# NCERT Solutions for Class 9 Science Chapter 12 Sound

These NCERT Solutions for Class 9 Science Chapter 12 Sound Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

## Sound NCERT Solutions for Class 9 Science Chapter 12

### Class 9 Science Chapter 12 Sound InText Questions and Answers

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way, vibrations produced by an object are transferred from one particle to another till it reaches the ear.

Question 2.
Explain how sound is produced by your school bell.
When the school bell vibrates, it forces the adjacent particles in air to vibrate. This disturbance gives rise to a wave and when the bell moves forward, it pushes the air in front of it. This creates a region of high pressures known as compression. When the bell moves backwards, it creates a region of low pressure know as rarefaction. As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. This makes the sound of a bell propagate through air.

Question 3.
Why are sound waves called mechanical waves?
Sound waves force the medium particles to vibrate. Hence, these waves are known as mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium.

Question 4.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Sound needs a medium to propagate. Since the moon is devoid of any atmosphere, you cannot hear any sound on the moon. Question 5.
Distinguish between loudness and intensity of sound.
Intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.

Question 6.
How are the wavelength and frequency of a sound wave related to its speed?
Speed, wavelength, and frequency of a sound wave are related by the following equation:
Speed (v) = Wavelength (λ) × Frequency (u)
v = λ × u

Question 7.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans. Frequency of the sound wave, υ = 220 Hz
Speed of the sound wave, v = 440 m s-1
For a sound wave,
Speed = Wavelength × Frequency
v = λ × u
∴ λ = $$\frac{v}{u}=\frac{440}{220}$$ = 2 m
Hence, the wavelength of the sound wave is 2m.

Question 8.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
The time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:
T = $$\frac{1}{\text { Frequency }}=\frac{1}{500}$$ = 0.002s

Question 9.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
The speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases.
Therefore, for a given temperature, sound travels fastest in iron. Question 10.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms-1?
Speed of sound, v = 342 ms-1
Echo returns in time, t = 3s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source = $$\frac{1026}{2}$$ m = 513 m

Question 11.
Why are the ceilings of concert halls curved?
Ceilings of concert halls are curved so that sound after reflection (from the walls) spreads uniformly in all directions.

Question 12.
What is the audible range of the average human ear?
The audible range of an average human ear lies between 20 Hz to 20,000 Hz. Humans cannot hear sounds having frequencies less than 20 Hz and greater than 20,000 Hz.

Question 13.
What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

Question 14.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in saltwater is 1531 m/s, how far away is the cliff?
Time taken by the sonar pulse to return, t = 1.02 s
Speed of sound in saltwater, v = 1531 ms-1
Total distance covered by the sonar pulse = Speed of sound × Time taken
Total distance covered by the sonar pulse
= 1.02 × 1531 = 1561.62 …..(i)
Let d be the distance of the cliff from the submarine.
Total distance covered by the sonar pulse = 2d
⇒ 2d = 1561.62 [From(i)]
⇒ d = 780.81 m

### Class 9 Science Chapter 12 Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Sound is produced by vibration. When a body vibrates, it forces the neighbouring particles of the medium to vibrate. This creates a disturbance in the medium, which travels in the form of waves. This disturbance, when reaches the ear, produces sound. Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
When a vibrating body moves forward, it creates a region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions (as shown in the following figure). Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Take an electric bell and hang this bell inside an empty bell-jar fitted with a vacuum pump (as shown in the following figure). Initially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell decreases. If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid of any air. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. When there is no air present inside, we can say that a vacuum is produced. Sound cannot travel through vacuum. This shows that sound needs a material medium for its propagation.

Question 4.
Why is sound wave called a longitudinal wave?
The vibration of the medium that travels along or parallel to the direction of the wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Hence, a sound wave is called a longitudinal wave.

Question 5.
Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Quality of sound is that characteristic which helps us identify a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different. Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
The speed of sound (344 m/s) is less than the speed of light (3 × 108 m/s). Sound of thunder takes more time to reach the Earth as compared to light. Hence, a flash is seen before we hear a thunder.

