# NCERT Class 9 Science Notes & Solutions – Chapter 12 – Sound

## Exercises Questions & Answers | Sound Class 9 Science Chapter 12

#### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 162

1. How does the sound produced by a vibrating object in a medium reach your ear?
Any vibrating object creates disturbance in the medium. This disturbance produces compression & rarefaction in the medium. As a result sound reaches to our ear.

1. Explain how does school bell produce sound.
When we hit the bell by a wooden hammer, it then starts vibrating. The vibration of the bell creates compression & rarefaction in air which results in production of sound.

2. Why sound waves are called mechanical waves?
Basically a mechanical wave always needs a material medium to propagate. Sound waves also need a material medium (solid/liquid/gas) to propagate. So  sound waves are called mechanical waves.

3. Suppose you and your friend are on the moon. Will you be able to hear any sound by your friend?
Sound needs a material medium for propagation. There is no atmosphere on the moon. So propagation of sound is not possible on the moon. Hence, I will not hear any sound produced by my friend on the moon.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 166

1. Which wave property determines ⇒ (a) loudness, (b) pitch?
A. Loudness ⇒ It is determined by the amplitude of the sound wave. Louder sound has greater amplitude & soft sound has lower amplitude.
B. Pitch ⇒ Frequency of a sound wave determines the pith of that sound.

2. Guess which sound has a higher pitch ⇒ guitar or car horn?
Pitch of a sound is proportional to its frequency. The frequency of the sound of a horn is more than the frequency of the sound of a guitar. So the car horn has a higher pitch than a guitar.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 166

1. What are wavelength, frequency, time period & amplitude of a sound wave?
Wavelength is the distance between two consecutive compression or rarefaction. Basically wavelength is the distance of one complete oscillation. Greek letter lamda λ represents wavelength of the wave. S.I. unit of wavelength is metre (m).

Frequency is the number of complete oscillation per unit time i.e. in one second. Greek letter nu ν represents frequency of the wave. S.I. unit of frequency is hertz (Hz).

Time period is the time taken for one complete oscillation i.e. time taken to cross a fixed point by two consecutive compression or rarefaction. It is represented by T. S.I. unit of time period is second (s).

The magnitude of the maximum disturbance or displacement of a vibrating particle of the medium from the mean position is the amplitude of the wave.

##### 2. How are the wavelength & frequency of a sound wave related to its speed?

Wavelength (λ) is the distance covered by sound wave in one complete oscillation i.e. in T seconds. So in one second no. of complete oscillation is (1/T) = ν. It is called frequency of the wave. We also express frequency by f. So wave propagates in one second = ν x λ = speed of the wave.
Therefore, speed of sound  v = ν x λ  = f x λ = frequency x wavelength.

3. Calculate the wavelength of a sound wave whose frequency is 220 Hz & speed is 440 m/s in a given medium?
In this case, f = 220 Hz, v = 440 m/s, wavelength = λ = v/f. So, λ = (440/220) m = 2 m.

4. 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?
Given that, frequency = f = 500 Hz. Time period = T = ?
We know, frequency = 1/Time period,
⇒ Time period = 1/frequency = 1/500 sec = 0.002 sec.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 166

1. Distinguish between loudness & intensity of sound.
The average energy transported per unit area perpendicular to the direction of propagation is the intensity. Loudness of sound depends on its intensity. But their relation is not linear. Loudness depends on the amplitude of sound wave. The more the amplitude, the louder is the sound. On the other hand intensity is the rate of power transferred per unit area & is determined by frequency of sound wave.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science |  Math related to speed of sound

1. Speed of sound at 760 mm Hg pressure and at 0oC is 332 m/s. What will be the speed of sound at  35oC? What will be the speed if pressure changed to 700 mm Hg?
We know Vt  =  V0 ( 1 + 0.00183t). Here V0 = 332 m/s.  t = 35oC.
So V35oC = 332 ( 1 + 0.00183 x 35) m/s = 353.26 m/s.

2. A man standing on the bank of a river fires a gun. Another man standing on the opposite bank of the river hears the sound after 5 seconds of watching the flash of the gun. Calculate the width of the river if the speed of sound is 310 m/s.
H
ere the distance travelled by the sound in 5 seconds is the width of the river. So width of the river = 5 x 310 m = 1550 m.

3. Sound of a thunder is heard after 7 seconds of the flash of the light is seen. Find the distance of the cloud from ground level if the speed of sound is 340 m/s.
Distance = 7 x 340 m = 2,380 m.

