Class 9 Science Chapter 12: Sound - NCERT Solutions

This page focuses on the complete NCERT solutions for Class 9 Science Chapter 12: Sound. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Sound question answers will offer you valuable insights and explanations.

Sensation of hearing is sound with our ears we can listen different kind of sound and even we understand the difference between them but we don't know actual science behind the sound so this chapter make us cleare how of sound originate, it propagates, its quality and many more. You may have also heard the animals like elephant dogs etc can feel natural calamities like earthquake before it commencement what the reason behind this lets understand.

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Exercise 1

Q1	How does the sound produced by a vibrating object in a medium reach your ear?
Ans:	When a disturbance is created on an object, it starts vibrating and sets the particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. 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 ears.

Exercise 2

Q1	Explain how sound is produced by your school bell.
Ans:	When the school bell is struck with a hammer, it starts vibrating. This disturbance gives rise to the bell moving forward, it pushes the air in front of it. As a result of these vibrations, sound waves are produced.

Q2	Why are sound waves called mechanical waves?
Ans:	Waves which need a material medium for propagation are called mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium. Mechanical waves are governed by Newton’s laws of motion.

Q3	Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Ans:	No we will not be able to hear the sound because sound needs a medium to propagate. On the moon is no atmosphere, you cannot hear any sound on the moon.

Exercise 3

Q1	Which wave property determines (a) loudness, (b) pitch?
Ans:	Our experts will give the answer soon.

Q2	Guess which sound has a higher pitch; guitar or car horn?
Ans:	Our experts will give the answer soon.

Exercise 4

Q1	What are wavelength, frequency, time period and amplitude of a sound wave?
Ans:	Our experts will give the answer soon.

Q2	How are the wavelength and frequency of a sound wave related to its speed?
Ans:	The speed is defined as the distance travelled by a wave per unit time. Speed, wavelength, and frequency of a sound wave are related by the following equation:

Q3	Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans:	Frequency of the sound wave, = 220 Hz Speed of the sound wave, = 440 m s−1 For a sound wave, Speed = Wavelength × Frequency Hence, the wavelength of the sound wave is 2 m.

Q4	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?
Ans:	The time period between successive compressions from the source is equal to the time period of the tone, i.e., (1/500). This time period is reciprocal of the frequency of the wave and is given by the relation:

Exercise 5

Q1	Distinguish between loudness and intensity of sound.
Ans:	Intensity of a sound wave is not a physical quantity which can be accurately measured. It does not depend upon the sensitivity of the ear. Loudness is not an entirely physical quantity. 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.

Exercise 6

Q1	In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Ans:	The speed of sound depends on the nature of the medium. The speed of sound is fastest in case of solids. Its speed decreases in liquids and it is the slowest in gases. Therefore, in iron, the speed of sound is the fastest at a given temperature.

Exercise 7

Q1	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?
Ans:	Speed of sound, v = 342 m/s Time taken by sound to travel from the source to reflecting surface, t = 3/2 = 1.5 s Distance travelled by sound to reflecting surface from the source = v × t = 342 × 1.5 = 513 m

Exercise 8

Q1	Why are the ceilings of concert halls curved?
Ans:	Concert halls are very big, so the sound might not reach every corner of the hall. Ceilings of concert halls are curved so that sound after reflection spreads uniformly in all parts of the hall.

Exercise 9

Q1	What is the audible range of the average human ear?
Ans:	In human beings, the audible range of an average human ear lies between 20 Hz to 20,000 Hz. Humans cannot hear sounds having frequency less than 20 Hz and greater than 20,000 Hz.

Q2	What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
Ans:	(a) Infrasound has frequencies less than 20 Hz. (b) Ultrasound has frequencies more than 20,000 Hz.

Exercise 10

Q1	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?
Ans:	Time taken by the sonar pulse to return, t = 1.02 s Speed of sound in salt water, v = 1531 m/s Total distance covered by the sonar pulse = Speed of sound × Time taken Total distance covered by the sonar pulse = 1.02 x 1531 = 1561.62 m ....(i) Let d be the distance of the cliff from the submarine. Total distance covered by the sonar pulse = 2d ⇒ 2d = 1561.62 [From (i)] ⇒ d = 780.81 m ⇒ 2d = 1561.62 d = 780.81 m

Exercise 11

Q1	What is sound and how is it produced?
Ans:	Sound is defined as vibration that travels through the air or another medium as an audible mechanical wave. This creates a disturbance in the medium. It is produced from a vibrating body. This disturbance, when it reaches the ear, produces sound.

Q2	Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Ans:	Compressions and rarefactions are produced because the disturbance in the medium is caused by sound waves. When an object vibrates then it moves forward, it pushes and compresses the air in front of it creating a region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction.

Q3	Cite an experiment to show that sound needs a material medium for its propagation.
Ans:	Take an electric bell and connect it to electrical supply. An airtight glass bell jar with a vacuum pump. When we press the switch, we will be able to hear the bell. One can hear the sound of the ringing bell. Now, take a jar and make a small hole in the bottom and insert a small pipe though it. Connect the other end of the pipe to a vacuum cleaner. 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. It will be observed that the sound of the ringing bell decreases. When there is no air present inside, we will not be able to hear the sound of the bell. This shows that sound requires a medium for its propagation.