Question 7.
A 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.
For a sound wave,
Speed = Wavelength × Frequency
v = λ × υ
Given that the speed of sound in air = 344 m/s
(j) For, υ = 20.Hz
λ1 = $$\frac{v}{v_{1}}=\frac{344}{20}$$ = 17.2m
(ii) For, υ2 = 20,000 Hz
λ2 = $$\frac{v}{v_{2}}=\frac{344}{20,000}$$ = 0.0172 m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Question 8.
Two 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 air and in aluminium to reach the second child.
Let the length of the aluminium rod be d.
Speed of sound wave in aluminium at 25°C,
vAl = 6420 ms-1
Therefore, time taken by the sound wave to reach the other end,
$$t_{\mathrm{Al}}=\frac{d}{v_{\mathrm{Al}}}=\frac{d}{6420}$$
Speed of sound wave in air at 25°C, vAir = 346 ms-1 .
Therefore, time taken by sound wave to reach the other end,
$$t_{\text {Air }}=\frac{d}{v_{\text {Air }}}=\frac{d}{346}$$
The ratio of time taken by the sound wave in air and aluminium: $$\frac{t_{\text {Air }}}{t_{\mathrm{Al}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55$$

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Frequency is defined as the number of oscillations per second. It is given by the relation:
Frequency = $$\frac{\text { Number of oscillations }}{\text { Total Time }}$$
Number of oscillations = Frequency × Total time
Given, Frequency of sound = 100 Hz
Total time = 1 min = 60 s
Number of oscillations / Vibrations = 100 × 60 = 6000
Hence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Sound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane. Question 11.
When 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 echo sound on a hotter day?
An echo is heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease. Therefore, an echo can be heard only if the time interval between the original sound and the reflected sound is greater than 0.1 s.

Question 12.
Give two practical applications of reflection of sound waves.

• Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
• Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.

Question 13.
A 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-2and speed of sound = 340 ms-1.
Height of the tower, s = 500 m
Velocity of sound, v = 340 ms-1
Acceleration due to gravity, g = 10 ms-2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t1

According to the second equation of motion:
s = ut1 + $$\frac {1}{2}$$gt12
500 = 0 × t1 + $$\frac {1}{2}$$ × 10 × t12
r12 = 100
t1 = 10 s
Now, time taken by the sound to reach the top from the base of the tower, t2 = $$\frac {500}{340}$$ = 1.47S
Therefore, the.splash is heard at the top after time, t
Where, t = t1 + t2 = 10 + 1.47 = 11.47 s

Question 14.
A 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?
Speed of sound, v = 339 ms-1
Wavelength of sound, λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × υ
∴ υ = $$\frac{v}{\lambda}=\frac{339}{0.015}$$ = 22600 Hz
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible. Question 15.
What is reverberation? How can it be reduced?
Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. As the source produces sound, it starts travelling in all directions. Once it reaches the wall of a room, it is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound.

To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, rough plastic, heavy curtains, and cushioned seats can be used. to reduce reverberation.

Question 16.
What is loudness of sound? What factors does it depend on?
A loud sound has high energy. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 17.
Explain how bats use ultrasound to catch a prey.
Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are reflected by objects such as preys and returned to the bat’s ear. This allows a bat to know the distance of his prey.

Question 18.
How is ultrasound used for cleaning?
Objects to be Cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Question 19.
Explain the working and application of a sonar.
SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction, and speed of underwater objects such as submarines and shipwrecks with the help of ultrasounds. It is also used to measure the depth of seas and oceans. A beam of ultrasonic sound is produced and transmitted by the transducer (it is a device that produces ultrasonic sound) of the SONAR, which travels through seawater. The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the under-water object is calculated from the time (f) taken by the echo to return with speed (v) is given by 2d = v × t. This method of measuring distance is also known as ‘echo-ranging’.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Time taken to hear the echo, t = 5s
Distance of the object from the submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Velocity of sovmd in water,
v = $$\frac{2 d}{t}=\frac{2 \times 3625}{5}$$ = 1450 ms-1 Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound. Question 22.
Explain how the human ear works.
Different sounds produced in our surroundings are collected by pinna that sends these sounds to the eardrum via the ear canal. The eardrum starts vibrating back and forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone hammer into vibration. The vibrations are, passed from the hammer to the second bone anvil, and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval window and passes its vibration to the liquid in the cochlea. This produces electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and we get a sensation of hearing. ### Class 9 Science Chapter 12 Sound Additional Important Questions and Answers