4. What is the frequency of a sound wave having wavelength 1.7 m and speed 340 m/s.
As speed = frequency x wavelength, so frequency = speed / wavelength. Therefore, frequency ν = V/ λ = 340/1.7 Hz  = 200 Hz.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 167

1. In which of the three media ⇒ air, water or iron, does sound travel the fastest at a particular temperature?
In this case, sound travel fastest in iron at a particular temperature.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 168

1. An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s.
Given that speed of sound = 342 m/s. Also echo is heard in 3 s. So the distance travelled by the sound in 3 s = 342 x 3 m = 1026 m. In 3 s, the sound travels from the source to the reflector & returns back to the source. So the sound covers twice the distance between the source & the reflector. Hence, the distance from the source to the reflecting surface = 1026/2 m = 513 m.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 169

1. Why are the ceilings of concert halls curved?
Generally we make curved ceiling in a big hall to reflect sound to spread sound all over the wall. In the case of curved ceilings in concert hall, sound after reflection reaches uniformly to every corner of the concert hall. As a result, the audience can listen the sound clearly.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Echo of Sound

1. A man standing on the bank of a river fires a gun. A hill on the opposite bank reflects the sound and the man hears echo after 6 seconds. Calculate the width of the river if speed of sound is 340 m/s.
Speed of sound = 340 m/s. Time taken to hear echo = 6 seconds. During this 6 seconds, sound travels from the man standing on river bank  to hill on the opposite bank and again returns to the man as echo. So total distance is actually twice of the distance between the man and the hill i.e. twice of river width.
So total distance covered by the sound = 6 x 340 m = 2040 m.
And width of the river = 2040/2  m = 1020 m.

###### 2. A man hears echo after 3 seconds when he produces a sound near a tall building. Calculate the distance of the building from the man. Speed of sound is 332 m/s.

Total distance covered by the sound = 3 x 332 m = 996 m. So distance of the building from the man = 996/2  m = 498 m.

3. A plane is flying horizontally at speed 720 Km/hr. The pilot makes a gunshot and hears echo after 6 seconds. Speed of sound is 330 m/s. Calculate the height in which the plane is flying.
At first, observe the following picture to understand the problem-
Speed of the plane = 720 Km/hr = [ 720000/3600 ] m/s = 200 m/s.
Therefore, distance covered by the plane in 6 second = 200 x 6 m = 1200 m.
And distance covered by the sound in 6 second = 6 x 330 m = 1980 m.
So height in which the plane is  flying = √ ( 9902  –  6002 ) m =  728.69 m. ##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 170

1. What is the audible range of the average human ear?
20 Hz to 20,000 Hz.

2. What is the range of frequencies associated with ⇒ (a) Infrasound? (b) Ultrasound?
a. Infrasound: < 20 Hz. b. Ultrasound: > 20,000 Hz.

##### In-Text Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page No. 172

1. A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Sound travelled in 1.02 s in salt water = 1531 x 1.02 m = 1561.62. So, the distance between submarine and cliff = 1561.62/2 m =780.81 m.

##### Exercises Questions & Answers | NCERT Science Chapter 12 | Sound Class 9 Science | Page Nos. 174 - 175
###### 1. What is sound & how is it produced?

Sound is the energy which is emitting from any vibrating object and propagating  through a material medium (solid or liquid or gas) and producing  hearing sensation in ear. Sound is also a form of energy like heat, light etc. Any vibrating object produces sound.

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

Let imagine propagation of sound wave in air. Vibrating object produces forward and backward motion. When it moves forward, it compresses adjacent air and creates a region of high pressure. We term that region as a compression (‘a’ in picture) and it starts to move away from the vibrating object. When the vibrating object moves backwards, it creates region of low pressure. We term that region as a rarefaction (‘b’ in picture). As the object produces forward and backward motion continuously, so it creates a series of compression and rarefaction which makes propagation of sound wave through the medium (‘c’ in picture). A high pressure region (compression)  is actually a region of high density of particle in the medium and a low pressure region (rarefaction)  is actually a region of low density of particle in the medium. So you may imagine propagation of sound wave as density or pressure variations in medium i.e.  series of high density region and low density region. ###### 3. Cite an experiment to show that sound needs a material medium for its propagation.

Write down in detail the answer with picture of Q4.

###### 4. Why is sound wave called a longitudinal wave?

Sound waves are longitudinal waves because it creates oscillation in the particles of the medium parallel to the disturbance in the direction of propagation of sound waves.

###### 5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Basically the characteristic that differentiates various sounds having same pitch and loudness is the quality of that sound. Actually quality of a note depends on the number of overtones present in it. If number of overtones present in a sound is more, then quality of tone is high and pleasant. Due to difference in quality, one can identify his friend sitting with others in a dark room.

###### 6. Flash & thunder are produced simultaneously. But thunder is heard in a few seconds after the flash is seen, why?

This happens due to the difference in the velocity of light & sound waves. The speed of light is 3×108 m/s. On the other hand, the speed of sound in air is  330 m/s. As a result, light travels faster than sound. So we see the flush first & after few seconds we hear the thunder.

###### 7. A person has a hearing range from 20 Hz to 20 KHz. What are the typical wavelengths in air corresponding to these two frequencies? Speed of sound = 344 m/s.

Speed = Frequency x wavelength. In this case, speed = V = 344 m/s.
⇒ V = ν x λ. So, λ = V/ν.
i) λ for 20 Hz =  (344/20) m = 17.20 m.
ii) λ for 20 KHz =  (344/20000) m = 0.0172 m.  [ 20KHz = 20 x 103 Hz = 20000 Hz ].

###### 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. (Speed of sound in 1. aluminium = 6420 m/s 2.air = 346 m/s).

