NCERT Solutions Class 9 Science
The NCERT Solutions in English Language for Class 9 Science Chapter – 12 (Sound) has been provided here to help the students in solving the questions from this exercise.
Chapter – 12 (Sound)
Questions
1. How does the sound produced by a vibrating object in a medium reach your ear?
Answer – Vibrations in an object create disturbance in the medium and consequently compressions and rarefactions. Because of these compressions and rarefactions sound reaches to our ear.
Questions
1. Explain how sound is produced by your school bell.
Answer – School bell starts vibrating when heated which creates compression and rarefaction in air and sound is produced.
2. Why are sound waves called mechanical waves?
Answer – Sound waves require a medium to propagate to interact with the particles present in them. Therefore, sound waves are called mechanical waves.
3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer – No, because sound waves needs a medium through which they can propagate. Since there is no material medium on the moon due to absence of atmosphere, you cannot hear any sound on the moon.
Questions
1. Which wave property determines
(a) loudness,
(b) pitch?
Answer –
(a) Amplitude – The loudness of the sound and its amplitude is directly related to each other. The larger the amplitude, the louder 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.
2. Guess which sound has a higher pitch: guitar or car horn?
Answer – Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has high frequency than that of car horn. High the frequency higher is the pitch.
3. What are the wavelength, frequency, time period and amplitude of a sound wave?
Answer –
(a) Wavelength – Wavelength can be defined as the distance between two consecutive rarefactions or two consecutive compressions. The SI unit of wavelength is metre (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.
4. How are the wavelength and frequency of a sound wave related to its speed?
Answer – Wavelength, speed, and frequency are related in the following way:
Speed = Wavelength × Frequency
v = λ ν
5. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer – Frequency of the sound wave, ν= 220 Hz
Speed of the sound wave, v = 440 m s-1
For a sound wave,
Speed = Wavelength × Frequencyv = λ x ν
∴ λ = v / ν
= 440 / 220
= 2m
Hence, the wavelength of the sound wave is 2 m.
6. 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 – The time interval between successive compressions from the source is equal to the time period, and the time period is reciprocal to the frequency. Therefore, it can be calculated as follows:
T= 1/F
T= 1/500
T = 0.002 s
7. Distinguish between loudness and intensity of sound.
Answer – The amount of sound energy passing through an area every second is called the intensity of a sound wave. Loudness is defined by its amplitude.
Questions
1. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer – 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.
Questions
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 –
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 between the reflecting surface and the source.
Therefore, the distance of the reflecting surface from the source =1026/2 = 513 m
Questions
1. Why are the ceilings of concert halls curved?
Answer – The ceilings of concert halls are curved to spread sound uniformly in all directions after reflecting from the walls.
Questions
1. What is the audible range of the average human ear?
Answer – 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.
2. What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer –
(a) 20 Hz
(b) 20,000 Hz
Questions
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?
Answer –
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 × 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.
Exercise
1. What is sound, and how is it produced?
Answer – Sound is a form of eneergy 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 the sound.
Answer – 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.
3. Cite an experiment to show that sound needs a material medium for its propagation.
Answer – 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 realise that the sound of the ringing bell decreases. If you keep on pumping the air out of the bell jar, then the 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 the bell prong is still vibrating. When there is no air present in the bell jar, a vacuum is produced. Sound cannot travel through a vacuum. Therefore, this experiment shows that sound needs a material medium for its propagation.
4. Why is a sound wave called a longitudinal wave?
Answer – Sound wave is called longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrates parallel to the direction of propagation.
5. Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Answer – 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?
Answer – 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.
Answer –
Speed = Wavelength × frequency
v = λ × v
Speed of sound wave in air = 344 m/s
(a) For v = 20 Hz
λ1 = v/v1 = 344/20 = 17.2 m
(b) For v2 = 20,000 Hz
λ2 = v/v2 = 344/20,000 = 0.0172 m
Therefore, for human beings, the hearing wavelength is in the range of 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 the air and in aluminium to reach the second child.
Answer –
Velocity of sound in air= 346 m/s
Velocity of sound wwave in aluminium= 6420 m/s
Let length of rode be 1
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 : 1
9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer –
Frequency = (Number of oscillations) / Total time
Number of oscillations = Frequency × Total time
Given,
Frequency of sound = 100 Hz
Total time = 1 min (1 min = 60 s)
Number of oscillations or vibrations = 100 × 60 = 6000
The source vibrates 6000 times in a minute and produces a frequency of 100 Hz.
10. Does sound follow the same laws of reflection as light does? Explain.
Answer – 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 an echo sound on a hotter day?
Answer – 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 the reflection of sound waves.
Answer – Two practical applications of reflection of sound waves are:
(i) Reflection of sound is used to measure the speed and distance of underwater objects. This method is called SONAR.
(ii) Working of a stethoscope – The sound of a patient’s heartbeat reaches the doctor’s ear through multiple reflections of sound.
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.
Answer –
Height (s) of tower = 500 m
Velocity (v) of sound = 340 m s−1
Acceleration (g) due to gravity = 10 m s−1
Initial velocity (u) of the stone = 0
Time (t1) taken by the stone to fall to the tower base:
As per the second equation of motion,
s= ut1 + (½) g (t1)2
500 = 0 × t1 + (½) 10 (t1)2
(t1)2 = 100
t1 = 10 s
Time (t2) taken by sound to reach the top from the tower base = 500/340 = 1.47 s
t = t1 + t2
t = 10 + 1.47
t = 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?
Answer –
Speed (v) of sound = 339 m s−1
Wavelength (λ) of sound = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × v
v = v / λ
= 339 / 0.015
= 22600 Hz
The frequency of audible sound for human beings lies between the ranges of 20 Hz to 20,000 Hz. The frequency of the given sound is more than 20,000 Hz; therefore, it is not audible.
15. What is reverberation? How can it be reduced?
Answer – 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 the loudness of sound? What factors does it depend on?
Answer – Loud sounds have high energy. Loudness directly depends on the amplitude of vibrations. It is proportional to the square of the amplitude of vibrations of sound.
17. Explain how bats use ultrasound to catch prey.
Answer – Bats have the ability to produce high-pitched ultrasonic squeaks. These squeaks get reflected by objects, like prey, and return to their ears. This helps a bat to know how far its prey is.
18. How is ultrasound used for cleaning?
Answer – 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.
19. Explain the working and application of a sonar.
Answer – SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction, and speed of under-water objects such as submarines and ship wrecks 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 sea water. 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 (t) 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’.
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.
Answer –
Time (t) taken to hear the echo = 5 s
Distance (d) of an object from submarine = 3625 m
Total distance travelled by SONAR during reception and transmission in water = 2d
Velocity (v) of sound in water = 2d/t
= (2 × 3625) / 5
= 1450 ms-1
21. Explain how defects in a metal block can be detected using ultrasound.
Answer – 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.
22. Explain how the human ear works.
Answer – The human ear consists of three parts – the outer ear, middle ear and inner ear.
Outer ear – This is also called ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
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.
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.