Multiple choice Question
Choose the correct option:

Question 1.
Note is a sound
(a) of mixture of several f.requencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) a ways unpleasant to listen
(c) of a single frequency

Question 2.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
(a) sound will be louder but pitch will not be different
(b) sound will be louder and pitch will also be higher
(c) sound will be louder but pitch will be lower
(d) both loudness and pitch will remain unaffected
(a) sound will be louder but pitch will not be different

Question 3.
In SONAR, we use
(a) ultrasonic waves
(b) infrasonic waves
(d) audible sound waves
(a) ultrasonic waves

Question 4.
Sound travels in air if
(a) particles of medium travel from one place to another
(b) there is no moisture in the atmosphere
(c) disturbance moves
(d) both particles as well as disturbance travel from one place to another.
(c) disturbance moves Question 5.
When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength
(b) amplitude

Question 6.
In the curve (Fig.12.1) half the wavelength is (b)

Question 7.
Earthquake produces which kind of sound before the main shock wave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above
(b) infrasound

Question 8.
Infrasound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings
(c) rhinoceros

Question 9.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound
(c) frequency of the sitar string with the frequency of other musical instruments

Question 1.
What is sound?
It is a form of energy that enables us to hear.

Question 2.
What is necessary for a body to produce sound?
The vibrating bodies can only produce sound. Question 3.
Which three parameters are used to describe sound?
The three parameter used to describe sound are amplitude, frequency and quality.

Question 4.
What determines the loudness of a sound wave?
Amplitude determines the loudness of a sound.

Question 5.
What determines the pitch of a sound.
The frequency of sound wave determines the pitch.

Question 6.
What are two types of mechanical wave motions?
The two types of mechanical wave motions ate transverse waves and longitudinal waves.

Question 7.
Name the wave motion in which the wave propagates in the direction of motion.
In longitudinal wave motion, the waves propagate in the direction of motion.

Question 8.
Why is sound wave called a mechanical wave?
Sound cannot travel through vacuum, it requires the presence of a medium for its propagation.

Question 9.
In which physical medium, the sound travels the fastest?
Sound travels the fastest in solids. Question 10.
What are curtains and furniture in a house, good reflectors or absorbers of sound?
They are good absorbers of the sound.

Question 11.
What do you understand by the reflection of sound?
It refers to the bouncing back of the sound in the same medium after striking a non-absorptive solid surface.

Question 12.
Which part of human body helps produce the sound?
The vocal cord also called Adam’s apple help produce the sound.

Question 13.
In which medium will the reflection of sound be faster, air or water?
In water, the density of water is more than that of air and hence, the speed of sound in water is more than in air.

Question 14.
Name the property of sound, human ears respond to after hearing the sound.
Human ears are receptive to the loudness therefore, they only respond to loudness.

Question 15.
Which property is used to distinguish two sounds from different sources but having the same amplitude and pitch ?
The property used to distinguish the sound waves having same pitch and frequency is called quality. Question 16.
In which form the sound waves travels across a medium.
The sound waves travel in the form of. compression and rarefaction.

Question 17.
How is an ultrasonic wave different from the infrasonic wave?
Infrasonic waves have the frequency less than 20Hz while the ultrasonic waves have the frequency of more than 20KHz.

Question 18.
If the speed of sound incident on a surface is doubled, will there be any effect on the angle of reflection of sound ?
No, there will be no effect on the angle of reflection of sound because it is dependent on the angle of incidence, not on the speed of sound.

