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You must have heard many types of sound and an echo as well. But have you ever thought of how it produces and travels? This chapter will take you to the journey of these thoughts and will give you answers to these as well. You will understand the concept of echo, propagation of sound waves, reverberation, etc.

NCERT Solution Class 9 Science of Chapter 12 includes all Intext exercises and NCERT exercises for you. The NCERT exercise includes solutions to all 22 questions given in your NCERT Textbook. These questions will test your understanding and numerical solving abilities. ## NCERT Solution for Class 9 Science Chapter 12 : Sound Ex 12.1

#### Qustion1.How the sound does produced by a vibrating, object in a medium reach your car?

Solution:
When a disturbance is created on an object, it starts vibrating and sets the particles of the medium to vibrate. The particle of the medium is displaced from its mean position and exerts a force on the adjacent particle. As a result, the adjacent particle is disturbed from its mean position and the original particle comes back to rest. This process continues till the disturbance reaches our ear.

Exercise 12.2.1

#### Qustion1.Explain how sound is produced by your school bell.

Solution:
When the school bell is struck with a hammer, it starts vibrating and as a result of these vibrations sound waves are produced.

#### Question2.Why sound waves are called mechanical waves?

Solution:
Waves which need a material medium for propagation are called mechanical waves. Since sound waves also need a material medium for propagation, these are called mechanical waves. Sound wave is called mechanical waves or elastic waves as these are produced in a deformable or elastic medium.

#### Question3.Suppose you and your friend are on the moon. Will you be able to hear anysound produced by your friend?

Solution:
No we are not be able to hear the sound because sound requires a medium for its propagation. In the moon there is no atmosphere i.e., there is vacuum.

#### Qustion1.Which wave property determines (a) loudness, (b) pitch?

Solution:
(a) Loudness of a sound depends on the amplitude of sound wave oscillations. Greater the amplitude louder the sound is.
(b) Pitch depends on the frequency of sound waves. As frequency increases, the Pitch of sound increases and the sound appears shriller one.

#### Question2.Guess which has a higher pitch: a guitar or a car horn?

Solution:
A guitar has a higher pitch than a car horn. Provided the guitar is properly turned.

#### Question3.What is wave length, frequency? Time period and amplitude of a sound wave?

Solution:
Wavelength: The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength, (unit meter).
Frequency: The number of oscillation per unit time is called frequency, unit (Hz).
Time period: Time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period.
Amplitude: Amplitude can be defined as the maximum height reached by the trough or crest of a sound wave.

#### Question4.How are the wavelength and frequency of a sound wave related to its speed?

Solution:
Speed = wavelength x frequency
v f  

Solution:

#### Question6.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:
Here frequency of sound v = 500 Hz.
Time interval between successive compressions from the source
Time period T =
1 1
f 500

s = 0.002 s
[Note: Time interval between successive compressions does not depend on the position of a
Person listing the sound.]

Question7.
Distinguish between loudness and intensity of sound.

Solution: Refer pdf.

#### Qustion1.In which of the three media: air, water or iron; does sound travel the fastest at a particular temperature?

Solution:
Sound travels the fastest in iron which is solid medium.

Exercise 12.3.2

#### Qustion1.An echo is returned in 38. What is the distance of the reflecting surface from the source, given that the speed is 342 m/s-1 ?

Solution:
Here time of echo t = 3 s,
Speed of sound v = 342 m/s-1
Distance of the reflecting surface from the source = vt
= 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.

#### Question2.Why are the ceilings of concert halls curved?

Solution:
The ceilings of the concert halls are curved to ensure that after reflection from the ceilings, sound reaches all comers of the hall.

Exercise 12.4

#### Qustion1.What is the audible range of the average human ear?

Solution:
20 Hz to 20,000 Hz.

#### Question2.What is the range of frequencies associated with?a) Infrasound (b) Ultrasound.

Solution:
(a)Range of frequencies associated with infrasound: 1 Hz to 20 Hz.
(b) Range of frequencies associated with ultrasound: 2 x104 Hz.

#### Qustion1.A submarine emits a sonar pulse, which returns from an underwater cliff in1.02 s. If the speed of sound in salt water is 1531 m/s, how far/away is the cliff. ?

Solution:
Speed of sound in salt water= 1531 m/s
Time period= 1.02 sec
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.

#### Qustion1.What is sound and how is it produced?

Solution:
Sound is a form of energy and it is produced due to vibrations of different types of object, e.g., a vibrating tuning fork, a bell, wires in a sitar and a violin etc.

#### Question2.Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Solution:
1) When a vibrating object moves forward, it pushes the air in front of it and compresses the air creating a region of high pressure called compression (C).
2) It starts moving away from the surface of the vibrating object.
3) As this 00ch the surface moves backward creating a region of low pressure called rarefaction (R).

#### Question3.Cite an experiment to show. That sound needs a material medium for itspropagation.

Solution:
Take an electric bell and the air tight glass bell jar.
The electric bell is suspended inside an air tight glass jar which is connected to a vacuum pump.
Working:
(i) When we press the switch, we will be able to hear the bell.
(ii) When the air in the jar is pumped out gradually, the sound becomes feeble although the same
amount of current is flowing through the bell.
(iii) When the air is removed completely, we will not be able to hear the sound of the bell.
Conclusion: This experiment shows that sound requires a medium for its propagation.

#### Question4.Why sound wave is called a longitudinal wave?

Solution:
A sound wave is called longitudinal wave as it travel in a medium in the form of compressions and rarefactions where the particles of the medium vibrate in a direction which is parallel to the direction of propagation of the sound wave.

