Sound

Class 9 – NCERT Science Solutions

Intext Questions 1

Question 1

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

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 2

Explain how sound is produced by your school bell.

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 3

Why are sound waves called mechanical waves?

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.

Question 4

Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer

No, I would not be able to hear any sound produced by my friend on the moon. Sound requires a medium, such as air, to travel through, and the moon lacks a significant atmosphere. Since there is no air or medium to transmit sound waves, any sounds my friend makes would not be heard by me on the moon. In the absence of a medium for sound propagation, communication through sound waves is not possible in the vacuum of space.

Intext Questions 2

Question 1

Which wave property determines

(a) loudness,

(b) pitch?

Answer

(a) Loudness is determined by amplitude. More the amplitude, louder is the sound and vice versa.

(b) Pitch is determined by frequency. The faster the vibration of the source, the higher is the frequency and the higher is the pitch.

Question 2

Guess which sound has a higher pitch: guitar or car horn?

Answer

Guitar sounds usually have a higher pitch than car horns. Guitars generally produce musical tones which are high pitched. On the other hand, car horns are designed to produce a loud and lower-pitched sound, often for alerting or signaling purposes.

Intext Questions 3

Question 1

What are the wavelength, frequency, time period and amplitude of a sound wave?

Answer

(a) Wavelength — The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength. The wavelength is usually represented by λ (Greek letter lambda). Its SI unit is metre (m).

(b) Frequency — The number of complete oscillations per unit time is the frequency of the sound wave. The SI unit of frequency is hertz (Hz). It is usually represented by the letter ν.

(c) Amplitude — The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave. It is usually represented by the letter A.

(d) Time period — The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave. It is represented by the symbol T. Its SI unit is second (s).

Question 2

How are the wavelength and frequency of a sound wave related to its speed?

Answer

Speed (v) = Wavelength (λ) x Frequency (ν)

Question 3

Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

Answer

Given,

Frequency (ν) = 220 Hz

Speed (v) = 440 mssm​

Wavelength (λ) = ?

We know,

Speed (v) = Wavelength (λ) x Frequency (ν)

Substituting we get,

440 = Wavelength × 220

Wavelength = 440220220440​ = 2

Hence, the wavelength of the sound wave = 2 m

Question 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?

Answer

Given,

Frequency (ν) = 500 Hz

The time interval between successive compressions from the source is equal to the time period.

We know,

time period (T) = 1νν1​

Substituting we get,

T = 15005001​ = 0.002 s

Hence, time interval between successive compressions from the source = 0.002 s.

Intext Questions 4

Question 1

Distinguish between loudness and intensity of sound.

Answer

Differences between loudness and intensity of sound:

Loudness	Intensity
Loudness is a measure of the response of the ear to the sound.	Intensity of sound is the amount of sound energy passing each second through unit area.
Loudness is a subjective quantity. It depends on the sensitivity of the human ear and individual auditory perception.	Intensity of sound provides an objective description of the strength of a sound wave based on its physical properties, such as amplitude and frequency.
Its unit is decible (dB)	Its unit is watt per metre2 (Wm-2)

Intext Questions 5

Question 1

In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature

Answer

Sound travels fastest in solids as compared to other medium. Hence, at a particular temperature, sound travels fastest in iron.

Intext Questions 6

Question 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 ms^-1?

Answer

Given,

Speed of sound (v) = 342 ms^-1

Time taken for hearing the echo (t) = 3 s

Distance travelled by sound = v × t = 342 × 3 = 1026 m

In 3 s, sound has to travel twice the distance between the reflecting surface and the source.

Hence, the distance between reflecting surface from the source

= 1026221026​ = 513 m

Intext Questions 7

Question 1

Why are the ceilings of concert halls curved?

Answer

The ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall as shown in the figure below:

Intext Questions 8

Question 1

What is the audible range of the average human ear?

Answer

20 Hz to 20,000 Hz.

Question 2

What is the range of frequencies associated with

(a) Infrasound?