Question 19.
What will happen, if a sound wave is made incident at 90° on a solid surface?
No reflection will occur but the development of resonance Will take place.

Question 20.
Can sound waves be transformed into » electrical impulses or vice versa?
Yes, the sound waves can be transformed into electrical impulses or vice versa as it happens in telephone.

Question 21.
What is the law of conservation of energy?
The law states that energy can neither be created nor be destroyed but one form of energy can be transformed into another form of energy.

Question 22.
State the energy transformations which take place when you clap your hands.
When dapping, the muscular energy is transformed into sound and heat energy. Question 23.
How is a light wave different from a sound wave?
Sound wave requires a medium for its propagation but not light wave. It can even pass through a vacuum.

Question 24.
The distance in between two adjoining crest and trough is ‘d’. What is the wavelength of the wave?
The wavelength is the distance in between two successive troughs or crests, therefore the wavelength would be 2d.

Question 25.
What do you mean by supersonic speed?
When the speed of an object/body exceeds the speed of sound, then the speed is called supersonic speed.

Question 1.
What is a wave? Can sound be visualized as a wave?
Wave is a disturbance that moves through a medium due to repeated oscillatory motion of the particles of the medium about their mean position. This oscillatory motion is passed over from one particle to another progressively. Hence, a wave only involves the transfer of energy not particles of the material medium.

Sound is visualized as a wave because the disturbance set by the sound in the medium travel through the medium in form of energy instead of the particle of the medium.

Question 2.
What are longitudinal waves? Give an example.
The waves in which the particles of the medium vibrate in the direction of the propagation of the wave are called longitudinal waves. During propagation of these waves, the particles of the medium only exhibit the vibratory motion along their mean position.

Sound waves propagate in the form of longitudinal waves. Question 3.
What are transverse waves? Give an example.
The waves in which the particles of the medium vibrate in the direction perpendicular to the direction of the propagation of wave is called transverse wave. The positive displacement from the mean position makes up the crest while the negative displacement from the mean position makes up the trough.

When a pebble is dropped in stagnant water, water ripples formed on the surface water represents the transverse waves.

Question 4.
What are mechanical and non¬mechanical waves? Give examples.
The waves which require the presence of a material medium for their propagation are called mechanical or elastic waves e.g. sound waves. These waves can propagate through all the three states of matter but not through a vacuum.

The waves which do not require the presence of a material medium for their propagation are called non-mechanical waves or non-elastic waves e.g. light waves, microwaves or radio waves. These waves can easily propagate across the vacuum.

Question 5.
How is the loudness or softness of a sound determined? Give the wave shape of a loud and soft sound of same frequency.
The loudness or softness of a sound depends upon its amplitude. A loud sound has higher amplitude than a soft sound when the amplitude describes the maximum displacement of the particle from its mean position. Question 6.
The sound produced by a vehicle and a flute travels through the same medium, air and arrive at the ear at same time. Are the two sounds different? If yes, give reasons.
No, the two sounds are different. The pitch of the sound is one of the characteristics responsible for the difference. The sound from the flute has higher frequency and has pleasing effect but the sound from vehicle has low pitch. It is cracking sound and causes irritation.

Question 7.
Different musical instruments produce different sounds. Why ?
The different musical instruments have different modes of vibrations. These instruments with different sizes under different conditions vibrate at different frequency producing sounds of different pitches and hence, they are different from one another. Question 8.
Give a graphical representation of for a sound wave having low pitch and high pitch.
The pitch of a sound is determined by its frequency. Therefore, a sound with low pitch has low frequency i.e. number of the waves produced per second and sound with higher pitch has higher frequency. (Fig. 12.6). Question 9.
In humans, whose voice is sharp, a male or female? Why?
In humans, the voice of a female is sharper than the voice of a male because the sound produced by a female has higher frequency i.e. pitch than the sound produced by a male.

