Question 1

Question 1What is sound, and how is it produced?

Accepted Answer

Sound is a form of energy that produces the sensation of hearing in our ears. Sound is produced when a body vibrates.For example — sound of morning alarm, a dog barking, music from different instruments etc.

Question 2

Question 2Describe, with the help of a diagram, how compressions and rarefactions are produced in the air near a source of the sound.

Accepted Answer

When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression as shown in the figure below:When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R), as shown in the above figure. As the object moves back and forth rapidly, a series of compressions and rarefactions are created in the air.

Question 3

Question 3Why is sound wave called a longitudinal wave?

Accepted Answer

In longitudinal waves the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. This is exactly how a sound wave propagates, hence sound waves are called longitudinal waves.

Question 4

Question 4Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Accepted Answer

Timbre or quality of sound makes it possible to recognize a person by his voice without seeing him. The vibrations produced by the vocal cord of each person have a characteristic wave form which is different for different persons.

Question 5

Question 5Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?

Accepted Answer

Thunder is heard a few seconds after the flash is seen because sound travels much slower than light.Light takes almost negligible time in comparison to sound in reaching us from the place of thunder because speed of light is much more (3 x 108 ms-1) than the speed of sound (= 330 ms-1).

Question 6

Question 6A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms-1.

Accepted Answer

Given,Speed (vair) = 344 ms-1We know,Speed (v) = Wavelength (λ) x Frequency (ν)orWavelength (λ) = vν\dfrac{	ext{v}}{	ext{ν}}νv​(a) For ν = 20 Hz, substituting we get,λ1 = 34420\dfrac{344}{20}20344​ = 17.2 m(b) For ν = 20,000 Hz, substituting we get,λ2 = 34420,000\dfrac{344}{20,000}20,000344​ = 0.0172 mHence, typical wavelengths of sound waves in air corresponding to 20 Hz and 20,000 Hz are 17.2 m and 0.0172 m respectively.

Question 7

Question 7Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in the air and in aluminium to reach the second child.

Accepted Answer

Let length of the aluminium rod be l.Speed of sound in air (vair) at 25°C = 346 ms-1Speed of sound in aluminium (valu) at 25°C = 6420 ms-1We know,Speed = DistanceTime\dfrac{	ext{Distance}}{	ext{Time}}TimeDistance​orTime (t) = DistanceSpeed\dfrac{	ext{Distance}}{	ext{Speed}}SpeedDistance​Substituting we get,tair = l346\dfrac{	ext{l}}{	ext{346}}346l​talu = l6420\dfrac{	ext{l}}{	ext{6420}}6420l​So, the ratio of time taken by sound in air and aluminium istairtalu\dfrac{	ext{t}_	ext{air}}{	ext{t}_	ext{alu}}talu​tair​​ = l346l6420\dfrac{\dfrac{	ext{l}}{	ext{346}}}{\dfrac{	ext{l}}{	ext{6420}}}6420l​346l​​ = 6420346\dfrac{6420}{346}3466420​ = 18.55Hence, the ratio of time taken by sound in air and aluminium = 18.55

Question 8

Question 8The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Accepted Answer

Given,Frequency (ν) = 100 HzTime (t) = 1 min = 60 sFrequency (ν) = Number of oscillationsTotal time (t)\dfrac{	ext{Number of oscillations}}{	ext{Total time (t)}}Total time (t)Number of oscillations​OrNumber of oscillations = ν × tSubstituting we get,Number of oscillations = 100 × 60 = 6000Hence, the source vibrates 6000 times in a minute.

Question 9

Question 9Does sound follow the same laws of reflection as light does? Explain.

Accepted Answer

Yes, the laws of reflection are same for light as well as sound.The law of reflection of sound states that the directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface at the point of incidence and the three lie in the same plane.

Question 10

Question 10When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear an echo sound on a hotter day?

Accepted Answer

Speed of sound increases with temperature. Hence, on a hotter day, speed of sound in air will be more.An echo is heard when the time interval between the reflected sound and the original sound is at least 0.1 seconds.Therefore, on a hotter day, as speed of sound in air will be more, for same distance, time taken by the reflected sound to reach the ear will be less. As chances of this time interval being less than 0.1 seconds is more so we might not hear an echo sound on a hotter day.

Question 11

Question 11Give two practical applications of the reflection of sound waves.

