Important Questions Class 11 Physics Chapter 15

Class 11 Physics Chapter 15 Important Questions 

Important Questions for CBSE Class 11 Physics Chapter 15 – Waves

Chapter 15 of Class 11 Physics, “Waves”, covers all pertinent topics about the different types and properties of waves. The chapter is divided into smaller segments to include topics such as electromagnetic waves, mechanical waves, etc. Students will also learn the meanings of key terminologies like wavelength, wave velocity, amplitude, frequency, etc.

On Extramarks, you can access authentic Important Questions for Class 11 Physics Chapter 15. Subject matter specialists have created these questions in accordance with the most recent CBSE Syllabus. The solutions are also provided so that you can double-check the correct answer for each question.

CBSE Class 11 Physics Chapter 15 Important Questions

Study Important Questions for Class 11 Physics Chapter 15 – Waves

Some of the Class 11 Physics Chapter 15 Important Questions are provided for your reference. You can also click on the link given here to view additional CBSE Extra Questions for Chapter 15 “Waves” of Class 11 Physics.

Very Short Questions and Answers

1 Mark

Q1. Why are pressure waves referred to as longitudinal waves?

Ans: In order for longitudinal waves to propagate through a medium, fluctuations in air volume and pressure must also occur. These variations in air volume and air pressure lead to the development of compressions and rarefactions, which is why longitudinal waves are also known as pressure waves.

Q2. On a cloudy day, the velocity of sound increases. Why?

Ans: On a cloudy day, the air is wet and contains a lot of moisture. This lowers the air density, and as velocity is inversely related to density, velocity increases on cloudy days.

Q3. Why are there two prongs on tuning forks?

Ans: A tuning fork has two prongs because they enable resonant vibrations, which prolong vibrations.

2 Marks  Questions and Answers

Q1. Sound waves generated in the air cannot be heard by someone submerged deep underwater. Why?

Ans: A person buried deep underwater cannot hear the sound made in the air because of the higher refractive index and faster sound propagation of water compared to that of air.

Index u=SiniSinr=VaVw=14=0.25

Refraction rmax=90°,imax=14°

Since imaxrmax

As a result, sounds can only be heard when they are reflected in the air and cannot be heard by those who are submerged.

Q2. Explain why one cannot hear an echo in a room. 

Ans: It is well known that an obstacle must be substantial and difficult for an echo to be audible.There must also be a distance to separate any blockage from the source. Since rooms are typically smaller than the conditions required for the formation of an echo, no echo can be heard in the space as a result.

Q3. Describe two situations in which the Doppler effect on sound does not exist.

Ans: There is no Doppler effect in sound (i.e., no change in frequency) under the following two conditions.

1. When the listener and the sound source are both moving at the same speed and direction
2. When one of the source listeners is moving around the circle at a steady speed while the other is in the circle’s centre.

3 Marks  Questions and Answers

Q1. A bat is flying around in an eave while using ultrasonic sounds for navigation. Assume that the bat emits sound at a frequency of 40 kHz. The bat is travelling at 0.03 times the speed of sound in the air during one quick swoop directly toward a flat wall surface. What frequency does the bat hear reflected off the wall?

Ans: Ultrasonic beep frequency emitted by the bat, V = 40kHz

Velocity of the bat, vb = 0.03v

Where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as,

v’ = vv-vbv

= vv-0.03v40

= 400.97kHz

This frequency is reflected by the stationary wall (v4 = 0) towards the bat.

The frequency (V’’) of the received sound is given by the relation,

v”= vv-vzv’

v+0.3vv400.97

= 1.03×400.97=42.47kHz

Q2. A stone that is thrown from the top of a tower that is 300 metres high splashes into a pond close to the tower’s base. Given that the speed of sound in air is 340 ms-1, when can we hear the splash at the top? (g = 9.8m,5−1) 

Ans: Given,

Height of the tower, s = 300m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8m5-1

Speed of sound in air = 340 m/s

The time taken (t1),  by the stone to strike the water in the pond can be calculated using the second equation of motion,

s = st1 + 12gt12

300 = 0 + 12×9.8xt12

∴t1 = 300×29.8=7.82s

Time taken by the sound to reach the top of the tower, t1 = 300340=0.88s

Therefore, the time after which the splash is heard, t = t1 + t2

= 7.82 + 0.88 = 8.7s

Q3. The fixed SONAR system of a submarine operates at a frequency of 40.0 kH. At a speed of 360 km/h-1, a hostile submarine approaches the SONAR. What sound frequency does the submarine reflect? Assume that the sound speed in water is 1450 ms-1.