Question 10.
A person standing on the railway platform of a rural area could neither see nor hear the sound of the train. With none to help him, how can he assess the arrival time of the train?
The person can assess the arrival time of the train by keeping his ear close to the rail line and sense the vibrations of the incoming train.

In steel, the speed of sound is 5960 m/s as compared to the speed of sound in air which is only 340 m/s. Hence, he can easily estimate the possible arrival time of the train. The fast vibration would indicate the early arrival while the slow vibrations would indicate the late arrival.

Question 11.
A person standing 1000 m away from a siren hears the sound. When will he hear the sound earlier and why?
(i) On a hot day or a calm day?
(ii) On a dry day or cloudy day having same temperature?
(i) The person will hear the sound earlier on a hot day than on a calm day because with increasing temperature, the speed of the sound increases. At 273 K, the speed of sound is 331 m/s while at 295 k, it is 344 ms.

(ii) The person will hear the sound earlier on a cloudy day than on a dry day because the speed of the sound depends upon the density of the medium of propagation. On a cloudy day because of the presence of the moisture in air, its density is more than on a dry day.

Question 12.
What is the audible range for the human beings?
The human ears respond to the sound waves having frequency of 20 Hz to 20 kHz. Therefore, this range of sound waves is called audible range. However, children can hear the sound up to the frequency of 25 kHz while with increasing age the range declines as old people become less sensitive to higher frequency of sound, their maximum frequency range goes up to 15,000 Hz. Question 13.
What are infrasound and ultrasound? Give examples.
The classification of sound waves into infrasound and ultrasound is based on the human audible range of 20 Hz to 20 kHz. The sound waves having frequency less than 20 Hz are called infra sounds and the sound waves having frequency greater than 20 kHz tire called ultrasounds.

Rhinoceros, whales and elephants can produce and respond to the infrasound waves.

Bats and rats can produce and receive ultrasound waves. Rats when playing use ultrasound waves while bats use ultrasound waves for flying in dark and capture their prey.

Question 14.
What is sonic boom? Is it harmful, if yes then give reason?
When sound producing source moves with the speed of sound or above such as some fighter aircrafts, it produces shock wave in air. These shock waves contain a large amount of energy. The air pressure variations associated with this type of shock wave sharp and loud sound called sonic boom.

The shock waves of sonic boom possess lot of energy. They can cause the shattering of glass windows and damage the buildings.

Question 15.
What do you mean by the reflection of sound? What are the laws of reflection of sound?
The reflection of the sound refers to the bouncing back of sound waves after being incident on a hard polished surface in the same medium.

According to the laws of the reflection of sound:

• The incident sound wave, reflected sound wave and the normal drawn at the point of incidence are in the same plane.
• The direction in which the sound is incident and is reflected make equal angle with the normal at the point of incidence.

Question 16.
What are megaphones? Why the loudness of sound is increased by megaphones?
A megaphone is a simple horn shape tube followed by a conical opening.
In a megaphone, the sound waves from a source are reflected successively from the conical surface of the tube and directed towards audience without spreading the sound. Moreover, the amplitude of the sound waves adds up increasing the loudness of sound.

Question 17.
What is an echo? State the conditions necessary for echo formation.
An echo refers to the reflected sound that is reheard by a speaker/listener himself. For the echo formation to occur

• there have to be a good reflective surface to reflect the sound.
• the minimum distance between the source of sound and the reflective surface has to be 17.2 m.
• there should not be any obstruction in the path of speaker and the reflective surface.

Question 18.
What is the minimum distance between the listener and the reflecting surface for hearing the distinct echo? Why?
In humans, the time period of the persistence of sound is 0.1 second. The speed of sound in air at 295 k (22°C) is 344 m/s.

For a listener to hear a distinct echo, the reflected sound should reach his ears after a period of 0.1s.

Hence, the total distance sound of echo has to cover is = 344 × 0.1 = 34.4 m.
Therefore, the minimum distance of the reflecting surface has to be 344/2 = 17.2 m

However, this difference is temperature dependent because with the increasing temperature, the speed of the sound increases.

Question 19.
Explain why the walls and roof of the auditorium are covered with sound absorbing material?
The walls and roof of a good auditorium are covered with sound absorbing material such as fibre boards and draperies to prevent the reverberation i.e., the repeated reflection of the sound waves. The reverberation makes the sound blurred and difficult to understand and interpret. Question 20.
Why is the flash of lightning seen much before the thunder is heard on a rainy day, although both occur simultaneously?
Although both lightning and thunder formation occur simultaneously at the same height in air yet the lightning flash is seen much before the hearing of thunder sound because of the difference in the speed of the sound and light The sound traveling with speed of 342 m/s takes a longer time than light traveling at the speed of 3 × 108 m/s to cover the same distance.

Question 21.
State some of the practical applications of echo.
An echo which is a reflected sound has its own applications such as

• In echo ranging or sonar, it is used to determine the depth of the oceans.
• The submarines floating in water not only measure the depth of the ocean but also the obstruction in their path in front if any.
• Bats use the echo or sound reflection in finding an obstruction free path for flying and capturing their prey.
• In medical sciences the ultrasound waves are used widely in diagnosis of structural disorders of body parts.

Question 22.
What is ultrasonography? State its use.
Ultrasonography is a technique in which the ultrasonic waves are used to asses the structure of a body part or tissue. The ultrasound waves are made to travel through tire body tissues. Where ever there is a change in tissue density, these waves get reflected, then the reflected waves are transformed into electrical signals to generate the picture of the tissue.

The technique is widely used in examination of the growing foetus in womb of a pregnant mother and in analysis of stones in different body organs such as gall bladder, liver, etc.

Question 23.
State some of the advantages of the ultrasounds in medical sciences.
Ultrasound consists of sound waves having frequency greater than 20 kHz. These waves are used in

• Echo-cardiography, a technique used to diagnose the blockage of heart valves or arteries.
• Creating pictures of different body organs to detect the presence of stone or any other structure like tumor.
• In breaking the small stones formed in kidney for their easy removal.

Question 24.
What is a stethoscope? On what principle does it work?
A stethoscope is a medical instrument used by a doctor to hear the heart sounds, ‘Lub- Dub’. It’s working is based on the repeated and multiple reflection of the sound waves, received by the broad round receiver.

Question 25.
How is pressure variation in a sound wave amplified in human ear?
The pressure vibrations produced by a vibrating object in form of compression and rarefaction reach inside the external ear and make the eardrum to vibrate. The three bones of the middle year: hammer, anvil and stirrup being solid causes the amplification of the waves by several times.

Question 26.
A longitudinal wave is produced on a toy slinky. The wave travels at the speed of 30 cm/s. If the frequency of the wave is 20 Hz. What is the minimum distance between the two consecutive compressions?
Velocity of wave, v = 30 cm/s = 0.3 m/s
Frequency of wave, n = 20 Hz
Distance between two consecutive compressions = λ.
We know that v = λn
λ = $$\frac{v}{n}$$
$$\frac {0.30}{20}$$ = 0.015 m = 15 cm Question 27.
A child hears an echo from a cliff 4 seconds later after the sound from a powerful cracker is produced. How far away is the cliff from the child?
Let the distance between the child and cliff = x
Total distance travelled by sound = 2x
Velocity of sound = 344 m/s
Velocity = $$\frac{\text { Distance travelled }}{\text { Time }}$$
344 = $$\frac{2 x}{4}$$
Or 2x = 688m

Question 28.
A sound wave have frequency of 2 kg Hz and wavelength of 35 cm. How long will it take to travel 1.5 km?
Frequency, v = 2k Hz = 2000 Hz
Wavelength λ = 35 cm = 0.35 m
We know that, velocity of wave = frequency × wavelength
v = v × A.
v = 20 × 0.35 = 700 m/s
Let time taken by wave = 5 km = 1500 m
t = $$\frac{\text { Distanec travedlled }}{\text { velocity of wave }}$$
$$\frac {1500}{700}$$ = 2.14 second

Question 29.
A person clapped hands near a mountain and heard the echo after 5s. What is the distance of the mountain from the person if the speed of sound at a given temperature is 346 m/s?
Given
Speed of sound, v = 346 m/s
Time taken for hearing the echo, t = s
Let the distance between the person and mountain = x
We know that
Velocity = $$\frac{\text { Distanec travedlled }}{\text { time }}$$
346 = $$\frac{2 x}{5}$$
2x = 346 × 5 = 1730
x = $$\frac {1730}{2}$$ = 865 m
Hence, distance between the person and mountain = 865 m Ans.