#### Question5.Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Solution:
The quality (or timbre) of sound is that characteristic which enables us to distinguish one sound from the other even when these are of the same pitch and loundness. Each person has its own quality of sound and it is this characteristic which enables us to identify a person from others even without looking at him (i.e., in a dark room).

#### Question6.Flash and thunder are produced simultaneously. But thunder is heard a fewseconds after the flash is seen why?

Solution:
In lightning process, flash and thunder are produced simultaneously. F lash is seen almost immediately because speed of light is extraordinarily large (c=3x108ms”).But thunder is heard a few seconds later because speed of sound is much less (about 346 ms” at 25°C) and requires time to cover up the distance from the site of thunder in sky to us.

#### Question7.A person has hearing range from 20 Hz to 20 kHz. What are the typicalwavelengths of sound waves in air corresponding to these two frequencies?Take the speed of ‘sound is /air as 344 m/s

Solution: Refer pdf.

#### Question8.Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times by the sound wave in air and in aluminum to reach the second child.

Solution: Refer pdf.

#### Question9.The frequency of a source of sound is 100 Hz. How many times does it vibrate ina minute?

Solution:
Frequency of sound = 100 Hz
Time taken = 1 minute = 60 sec
We know frequency = number of oscillation / time taken
No. of oscillations = vt
= 100x 60 = 6000 times.

#### Question10.Does sound follow the same laws of reflection as light does? Explain.

Solution:
Yes, sound follows the same laws of reflection as light does. To explain it, we taken two identical pipes and arrange them on a table near a wall. Keep a clock near the open end of one pipe. Try to hear the sound of clock by putting our ear near the open end of second pipe. Adjust the orientations of the pipes so that sound of clock is clearly heard at the open end of second pipe. Measure the
angle of incidence ‘z” and the angle of reflection ‘r’. We find that angle of incidence is equal to angle of reflection i.e.
< i = < r.

#### Question12.Give two practical applications of reflection of sound waves.

Solution:
Ear Trumpet: It is a sort of machine used by persons who are hard of hearing. The sound energy received by the wide end of the trumpet is connected into a much smaller area at the narrow end by multiple reflections. The narrow end of the trumpet which is inserted in the ear delivers the entire amount of energy falling on the wide end which makes the inaudible sound audible to the (user. Stethoscope: It is a medical instrument used frequently by doctors for making a rough diagnosis of the diseases existing inside the body at places which are either inaccessible or accessible only through major operations.

#### Question13.A stone is dropped from the top of a tower500m height 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 is 340 m/s.

Solution: Refer pdf.

#### Question14.A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Solution: Refer pdf.

#### Question15.What is reverberation? How can it be reduced?

Solution:
The persistence of audible sound due to the successive reflections from the surrounding objects even after the source has stopped to produce that sound is called reverberation. The reverberation of time is the time required for the intensity of sound to fall to 106 of its initial value.

How is reverberation reduced?
Since reverberation is due to repeated reflection of sound waves from ceiling , floor ,walls etc.  Of a hall or an auditorium, we can reduce it may increasing the absorption of sound energy. To achieve this:
1) The walls are covered with some sound ‘ absorbing material like felt, fiberboard, glass wool etc. or by heavy curtains with folds.
2) The floor is carpeted
3) The furniture is upholstered.
4) False ceiling of a suitable sound absorbing material is used.

#### Question16.What is loudness of sound? What factors does it on?

Solution:
Loudness of sound is that Characteristics of given sound due to which we identify the sound to be loud or soft. Within audible frequency range the loudness of sound depends on
(i) the amplitude of sound oscillations, and
(ii) Response of the ear of listener to the sound.

#### Question17.Explain how bats use ultrasound to catch a Prey.

Solution:

Bats search out and catch a prey by emitting and detecting reflections of ultrasonic waves. The high frequency ultrasonic squeaks of the bat are reflected from the prey and returned to. Bat’s ear. The nature of reflection tells the bat where the prey is and what it is like. Thus, he can catch the prey.

#### Question18.How is ultrasound used for cleaning?

Solution:
Cleansing instruments and electronic Components:
The cleaning is done by the method called cavitation or cold boiling. An instrument that needs cleaning but hose parts can be reached directly is placed in a liquid. The ultrasonic waves passing through the liquid produced tiny bubbles here the rarefaction of the ultrasonic wave reaches. When the compression of me wave reaches these bubb les, the bubbles are compressed until they implode this leads to the creation of several localized shock waves. These blastaw ay any dirt contamination from the unreachable portions. Usually, frequencies in the range 20 kH z to 30 kH are used for this purpose.

#### Question19.Explain the working and application of SONAR.

Solution:  Refer pdf.

#### Question20.A sonar device on a submarine semi; out a signal and receives an echo 5 3 later.Calculate the speed of sound in water if the distance of the object fr-hesubmarine is 3625 m.

Solution: Refer pdf.

#### Question21.Explain how defects in a metal block can be detected using ultrasound.

Solution:
To detect defects in a metal block, ultrasonic waves are passed through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect in the block, the ultrasonic waves are reflected back indicating the presence of defect of flaw in the object as shown in figure.

Question22.
Explain how the human ear works.

Solution:
The function of the human ear is shown in fig
(1) The outer ear collects sound wave which are conducted through the auditory canal.
(2) These waves fall on the ear drum and set it into vibrations.
(3)The middle eai which is of the size of a small marble and houses ossicles (three bones: hammer, anvil and stirrup) amplifies these oscillations about 60 times.
(4)The inner ear which contains cochlea and is filled with a fluid converts these pressure variations into the electrical signals.
(5) These electrical signals are conveyed to the brain via auditory nerve for interpretation. 