# NCERT Solutions for Class 9 Science Chapter 12 Sound

NCERT Solutions for Class 9 Science Chapter 12 Sound are a great resource for students who want to get ahead in their studies. The questions and answers are provided in a detailed manner which helps the students understand the concept better.

Through Chapter 12 CBSE Class 9 Science NCERT Solutions will be able to understand the concepts better and prepare your answers more accurately. It will give good experience and provide opportunities to learn new things.

## Chapter 12 Sound Class 9 Science NCERT Solutions

### In Text Questions

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

Solution

When an object vibrates, it necessitates the surrounding particles of the medium to vibrate. The particles that are adjacent to vibrating particles are forced to vibrate. Hence the sound produced by a vibrating object in a medium is transferred from particle to particle till it reaches your ear.

1. Explain how sound is produced by your school bell.

Solution

When the school bell is struck by a hammer, it starts moving forward and backward due to elasticity and inertia. Whenever a bell moves forward, a wave of increased pressure, also known as compression, is produced. When the bell moves backward, a wave of rarefaction, or decrease in pressure, is produced. Sound is produced when a series of compressions and rarefactions reach our ears.

2. Why are sound waves called mechanical waves?

Solution

Sound waves are called mechanical waves because these are produced due to the forward and backward motion of the material particles. Sound cannot travel in vacuum. A material medium is essential for the production of sound waves.

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

Solution

No, you will not be able to hear any sound when you and your friend are on moon. This is because there is no atmosphere on the moon. For transmission of sound, a material medium is must.

1. Which wave property determines
(a) loudness
(b) pitch?

Solution

(a) Amplitude
(b) Frequency

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

Solution

High the frequency higher is the pitch. So, guitar has a higher pitch than car horn, because sound produced by the strings of guitar has high frequency than that of car horn.

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

Solution

Wavelength is the distance between two consecutive compressions or two consecutive rarefactions. It is usually represented by λ and its SI unit is metre.

Frequency is the number of oscillations completed in a unit time. It is represented by v and its SI unit is s-1 or Hertz (Hz).

Time period is the time taken to complete one oscillation. It is represented by T and its SI unit is second.
T=1/v or v = 1/T

Amplitude is the maximum displacement of the particles of the medium on either side of the mean position. It is usually represented by A and its SI unit is metre.

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

Solution

The relation is v=vλ, where v is the speed, v is the frequency and λ is the wavelength.

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

Solution

Frequency of the sound wave, n = 220 Hz
Speed of the sound wave, v = 440 ms–1
For a sound wave,
Speed = Wavelength × Frequency;
v = λ × ν
λ = v/ν = 440/220 = 2m
Hence, the wavelength of the sound wave is 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?

Solution

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 = 1/Frequency = 1/500 = 0.002 s.

1. Distinguish between loudness and intensity of sound.

Solution

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.

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

Solution

The speed of sound depends on the nature of the medium. Sound travels 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.

2. An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m s–1?

Solution

Speed of sound, v = 342 m s–1
Echo returns in time, t = 3 s
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 = 1026/2 = 513 m.

1. Why are the ceilings of concert halls curved?

Solution

Ceilings of concert halls are curved so that sound after reflection from the walls spreads uniformly in all directions.

1. What is the audible range of the average human ear?

Solution

The audible range of an average human ear lies between 20 Hz to 20,000 Hz.

2. What is the range of frequencies associated with :
(a) Infrasound
(b) Ultrasound

Solution

(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

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 from the submarine?

Solution

Time taken by the sonar pulse to return, t = 1.02 s
Speed of sound in salt water, v = 1531 ms–1
Distance of the cliff from the submarine = Speed of sound × Time taken
Distance of the cliff from the submarine = 1.02 × 1531 = 1561.62 m
Distance travelled by the sonar pulse during its
transmission and reception in water = 2 × Actual distance = 2d
Actual Distance, d = Distance of the cliff from the submarine/2
= 1561.62/2 = 780.81 m

### Exercises

1. What is sound and how is it produced?

Solution

Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in air by vibrating objects.

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

Solution

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. It creates a region of low pressure in its vicinity when it moves backward. 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 below figure.

3. Cite an experiment to show that sound needs a material medium for its propagation.

Solution

Take an electric bell and hang it inside an empty bell-jar which is fitted with a vacuum pump. Suspend the bell inside the jar and press the switch of the bell. You will be able to hear the bell ring. Now, pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out. This shows that sound needs a material medium to travel.