Question 30.
A ship sends out ultrasound produced by the transmitter that return from the sea bed and detected after 3.42s. If the speed of ultrasound through seawater is 1531 m/s. What is the distance of sea bed from the ship?
Given
Time between transmission and detection, t = 3.42 s
Speed of the ultrasound in sea water, v = 1531 m/s
Let the distance of sea bed from ship = d
Distance travelled by ultrasound = 2 × d
We know that
v = $$\frac{\text { distance wavelength }}{\text { time taken }}$$
v = $$\frac{2 d}{t}$$
1531 = $$\frac{2 \times d}{3.42}$$
Hence, the distance of sea bed from the ship = 2618 m.

Question 31.
The frequency of a sound wave is 550 Hz. What is its wavelength? Sound travels with the a speed of 330 m/s. Calculate the time period of the wave also.
Given
Frequency of sound wave, υ = 550 Hz
Velocity of sound wave, v = 330 m/s
Wavelength, λ = ?
λ = $$\frac{v}{υ}=\frac{330}{550}=\frac{3}{5} m$$ = 0.6m
Time period = $$\frac{1}{υ}=\frac{1}{550}$$
= 0.0018 second. Question 32.
If the velocity of sound, in a medium is 1400 m/s and its wavelength is 100m. What is its frequency? Can you hear this sound?
Velocity of sound, v = 1400 m/s
Wavelength, λ = 100m
Frequency υ = ?
We know that, v = υ × λ
υ = $$\frac{v}{\lambda}=\frac{1400}{100}$$ = 14 Hz
We can hot hear this sound, because we can hear the sound having frequency range from 20 Hz to 20000 Hz.

Question 33.
A longitudinal wave travels in a coiled spring or slinky at the rate of 4m/s. The distance between two consecutive compression is 20 cm. Find (i) wavelength of longitudinal wave (ii) frequency of longitudinal wave.
Velocity of the longitudinal wave, v = 4 m/s
Distance between two consecutive compressions, wavelength, λ = 20 cm
= 0.2m
Frequency υ = ?
We know that
v = υ × λ.
υ = $$\frac{v}{\lambda}=\frac{4}{0.2}$$ = 20 Hz

Question 34.
A body vibrating with a time period of 2 milli seconds produces a wave travelling with a velocity of 1250 m/s. What is the frequency of vibrating body? (ii) What is the wavelength of the travelling wave?
Ans.wer:
Given, Time period, T = 2 miliseconds
= 2 × 10-3 S
Frequency, υ = $$\frac{1}{\mathrm{~T}}=\frac{1}{2 \times 10^{-3}}$$ = 500Hz
Wavelenght of the wave, λ = ?
Velocity of the Wave, v = 1250 m/s
We know that
v = $$\frac{v}{υ}=\frac{1250}{500}$$ = 2.5m

Question 35.
A Sonar echo takes 2.2 s to return from a whale. How far away is the whale? (Take speed of ultrasound to be 1531 m/s in seawater).
Time taken by ultra sound between Transmission and reception, t = 2.2s
Speed of ultra sound wave, v = 1531 m/s
Depth of the whale from sea level = h
Distance travelled by ultrasound = 2 × h
We know that,
distance travelled = speed × time
2h = 1531 × 2.2
h = $$\frac{1531 \times 2.2}{2}$$
h = $$\frac{1531 \times 2.2}{2}$$ = 1684.1 m
Hence, whale is 1684.1 away from the sea level