Q4	Why is sound wave called a longitudinal wave?
Ans:	In longitudinal waves, the motion of the individual particles of the medium is in a direction that is parallel to the direction of energy transport. A longitudinal wave can be created in a slinky if the slinky is stretched out in a horizontal direction and the first coils of the slinky are vibrated horizontally. This is known as longitudinal wave.

Q5	Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Ans:	Quality of sound is that characteristic which helps us to distinguish one sound from another. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

Q6	Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Ans:	The speed of light is greater than the speed of sound. Sound of thunder takes more time to reach the earth as compared to light. Hence, a flash is seen before we hear a thunder. The speed of light is (3x10⁸ m/s) and the speed of sound is 330 m/s.

Q7	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.
Ans:	Hearing range for humans = 20 Hz to 20 kHz Speed of sound = 344 m/s For a sound wave, Speed = Wavelength × Frequency i. For = 20 Hz ii. For = 20,000 Hz Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Q8	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.
Ans:	Speed of sound in air = 346 m/s Speed of sound in Aluminium = 6420 m/s Let the length of the aluminium rod = d. Therefore, time taken by the sound wave to reach the other end, Therefore, time taken by sound wave to reach the other end, The ratio of time taken by the sound wave in air and aluminium:

Q9	The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Ans:	The no. of oscillations produced in one second is called frequency . Frequency of sound = 100 Hz Total time = 1 min = 60 s We know that, Number of oscillations = Frequency × Total time Number of oscillations/Vibrations = 100 × 60 = 6000 Therefore , the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.

Q10	Does sound follow the same laws of reflection as light does? Explain.
Ans:	Yes, sound follows the same laws of reflection of light. The incident sound wave, reflected sound wave and normal sound wave all lie on the same plane. Also, the angle of incidence of sound is equal to angle of reflection of sound.

Q11	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?
Ans:	As the temp. increases, the speed of sound also increases. Therefore, the speed of sound on a hotter day is more. An echo is heard when the time interval between the original sound and the reflected sound is at least 0.1 s.

Q12	Give two practical applications of reflection of sound waves.
Ans:	(i) Sound navigation and ranging which is used to find the depth of the ocean. This method is known as SONAR. (ii) Stethoscope is used by doctors to measure heartbeat. Dr. is able to hear the sound due to multiple reflection of sound in the stethoscope.

Q13	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.
Ans:	Height of the tower, s = 500 m Velocity of sound, v = 340 m/s Acceleration due to gravity, g = 10 m/s² Initial velocity of the stone, u = 0 Time taken by the stone to fall to the base of the tower, t₁ According to the second equation of motion: Now, time taken by the sound to reach the top from the base of the tower, Hence , the splash is heard at the top after time, t Where, t = t₁ + t₂ = 10 + 1.47 = 11.47 s

Q14	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?
Ans:	Speed of sound, = 339 m/s Wavelength of sound, = 1.5 cm = 1.5 / 100 = 0.015 m Speed of sound = Wavelength × Frequency Frequency of of sound is 22600 Hz. The audible range of the human ear is 20 Hz to 20000 Hz. It would not be audible.

Q15	What is reverberation? How can it be reduced?
Ans:	Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. Reverberation can be reduced by absorbing the sound using some materials as it reaches the wall and ceiling of the room and thus prevents the sound from getting reflected. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound. Some materials which are used to reduce reverberation are fibreboard, heavy curtains, plastics etc.

Q16	What is loudness of sound? What factors does it depend on?
Ans:	Loudness is defined as a measure of the response of the ear to the sound. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Q17	Explain how bats use ultrasound to catch a prey.
Ans:	Sound frequencies greater than 20,000 Hz are called ultrasounds. Bats produce high-pitched ultrasonic squeaks. These squeaks reflect on prey and return back to the bats ear. This allows a bat to know the distance of his prey.

Q18	How is ultrasound used for cleaning?
Ans:	The objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency ultrasonic waves are capable of removing the dirt from the objects very easily.

Q19	Explain the working and application of a sonar.
Ans:	SONAR is an acronym for Sound Navigation And Ranging . SONAR is a device that uses ultrasonic waves to measure the distance , direction and speed of underwater objects . It is also used to measure the depth of seas and oceans. A wave of ultrasonic sound is produced and transmitted by the transducer of the SONAR, which travels through sea water and strikes the object on the seabed, gets reflected back and is sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. 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’.

Q20	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.
Ans:	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,

Q21	Explain how defects in a metal block can be detected using ultrasound.
Ans:	Ultrasounds can be used to detect cracks and flaws in metal blocks. Defects in metal blocks occur due to crack in them. Cracks have air in them. The speed of sound in metal is greater than that in air. Thus, air acts as a rarer medium for sound waves. Whenever ultrasound is transmitted through a metal block that has defects, these waves get reflected in encountering such cracks. Therefore, these waves are not detected at the detector placed on the other side of the block, opposite to that of the transmitter. This indicates that there is a defect in the block.

Q22	Explain how the human ear works.
Ans:	Different sounds produced in our surroundings are collected by pinna. The collected sound passes through the auditory canal. At the end of the auditory canal, there is a thin membrane called the eardrum or tympanic membrane. When compression the medium reaches the eardrum the pressure on the outside of the membrane increases and force the eardrum inward. Similarly, the eardrum moves outward when the rarefaction reaches it. The eardrum starts vibrating back and forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone hammer into vibration. The vibrations are passed from the hammer to the second bone anvil, and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval window and passes its vibration to the liquid in the cochlea. This produces electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and we get a sensation of hearing.