class 11 Physics Chapter 15 Waves

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 $(a)\; 4\; m$

Answer:

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

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

$(b)\; 0.5\; m$

Answer:

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

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 $(c)\lambda /2$

Answer:

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

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

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 $(d)3\lambda /4$

Answer:

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

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

Q.15.11 The transverse displacement of a string (clamped at its both ends) is given by

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

Answer the following :

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

Answer:

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 $3.0\times 10^{-2}\; kg$ .

Answer the following :

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

Answer:

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

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

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

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

Q.15.11 (c) 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 $3.0\times 10^{-2}\; kg$ .

Answer the following :

(c) Determine the tension in the string

Answer:

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

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

15.12) For the wave on a string described in Exercise 15.11, 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) (a) All the points vibrate with the same frequency of 60 Hz.

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

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

(ii)

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

Q.15.13 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 (i) a travelling wave, (ii) a stationary wave or (iii) none at all:

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

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

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

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

Answer:

(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.

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

Q.15.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\: Hz$ . The mass of the wire is $3.5\times 10^{-2}\: kg$ and its linear mass density is $4.0\times 10^{-2}\: kg\: m^{-1}$ . What is (a) the speed of a transverse wave on the string?

Answer:

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

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

Q.15.14 (b) A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\: Hz$ . The mass of the wire is $3.5\times 10^{-2}\: kg$ and its linear mass density is $4.0\times 10^{-2}\: kg\: m^{-1}$ . What is (b) the tension in the string?

Answer:

Tension in the string is given by

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

Q.15.15 A metre-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:

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 l ₁ = 25.5 cm

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

For l ₂ = 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)

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

Q.15.16 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 the speed of sound in steel?

Answer:

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

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

Q.15.17 A pipe $20\: cm$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 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:

Let the n ^th 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 m ^th 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.

Q.15.18 Two sitar strings A and B playing the note ‘Ga’ are slightly 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:

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

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,

Answer:

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”,

Answer:

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,

Answer:

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,

Answer:

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.

Answer:

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.

Q.15.20 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 $340\: m\; s^{-1}$ .

Answer:

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

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, v _o is the speed of the observer and v _s is 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 .

Q.15.21 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’s 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:

Speed of the wind v _w = 10 m s ^-1

Speed of sound in still air v _a = 340 m s ^-1

Effective speed with which the wave reaches the observer = v = v _w + v _a = 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

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

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

Q.15.22 (a) A travelling harmonic wave on a string is described by

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

Answer:

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

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

Q.15.22 A travelling harmonic wave on a string is described by

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

(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.

Answer:

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.

Q.15.23 A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) 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) 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,

Q.15.24 One end of a long string of linear mass density $8.0\times 10^{-3}kg\: m^{-1}$ 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:

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

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 displacements (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.

Q.15.25 A SONAR system fixed in a submarine operates 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:

Frequency of SONAR $(\nu )$ =40 kHz

Speed of enemy submarine v _o =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

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

Q.15.26 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 $4.0\: km\: s^{-1}$ , 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 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.

Q.15.27 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 toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall ?

Answer:

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 15 Waves Free PDF Download

The ncert solutions waves class 11 pdf includes in-depth explanations of all the crucial topics and subtopics, making it easier for students to comprehend the concepts.

The NCERT solutions for Waves Physics class 11 are crucial in establishing the groundwork for the Class 12 CBSE board exams, which determine students' admission to higher education institutes. Emphasis on the fundamental concepts and theoretical aspects of this chapter is critical for building a strong foundation for the Class 12 CBSE board exam. Moreover, Class 11 Physics Ch 15 NCERT solutions have been developed in accordance with the latest CBSE and NCERT guidelines and syllabus.

NCERT Solutions for Class 11 Physics Chapter Wise

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

Class 11 physics waves ncert solutions: Important Formulas and Diagrams

Chapter 15: Waves

Wave velocity

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Wave velocity(W) = Frequency(f) * wavelength (λ) [SI unit- m/s]

velocity of the transverse wave in a stretched string

$V= \sqrt{\frac{T}{\mu }}$

where:

v = Velocity of the wave (in meters per second, m/s), T = Tension in the string (in Newtons, N) and μ = Linear mass density of the string (in kilograms per meter, kg/m)

Velocity of a longitudinal wave in an elastic medium

$V= \sqrt{\frac{E}{\rho }}$

Where:

E = Young's modulus of the material (in Pascals, Pa), ρ = Density of the material (in kilograms per cubic meter, kg/m³)

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 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.

Key features of Class 11 Physics Waves NCERT Solutions

Comprehensive Coverage: These class 11 physics ch 15 ncert solutions cover all the important topics and questions presented in the chapter, ensuring a thorough understanding of wave phenomena.
Exercise and Additional Exercise Solutions: Detailed solutions are provided for exercise questions (15.1 to 15.21) and additional exercise questions (15.22 to 15.27), allowing for comprehensive practice and self-assessment.
Clarity and Simplicity: The solutions are explained in clear and simple language, making complex wave concepts more accessible to students.
Problem-Solving Skills: By practising with these solutions and class 11 physics chapter 15 waves notes, students develop strong problem-solving skills, which are essential in physics and other subjects.
Exam Preparation: These wave class 11 solutions are designed to help students prepare effectively for their exams, including board exams and competitive exams like NEET and JEE Mains.

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Frequently Asked Questions (FAQs)

1. What are the features of NCERT solutions for Class 11 Physics Chapter 15?

Waves is one of the important chapters of Class 11 NCERT syllabus. The questions given in the NCERT books are important to understand the concepts well. There are two exercise and 27 questions discussed in the Waves Class 11 NCERT chapter. For more questions refer to NCERT exemplar.

2. What are the topics covered in Waves Class 11 NCERT chapter?

The main topics covered in the NCERT Class 11 textbook chapter 15 are

Transverse and longitudinal waves
Displacement relation in a progressive wave
The speed of a traveling wave
The principle of superposition of waves
Reflection of waves
Beats
Doppler effect

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NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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