NCERT Solutions for Class 9 Science Chapter 12 Sound
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.
2. Explain how sound is produced by your school bell.
Solution:
When the school bell is hit with a hammer, it moves forward and backwards producing compression and rarefaction due to vibrations. This is how sound is produced by the school bell.
3. Why are sound waves called mechanical waves?
Solution:
Sound waves require a medium to propagate to interact with the particles present in it. Therefore, sound waves are called mechanical waves.
4. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Solution:
No. Sound waves require a medium to propagate. Due to the absence of atmosphere on the moon and since sound cannot travel in vacuum, I will not be able to hear any sound produced by my friend.
5. Which wave property determines (a) loudness, (b) pitch?
Solution:
(a). Amplitude – The loudness of the sound and its amplitude is directly related to each other. Larger the amplitude, louder is the sound.
(b). Frequency – The pitch of the sound and its frequency is directly related to each other. If the pitch is high then the frequency of sound is also high.
6. Guess which sound has a higher pitch: guitar or car horn?
Solution:
The pitch of a sound is directly proportional to its frequency. Therefore, the guitar has a higher pitch when compared to a car horn.
7. What are wavelength, frequency, time period and amplitude of a sound wave?
Solution:
(a) Wavelength – Wavelength can be defined as the distance between two consecutive rarefactions or two consecutive compressions. The SI unit of wavelength is meter (m).
(b) Frequency – Frequency is defined as the number of oscillations per second. The SI unit of frequency is hertz (Hz).
(c) Amplitude – Amplitude can be defined as the maximum height reached by the trough or crest of a sound wave.
(d) Time period – The time period is defined as the time required to produce one complete cycle of a sound wave.
8. How are the wavelength and frequency of a sound wave related to its speed?
Solution:
Wavelength, speed, and frequency are related in the following way:
Speed = Wavelength x Frequency
v = λ ν
9. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Solution:
Given that,
Frequency of sound wave = 220 Hz.
Speed of sound wave = 440 m/s.
Calculate wavelength.
We know that,
Speed = Wavelength × Frequency
v = λ ν
440 = Wavelength × 220
Wavelength = 440/220
Wavelength = 2
Therefore, the wavelength of the sound wave = 2 meters.
10. 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 successive compressions from the source is equal to the time period and time period is reciprocal of the frequency. Therefore, it can be calculated as follows:
T= 1/F
T= 1/500
T = 0.002 s.
11. Distinguish between loudness and intensity of sound.
Solution:
The amount of sound energy passing through an area every second is called intensity of a sound wave. Loudness is defined by its amplitude.
12. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Solution:
Sound travels faster in solids when compared to any other medium. Therefore, at a particular temperature, sound travels fastest in iron and slowest in gas.
13. 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?
Solution:
Speed of sound (v) = 342 ms-1
Echo returns in time (t) = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given interval of time, sound must travel a distance which is twice the distance of reflecting surface and source.
Therefore, the distance of reflecting surface from the source =1026/2 = 513 m.
14. Why are the ceilings of concert halls curved?
Solution:
Ceilings of concert halls are curved to uniformly spread sound in all directions after reflecting from the walls.
15. What is the audible range of the average human ear?
Solution:
20 Hz to 20,000 Hz. Any sound less than 20 Hz or greater than 20,000 Hz frequency is not audible to human ears.
16. What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
Solution:
(a). 20 Hz
(b). 20,000 Hz.
17. 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?
Solution:
Time (t) taken by the sonar pulse to return = 1.02 s
Speed (v) of sound in salt water = 1531 m s-1
Distance travelled by sonar pulse = Speed of sound × Time taken
= 1531 x 1.02 = 1561.62 m
Distance of the cliff from the submarine = (Total distance travelled by sonar pulse) / 2
= 1561.62 / 2
= 780.81 m.
18. What is sound and how is it produced?
Solution:
Sound is produced due to vibrations. When a body vibrates, it forces the adjacent particles of the medium to vibrate. This results in a disturbance in the medium, which travels as waves and reaches the ear. Hence, sound is produced.
19. Describe with the help of a diagram, how compressions and rarefactions are produced in the air near a source of sound.
Solution:
When the school bell is hit with a hammer, it moves forward and backwards producing compression and rarefaction due to vibrations. When it moves forward, it creates high pressure in its surrounding area. This high-pressure region is known as compression. When it moves backwards, it creates a low-pressure region in its surrounding. This region is called rarefaction.
20. 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 (as shown in the figure below).
Initially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. You will realize that the sound of the ringing bell decreases. If you keep on pumping the air out of the bell-jar, then glass-jar will be devoid of any air after some time. Now try to ring the bell. No sound is heard but you can see bell prong is still vibrating. When there is no air present in the bell jar, a vacuum is produced. Sound cannot travel through vacuum. Therefore, this experiment shows that sound needs a material medium for its propagation.