Question 1.
Give a simple activity to show that sound is produced by vibration.
Suspend a smallplastic ball by a thread from a support. (Fig. 12.7) Take a tuning fork just touch, the ball with prong of the turning fork without setting it into vibration. There is no displacement in the ball. Now set the tuning fork into vibration by striking its prong with a rubber pad. Now touch the suspended ball with the prong of tuning fork. The ball get displaced from its position. Now, bring the tuning fork near your ear. You will hear sound. The ball get displaced due to thee vibration of the prong of the tuning fork. This activity shows that sound is produced due to vibration. Fig.: Vibrating tuning fork just touching the suspended ball Question 2.
Define speed, frequency and wavelength of a wave and derive the relation between them.
Speed: The speed of a sound wave is defined as foe distance travelled by a point on a wave such as compression or a rarefaction in a unit time.
Speed = $$\frac{\text { distance }}{\text { time }}$$
Frequency: The number of compressions or rarefactions or crest and trough produced per unit time is known as foe frequency. It is denoted by Wave length: The distance two consecutive compressions or rarefactions is called wavelength. It is denoted by X.
Or the distance between two consecutive crests or troughs is called wavelength.

Relation between speed, frequency and wavelength: We know that foe speed of sound wave is given by
Speed = $$\frac{\text { distance travelled by wave }}{\text { time taken }}$$
Now we known that foe wavelength (λ) is equal to foe distance travelled in one complete oscillation and foe time taken to complete one oscilaltion is called time period and is denoted by T.

Now distance travelled in time T = λ
Therefore, Speed = $$\frac{\lambda}{\mathrm{T}}$$
or v = $$\frac{\lambda}{\mathrm{T}}$$ ………(i)
But we know that the frequency is the number of oscillation per unit time.
or frequency = $$\frac{1}{\text { Time Period }}$$
v = $$\frac{1}{\mathrm{~T}}$$ ……….(ii)
Now putting foe value of $$\frac{1}{\mathrm{~T}}$$ in equation (i) we get v = λ × $$\frac{1}{\mathrm{~T}}$$
v = λ × υ (∵ υ = $$\frac{1}{\mathrm{~T}}$$)
Hence, Speed = wavelength × frequncy.

Question 3.
(i) What is reflection of sound? Prove that the reflection of sound follows the same law of reflection of light.
(ii) What are the uses of multiple reflection of sound?
(i) Reflection of sound: When a sound wave strikes a solid or liquid surface, it is reflected back according to the laws of reflection i.e. the direction of incident and reflected sound makes equal angle with the normal to the reflecting surface and the three are in the same plane. These laws can be proved by the following activity.

Activity: Take two identical pipes as shown in the fig. 12.8. The length of the pipe should be sufficiently long (about 75 cm). Arrange them on a table near wall. Keep a clock near the open end of one of the pipes and try to hear the sound of the clock through the other pipe. Now adjust the pipes so that you can best hear the sound of the clock. Now, measure the angle between the incident sound and the normal and the reflected sound and the normal to the reflecting surface. Now repeat the activity by changing the angles of the pipes. We find that the angle of indicent of sound wave is always equal to the angle of reflection of sound wave.

Now, becasue both the tubes are placed on the table, it also proves that incidient sound, normal and relfected sound lie th one plane.

(ii) Uses of multiple reflection of sound : (1) Megaphones or loudspeaker, horns, musical instrument such as trumpets and shehanais all designed in such a way to send sound in a particular direction. In these instruments the conical opening reflects sound successively and the amplitude of the sound wave adds up and the loudness of sound increases.
(ii) In stethoscope, the sound of heartbeat and lungs reaches the doctor’s ears by multiple reflection.
(iii) The ceilling of the conference hall and cinema halls are made curved so that sound after reflection reaches to all the corners of the halls.
(iv) Sometimes curved sound board is placed behind the stage so that sound after reflection from the sound board spread evenly across the width of the hall. Question 4.
What are the application of ultrasound?