Accepted Answer

Two practical applications of the reflection of sound waves are:Stethoscope — It is a medical device used to hear sounds within the body mainly heartbeat. In stethoscopes the sound of the patient's heartbeat reaches the doctor's ears by multiple reflection of sound.Curved ceilings of concert halls — Generally the ceilings of concert halls, conference halls and cinema halls are curved so that sound after reflection reaches all corners of the hall.

Question 12

Question 12A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms-2 and speed of sound = 340 ms-1

Accepted Answer

Given,Height of tower (s) = 500 mVelocity of sound (v) = 340 ms-1Acceleration due to gravity (g) = 10 ms-2Initial velocity of the stone (u) = 0Time taken by the stone to fall to the tower base (t1):According to the second equation of motion,s = ut1 + 12\dfrac{1}{2}21​gt12Substituting we get,500 = (0 x t1) + (12\dfrac{1}{2}21​ x 10 x t12)500 = 12\dfrac{1}{2}21​ x 10 x t121000 = 10 x t12t12 = 100010\dfrac{1000}{10}101000​ = 100Hence, t1 = 100\sqrt{100}100​ = 10 sTime taken by sound to reach the top from the tower base (t2)= ?Using,Speed = DistanceTime\dfrac{	ext{Distance}}{	ext{Time}}TimeDistance​orTime = DistanceSpeed\dfrac{	ext{Distance}}{	ext{Speed}}SpeedDistance​Substituting we get,t2 = 500340\dfrac{500}{340}340500​ = 1.47 st = t1 + t2t = 10 + 1.47t = 11.47 sHence, the splash will be heard at the top after 11.47 s.

Question 13

Question 13A sound wave travels at a speed of 339 ms-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Accepted Answer

Given,Speed of sound (v) = 339 ms-1Wavelength of sound (λ) = 1.5 cmConverting cm to m100 cm = 1 m1.5 cm = 1.5100\dfrac{1.5}{100}1001.5​= 0.015 mWe know,Speed (v) = Wavelength (λ) x Frequency (ν)orFrequency (ν) = Speed (v)Wavelength (λ)\dfrac{	ext{Speed (v)}}{	ext{Wavelength (λ)}}Wavelength (λ)Speed (v)​Substituting we get,ν = 3390.015\dfrac{339}{0.015}0.015339​ = 22,600 HzHence, the frequency of the sound = 22,600 HzNo, it will not be audible as it is outside the audible range for human ear i.e. 20 Hz to 20,000 Hz.

Question 14

Question 14What is reverberation? How can it be reduced?

Accepted Answer

The persistence of sound in an auditorium as a result of repeated reflections is known as reverberation.To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties

Question 15

Question 15What is loudness of sound? What factors does it depend on?

Accepted Answer

Loudness is a physiological response of the ear to the intensity of sound.The loudness of sound heard by a listener depends on the following factors:Amplitude — Loudness is proportional to the square of the amplitude.Distance from source — Loudness varies inversely as the square of the distance from the source.Surface area of the vibrating body — Loudness depends on the surface area of the vibrating body.

Question 16

Question 16How is ultrasound used for cleaning?

Accepted Answer

Ultrasound is used to clean parts located in hard-to-reach places, for example, spiral tube, odd shaped parts, electronic components, etc. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. Thus, the object is thoroughly cleaned.

Question 17

Question 17Explain how defects in a metal block can be detected using ultrasound.

Accepted Answer

Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect, as shown below:

Question 18

Question 1How does the sound produced by a vibrating object in a medium reach your ear?

Accepted Answer

Sound moves through a medium from the point of generation to the listener. When an object vibrates, it sets the particles of the medium around it vibrating. A particle of the medium in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle. As a result of which the adjacent particle gets displaced from its position of rest. After displacing the adjacent particle the first particle comes back to its original position. This process continues in the medium till the sound reaches our ear.

Question 19

Question 2Explain how sound is produced by your school bell.

Accepted Answer

When struck by a hammer, the school bell is set into motion, causing it to oscillate back and forth creating compressions and rarefactions in the air. This alternation of compressions and rarefactions forms a sound wave that travels to our ears, allowing us to hear the bell's sound.

Question 20

Question 3Why are sound waves called mechanical waves?

Accepted Answer

Sound waves are called mechanical waves because they require a material medium, like air or water or solids, for propagation. They propagate by causing particles in the medium to oscillate, transmitting energy through mechanical vibrations.

Sound

Exercises

Intext Questions 1

Intext Questions 2

Intext Questions 3

Intext Questions 4

Intext Questions 5

Intext Questions 6

Intext Questions 7

Intext Questions 8