Take length of the aluminium rod = d metre. Here sound propagates through two medium i.e. through aluminium and air. So second child will hear two sounds.
i) Time taken by the sound to reach from 1st child to 2nd child through air = (d/346) s.
ii) Time taken by the sound to reach from 1st child to 2nd child through aluminium = (d/6420) s.
So ratio two times = (d/346) : (d/6420) = 18.55 : 1 = 18.55.

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

At first go through Q7. Frequency = 100 Hz = 100 nos. complete oscillation in a second = 100 nos. complete vibration.
So in 01 second, source of sound makes 100 nos. vibration.
⇒ in 01 minute i.e. 60 seconds, source of sound makes 60 x 100 = 6,000 nos. vibration.

###### 10. Does sound follow the same laws of reflection as light does? Explain.

Of course yes. Sound also follows the same law of reflection as light does. Sound needs a widespread surface such as building, wall, hilly terrain etc. for its reflection. A small mirror can reflect light but sound needs big surface. Because sound’s wavelength is much higher than light. It follows same rule of reflection like others where direction of incidence and direction of reflection make equal angles with the normal to the reflecting surface at the point of incidence (i.e. ∠angle of incidence = ∠angle of reflection). Also the incident & reflected sound wave & the normal at the point of incidence lie in the same plane. Read Q16 in detail for its experiment.

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

The speed of sound increases with the increase in temperature of the medium. So sound will take less time on a hotter day to cover the same distance. Therefore, on a hotter day, sound takes less time to cover the distance between source & reflector. As a result, we hear echo sooner.

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

1. Stethoscope: Doctors use it to listen sound from heart or lungs of patient. Sound produced by heart or lungs reflects multiple times inside the tube. 2. Bulb horn: In this case sound is amplified & sent to the desired direction because of reflection. 3. Speaking tube: Reflection of sound also happens in speaking tube which is used for conversation between driver and passenger in a big car.

###### 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-2, speed of sound = 340 m/s.

There are two times here. T1 time taken by stone to reach pond from top of tower and T2 time taken by sound to reach from pond to top of tower.
T1 time: In motion chapter we get  S = ut + ½  at2. Here S = 500 m, u = 0, a = 10 ms-2, t = T1 = ?
⇒ 500 = 0.t  + ½  . 10. t2, ⇒ 5 t2 = 500, ⇒ t = 10 second.
T2 time: Time taken by sound to reach from pond to top of tower = ( 500/340 ) s = 1.47 s.
So total time = T1 + T2 = ( 10 + 1.47 ) s = 11.47 s.

###### 14. A sound wave travels at a speed of 339 m/s. If it’s wavelength is 1.5 cm. What is the frequency of the wave? Will it be audible?

We know V = ν x λ. In this case speed = V = 339 m/s , λ = 1.5 cm = 0.015 m.
⇒  ν = V/ λ = ( 339 / 0.015 ) Hz = 22,600 Hz.
Human range of hearing is 20 Hz to 20 KHz. So 22,600 Hz is not audible.

###### 15. What is reverberation? How can it be produced?

Sound produced in auditorium or big hall persists for sometime due to multiple reflection from the wall and ceiling. We term it as reverberation. One can keep hearing reverberation until sound reaches inaudible level. Sometimes thunder sound persists for sometime due to multiple reflection at various layer in clouds. We especially use sound absorbing material in walls, roofs, seats to reduce reverberation in hall.

###### 16. What is loudness of sound? What factors does it depend on?

Write down Q10 in detail.

###### 17. Explain how bats use ultrasound to catch a prey.

Bats use the echolocation technique to catch a prey. This method uses waves & echoes to determine objects in space. Bats emit ultrasonic waves which are reflected from the prey & are detected by its ears. By analyzing the reflected wave, the bats guess the location & nature of the prey.

###### 18. How is ultrasound used for cleaning?

At first take the objects to be cleaned. Then put that object in a cleaning solution. After that pass ultrasonic waves through the cleaning solution. The high frequency of the ultrasonic waves detach the dirt from the objects.

###### 19. Explain the working & application of a sonar?

Write down in detail SONAR from the question Q24.

###### 20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of the sound in water if the distance of the object from submarine is 3625 m.

Sound created in submarine travels from submarine to object = 3625 m. It reflects back to submarine from object to submarine = 3625 m. So total distance covered by the sound is (3625+3625) m =7250 m. And sound covers it in 5 seconds. So, the speed of sound = (7250/5) m/s = 1450 m/s.

###### 21. Explain how defects in a metal block can be detected can be detected using ultrasound.

We may use ultrasound to detect defect in metal blocks. Defects in metal blocks do not allow ultrasound to pass through it. It reflects back the ultrasonic waves. At first place a detector on one side of the metal block. Then pass the ultrasonic waves through the metal block from the other side. The defect reflects back the ultrasound. Even a small defect can reflect back the ultrasonic waves. Then the detector detects the reflected waves & indicates the nature of the defect.

###### 22. Explain how the human ear works.

Write down Q25 in detail.

Sound Class 9 Science – To Be continued – Sound Class 9 Science