4. Why sound wave is called a longitudinal wave?

Solution

The vibration of the medium that travels parallel to the direction of the wave or along in the direction of the wave, is called a longitudinal wave. The direction of particles of the medium vibrates parallel to the direction of the propagation of disturbance. Therefore, a sound wave is called a longitudinal wave.

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

Solution

The quality or timber of sound enables us to identify our friend by his voice.

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

Solution

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.

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 m s–1.

Solution

For a sound wave,
Speed = Wavelength × Frequency
v = λ × ν
Speed of sound in air = 344 m/s (Given)
(i) For, ν = 20 Hz, λ1 = v/ν = 344/20 = 17.2 m
(ii) For ν = 20000 Hz = λ2 = 344/20000 = 0.0172 m.

Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

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.

Solution

Velocity of sound in air = 346 m/s
Velocity of sound wave in aluminium = 6420 m/s
Let length of rode be 1 m. Time taken for sound wave in air, t1 = 1/Velocity in air
Time taken for sound wave in aluminium, t2 = 1/Velocity in aluminium
Therefore, t1/t2 = Velocity in aluminium/Velocity in air
= 6420/346 = 18.55

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

Solution

Frequency = 100 Hz (given)
This means the source of sound vibrates 100 times in one second. Therefore, number of vibrations in 1 minute, i.e., in 60 seconds = 100 × 60 = 6000 times.

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

Solution

Yes. Sound follows the same laws of reflection as light. The reflected sound wave and the incident sound wave make an equal angle with the normal to the surface at the point of incidence. Also, the reflected sound wave, the normal to the point of incidence, and the incident sound wave all lie in the same plane.

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?

Solution

An echo is heard when the time for the reflected sound is heard after 0.1 s
Time Taken = Total Distance/Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.

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

Solution

(i) Stethoscope is used for listening to heartbeats produced by multiple reflections of sound.
(ii) Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.

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 m s–2 and speed of sound = 340 m s–1.

Solution

Height of the tower, s = 500 m
Velocity of sound, v = 340 m s–1
Acceleration due to gravity, g = 10 m s–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 + 1/2 gt12
500 = 0 × t1 + 1/2 × 10 × t12
t12 = 100
t1 = 10 s
Now, time taken by the sound to reach the top from the base of the tower, t2 = 500/340 = 1.47 s
Therefore, the splash is heard at the top after time, t
where, t = t1 + t2 = 10 + 1.47 = 11.47 s

14. A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Solution

Speed of sound, v = 339 m s–1
Wavelength of sound, λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × ν
v = ν/λ = 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.

15. What is reverberation? How can it be reduced?

Solution

The repeated multiple reflections of sound in any big enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

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

Solution

Loudness of sound is a measure of response of the ear to the sound. It depends upon the intensity of sound (sound energy per unit area per unit time) and the sensitivity of the ear.

17. Explain how bats use ultrasound to catch prey.

Solution

Bats have the ability to produce high-pitched ultrasonic squeaks which are reflected from prey and returned to bat’s ear. The amount and nature of reflected wave helps bat in estimating the size and distance of prey.

18. How is ultrasound used for cleaning?

Solution

Objects to be cleaned 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.

19. Explain the working and application of a sonar.

Solution

SONAR is an abbreviation of Sound Navigation and Ranging. It is an acoustic device used in measuring the direction, speed, and depth of under-water objects viz. ship wrecks and submarines using ultrasound.
(i) Sonar consists of a transmitter and a detector and is installed in a boat or a ship.
(ii) The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.
(iii) The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted.
(iv) The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.
Let the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 2d travelled by the ultrasound is then, 2d = v×t.

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.

Solution

Time taken to hear the echo, t = 5 s
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 sound in water, v = 2d/t = 2 ×3625/5 = 1450 ms-1

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

Solution

Ultrasounds can be used to detect minor cracks or flaws in metal blocks. Ultrasound is allowed to pass through the metal blocks and detectors are used to detect the transmitted waves. If there is a crack in the metal block, the ultrasound gets reflected back indicating the presence of the cracks or flaws in the metal block.

22. Explain how the human ear works.

Solution

The human ear consists of three parts – the outer ear, middle ear and inner ear.
(i) Outer ear : This is also called ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
(ii) Middle ear : The sound reaches the end of the auditory canal where there is a thin membrane
called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones, hammer, anvil and stirrup.
(iii) Inner ear : These vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.