Ans: Given,

Operating frequency of the SONAR system, V = 40kHz

Speed of the hostile submarine, vA = 360 km/h = 100 m/s

Speed of sound in water, v = 1450 m/s

The source is at rest, and the observer (hostile submarine) is moving toward it.

Hence, the apparent frequency (v’) received and reflected by the submarine is

v’ = vv-vAv

= 1450+1001450×40=42.76kHz

The frequency (v”) received by the enemy submarine is

v” = vv-v2v’

where, v2 = 100 m/s

∴v” = 14501450-100×42.76=45.93kHz

4 Marks  Questions and Answers

Q1. A 20 cm long pipe has a closed end. Which harmonic mode of the pipe is resonantly stimulated by a 430 Hz source? If both ends of the pipe are open, will the same source be in resonance with it? (Sound travels at 340 m/s in air.)

Ans: Given,

Length of the pipe I = 20 cm = 0.2m

Source frequency = n

The normal mode of frequency, vn = 430 Hz

Speed of sound, v = 340 m/s

In a closed pipe, the normal mode of frequency is given by the relation,

vn = (2n – 1)v4; n is an integer = 0, 1, 2, 3, …

430 = (2n – 1) 3404×0.2

2n – 1 = 430×4×0.2340=1.01

2n = 2.01

n = 1

Hence, the first mode of vibration frequency is resonantly stimulated by the given source.

In a pipe that opens at both ends, the nth mode of vibration frequency is given by the relation,

vn = Nv2l

vn = 2VAv

= 2×0.2×430340=0.5

Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Q2. Justify (or describe):

(a) A displacement node in a sound wave is also a pressure antinode,

Ans: A node is a location where the pressure is greatest and the vibration amplitude is lowest. On the other hand, an antinode is a location where the pressure is the least and the vibration amplitude is the largest.

As a result, a pressure antinode is simply a displacement node, and vice versa.

(b) Without using their “eyes,” bats can determine the sizes, shapes, and distances between obstacles.

Ans: Bats generate extremely high-frequency ultrasonic sound waves. These waves are deflected by obstacles and returned to them. A bat uses its cerebral sensors to calculate the distance, direction, nature, and size of an obstacle after receiving a reflected wave (frequency).

(c) A note on a violin and a note on a sitar could share the same frequency even though the two notes are distinct.

Ans: The overtones of a violin and a sitar, as well as their intensity, are distinct. Therefore, even if a sitar and a violin have the same vibrational frequency, their notes can be distinguished.

(d) While transverse and longitudinal waves can both travel through solids, only longitudinal waves can do so in gases.

Ans: One characteristic of solids is their shear modulus. They can endure shearing pressure. Transverse waves move in a way that shears the medium they pass through. This type of wave cannot pass through gases and can only move through solids. Gases and solids may both withstand compressive stress. Longitudinal waves can therefore move through solids and through gases.

(e) As a pulse travels through a dispersive material, its shape is distorted.

Ans: Because a pulse is a mix of waves with varying wavelengths, its shape is altered during transmission through a dispersive medium. Pulse distortion occurs during propagation because waves of different wavelengths move at varying speeds.

Q3. An electrically operated tuning fork with a frequency of 256 Hz is linked to one end of a long string with a linear mass density of 8.010-3kgm-1. The other end is linked to a pan with a 90 kg mass and travels over a pulley. The pulley end completely absorbs all incoming energy, resulting in waves that are reflected at this end having minimal amplitude. The string’s left end (the fork end) has zero transverse displacement (y=0) and is travelling in a positive y direction at time t = 0. The wave has an amplitude of 5.0 cm. As a function of x and t, calculate the transverse displacement y that characterises the wave on the string.

Ans: The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation,

y(x, t) = a sin(wt – kx) – (0)

Linear mass density, = 8.0 10-3 kg m-1

Frequency of the tuning fork, v = 256 Hz

Amplitude of the wave, a = 5.0 cm = 0.05 m

Mass of the pan, m = 90 kg

Tension in the string, T = mg = 90 9.8 = 882

The velocity of the transverse wave v, is given by the relation,

v = T

= 8828.0×10-1=332m/s

Angular frequency = 2πυ

= 2×3.14×256

= 1608.5 = 1.6 103 rad/s – (ui)

Wavelength, = vv = 332256m

∴ Propagation constant k = 2π

= 2×3.14332256=4.84w-1

Displacement equation:

y(x,t) = 0.05sin(1.6103t – 4.84x)m

5 Marks  Questions and Answers

Q1. The following formula describes a moving harmonic wave on a string:

y(x,t) = 7.5sin (0.0050x+12t+4).

(a) What are the oscillation displacement and velocity at x = 1 cm and t = 1 s? Is this velocity the same as the velocity of wave propagation?

(b) Find the string points at t = 2 s, t = 5 s, and t = 11 s that have the same transverse displacements and velocities as the x = 1 cm point.

Ans: (a) The given harmonic wave is,

y(x,t) = 7.5 sin (0.0050x + 12t + 4)

For x = 1 cm, and t = 15

y(1, 1) = 7.5 sin(0.0050x + 12t + 4)

= 7.5 sin (12.0050 + 4)

Where, = 12.0050 + 4 = 12.0050 + 3.144 = 12.79 rad

= 1803.14×12.79=732.81”

∴y=1,1=7.5 sin⁡(732.81″)

= 7.5 sin (908+12.81) = 7.5 sin 12.81°

= 7.5 0.2217

= 1.6629 1.663 cm

The velocity of the oscillation at a given point and time is,

v=ddtyx,t=ddt7.5sin 0.0050x+12t+4

7.512cos 0.0050x+12t+4

At tx = 1 cm and t = 1 1:

v=y1,1 = 90 cos(12.005+4)

= 90 cos 73281° = 90 cos(90 cos8+12.82”)

= 90 cos (12.81”)

= 90 0.975 = 87.75 cm/s

Now, the equation of a propagating wave is,

y(xnt) = asin(kx+wt+∅)

Where,

k=2π

And, = 2πv

∴v=2π

Speed, v=vλ=k

Where,

ω=12rads

K = 0.0050 m-1

∴v=120.0050=2400 cm/s

Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as,

k=2π

∴A=2πk=2×3.140.0050

= 1253 cm = 12.56 m

Therefore, all the points at distances (n = ±1, 2, …, and so on) i.e., 12.56m, 25.12m and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5s, and 11 s.

Q2. Using the equation v=yP, explain why the speed of sound in air 

1. is pressure independent,
2. rises as the temperature rises, and

iii. rises as humidity rises.

Ans: Given,

1. Take the relation:

v=yP … (1)

Where,

Density, = MassVolume=MV

M = Molecular weight of the gas

V = Volume of the gas

Hence, equation (1) reduces to

v = yPVM … (2)

Now from the ideal gas equation for n = 1

PV = mT

For constant T, PV =

Since both M and Y are constants, v = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent at the change in the pressure of the gas.

1. Now,

Take the relation:

v=yP  … (1)

For one mole at an ideal gas, the gas equation can be written as,

PV = FT

P = RTV … (2)

By substituting equation (2) in (1),

v=yRTV = yRTM … (3)

Where,

M = pv, which is constant

Y and F are also constants

We conclude from equation (3) that v∝T.

As a result, the square root of the temperature of the gaseous medium, i.e., is inversely related to the speed of sound in a gas. In other words, the temperature of the gaseous medium rises as the speed of sound increases, and vice versa.

iii. Now assume that,

vm and vA be the speeds of sound in moist air and dry air respectively.

And m and A be the densities of moist air and dry air respectively.

Take the relation:

v=yP

Hence, the speed of sound in moist air is,

vm=yPm … (1)

And the speed of sound in dry air is,

vA=yPd … (2)

On dividing equations (1) and (2), we obtain,

vmvA=1Pxy=Am

However, the presence of water vapour reduced the density of air, i.e.,

d<m

∴Vk>V6

As a result, sound moves through moist air more quickly than through dry air. As a result, in a gaseous medium, humidity causes an increase in the speed of sound.

Conclusion

Studying a topic once may help you grasp the concepts, but reviewing it again will help you remember it. Therefore, your chances of getting high grades on tests increase as you study more. By solving these questions, you can thoroughly review everything you have learned from Chapter 15 Waves of Class 11 Physics. Extramarks offers Class 11 Physics Chapter 15 Important Questions curated by subject matter experts according to the latest exam patterns and frequently asked questions that will help you ace the exams.