(b) Ultrasound?

Answer

(a) Sounds of frequencies below 20 Hz are called infrasonic sound or infrasound.

(b) Sounds of frequencies higher than 20,000 Hz are called ultrasonic sound or ultrasound.

Exercises

Question 1

What is sound, and how is it produced?

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

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

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

Why is sound wave called a longitudinal wave?

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

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

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

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

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

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.

Answer

Given,

Speed (v_air) = 344 ms^-1

We know,

Speed (v) = Wavelength (λ) x Frequency (ν)

or

Wavelength (λ) = vννv​

(a) For ν = 20 Hz, substituting we get,

λ₁ = 3442020344​ = 17.2 m

(b) For ν = 20,000 Hz, substituting we get,

λ₂ = 34420,00020,000344​ = 0.0172 m

Hence, 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

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 the air and in aluminium to reach the second child.

Answer

Let length of the aluminium rod be l.

Speed of sound in air (v_air) at 25°C = 346 ms^-1

Speed of sound in aluminium (v_alu) at 25°C = 6420 ms^-1

We know,

Speed = DistanceTimeTimeDistance​

or

Time (t) = DistanceSpeedSpeedDistance​

Substituting we get,

t_air = l346346l​

t_alu = l64206420l​

So, the ratio of time taken by sound in air and aluminium is

tairtalutalu​tair​​ = l346l64206420l​346l​​ = 64203463466420​ = 18.55

Hence, the ratio of time taken by sound in air and aluminium = 18.55

Question 8

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

Answer

Given,

Frequency (ν) = 100 Hz

Time (t) = 1 min = 60 s

Frequency (ν) = Number of oscillationsTotal time (t)Total time (t)Number of oscillations​

Or

Number of oscillations = ν × t

Substituting we get,

Number of oscillations = 100 × 60 = 6000

Hence, the source vibrates 6000 times in a minute.

Question 9

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

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

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 an echo sound on a hotter day?

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

Give two practical applications of the reflection of sound waves.

Answer

Two practical applications of the reflection of sound waves are:

1. 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.
2. 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

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 and speed of sound = 340 ms^-1

Answer

Given,

Height of 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

Time taken by the stone to fall to the tower base (t₁):

According to the second equation of motion,

s = ut₁ + 1221​gt₁²

Substituting we get,

500 = (0 x t₁) + (1221​ x 10 x t₁²)

500 = 1221​ x 10 x t₁²

1000 = 10 x t₁²

t₁² = 100010101000​ = 100

Hence, t₁ = 100100​ = 10 s

Time taken by sound to reach the top from the tower base (t₂)= ?

Using,

Speed = DistanceTimeTimeDistance​

or

Time = DistanceSpeedSpeedDistance​

Substituting we get,

t₂ = 500340340500​ = 1.47 s

t = t₁ + t₂

t = 10 + 1.47

t = 11.47 s

Hence, the splash will be heard at the top after 11.47 s.

Question 13

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?

Answer

Given,

Speed of sound (v) = 339 ms^-1

Wavelength of sound (λ) = 1.5 cm

Converting cm to m

100 cm = 1 m

1.5 cm = 1.51001001.5​= 0.015 m

We know,

Speed (v) = Wavelength (λ) x Frequency (ν)

or

Frequency (ν) = Speed (v)Wavelength (λ)Wavelength (λ)Speed (v)​

Substituting we get,

ν = 3390.0150.015339​ = 22,600 Hz

Hence, the frequency of the sound = 22,600 Hz

No, 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

What is reverberation? How can it be reduced?

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

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

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:

1. Amplitude — Loudness is proportional to the square of the amplitude.
2. Distance from source — Loudness varies inversely as the square of the distance from the source.
3. Surface area of the vibrating body — Loudness depends on the surface area of the vibrating body.

Question 16

How is ultrasound used for cleaning?

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

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

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: