NCERT Solutions for Class 9 Science, Chapter 12 : Sound

Class 9 Science, Chapter 12 : Sound

Class 9 is the first stepping stone for a student in the competitive world. With the introduction of the CBSE Board Exam for class 10 a few years back, this has become an important gateway for a student. Based on the results of class 9th a student selects his future stream of Science, Commerce or Arts suiting his interest.

Takshila Learning is providing NCERT Solutions for Class 9 Physics as per the latest syllabus by CBSE. Class 9 is the building block for the CBSE Class 10 Board Exams, not only for your exams but also for your higher studies and career. Physics is the most essential subject and the knowledge in this field opens up wider career opportunities for the students.

Below you can find the NCERT solution for Class 9th Science. You can get a Solution for the all-important question of Class 9 Science, Chapter 12 : Sound

Important Formulae

• Speed = Wavelength × Frequency
1. Velocity=Frequency×Wavelength
2. Frequency=1/Time period
3. For an echo,
4. Velocity=(2×Distance)/Time

Pg 162

Q1.How does the sound produced by a vibrating object in a medium reach your ear?

Solution:

When an object vibrates, it forces the neighboring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. during this way, vibrations produced by an object are transferred from one particle to a different till it reaches the ear.

Pg 163

Q1. Explain how sound is produced by your school bell.

Solution:

When the bell vibrates, it forces the adjacent particles in air to vibrate. This disturbance gives rise to a wave and when the bell moves forward, it pushes the air ahead of it. This creates an area of high pressures called compression. When the bell moves backwards, it creates an area of depression know as rarefaction. As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. This makes the sound of a bell propagate through air.

Q2. Why are sound waves called mechanical waves?

Solution:

Sound waves force the medium particles to vibrate. Hence, these waves are referred to as mechanical waves. Sound waves propagate through a medium thanks to the interaction of the particles present therein medium.

Q3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Solution:

Sound needs a medium to propagate. Since the moon is devoid of any atmosphere, you can’t hear any sound on the moon.

Pg 166

Q1. Which wave property determines (a) loudness, (b) pitch?

Solution:

(a) Amplitude (b) Frequency

(a) The loudness of a sound depends on its amplitude. If the amplitude of a sound is large, then the sound produced will be loud.

(b) The pitch of a sound depends on its frequency. A sound are considered a high pitched sound, if its frequency is high.

Q2. Guess which sound has a higher pitch: guitar or car horn?

Solution:

The frequency of the vibration of a sound produced by a guitar is greater than that produced by a car horn. Since the pitch of a sound is proportional to its frequency, the guitar has a higher pitch than a car horn.

Pg 166

Q1. What are wavelength, frequency, time period and amplitude of a sound wave?

Solution:

Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is understood because the wavelength. Its SI unit is metre (m).

Frequency: The number of complete oscillations per second is understood because the frequency of a acoustic wave. it’s measured in hertz (Hz).

Amplitude: The maximum height reached by the crest or trough of a sound wave is termed its amplitude.

Q2. How are the wavelength and frequency of a sound wave related to its speed?

Solution

Speed, wavelength, and frequency of a sound wave are related by the following equation:

Speed = Wavelength  × Frequency

Q3. 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, = 220 Hz

Speed of the sound wave, v = 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?

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.

Here,

Pg 166

Q1. 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.

Pg 167

Q1. 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 the fastest in solids. Its speed decreases in liquids and it is the slowest in gases.

So in this case, for a given temperature, sound travels fastest in iron.

Pg 168

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?

Solution:

Speed of sound,  = = 342 m s−1

Echo returns in time, t = 3 s

Distance travelled by sound =  × 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 = 513m

Pg 169

Q1. Why are the ceilings of concert halls curved?

Solution:

Ceilings of concert halls are curved to allow sound after reflection (from the walls) to spread uniformly in all directions.

Pg 170

Q1. 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. 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?

Solution: (a) Infrasound has frequencies less than 20 Hz.

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

Pg 172

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?

Pg 174

Q1. What is sound and how is it produced?

Solution:

Sound is produced by vibration. When a body vibrates, it forces the neighboring particles of the medium to vibrate. This creates a disturbance in the medium, which travels in the form of waves. When this disturbance reaches the ear, it produces sound.

Q2. Describe with the help of a diagram, how compressions and rarefactions are produced in 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. When it moves backward, it creates a region of low pressure in its vicinity. 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 the following figure).

Q3. Cite an experiment to show that sound needs a material medium for its propagation.

Solution:

Take an electric bell and hang this bell inside an empty bell-jar fitted with a vacuum pump (as shown in the following figure).

Initially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell decreases. If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid of any air. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. When there is no air present inside, we can say that a vacuum is produced. Sound cannot travel through vacuum. This shows that sound needs a material medium for its propagation.

Q4. Why is sound wave called a longitudinal wave?

Solution:

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

Q5. Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Solution:

Quality of sound is that characteristic which helps us identify a particular person. 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?

Solution:

In lightning process, flash and thunder are produced simultaneously. Flash is seen almost immediately as speed of light is very large. But thunder is heard a few seconds later as speed of sound << speed of light.

Speed of sound increases with the increase in density of the medium.

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.

Solution:

Speed=wave length * frequency

Wavelength = speed / frequency

Therefore:

1. Frequency of sound = 20 HZ

Speed of sound = 344

Wave length  = 344/20= 17.2 m

1. Frequency of sound=20khz (20000 hz)

Speed of sound = 344

Wave length  =  344/20000 = 0.0172

Q8. 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 taken by the sound wave in air and in aluminum to reach the second child.

Solution:

speed of sound in air = 346m/s

speed of sound in aluminium rod = 6420m/s

let the length of the rod be x

we know,

time = distance/speed

time taken in air = x/346

time taken in aluminium rod = x/6420

ratio =(x/346)/(x/6420)

= x/346 * 6420/x    [reciprocal]

x would get cancelled

= 6420/346

after cancelling

= 18.55

Q9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Solution:

Frequency is defined as the number of oscillations per second. It is given by the relation:

Frequency = Number of oscillations / Total time

Number of oscillations = Frequency × Total time

Given, Frequency of sound = 100 Hz

Total time = 1 min = 60 s

Number of oscillations/Vibrations = 100 × 60 = 6000

Hence, 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.

Solution:

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

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?

Solution:

An echo is heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease. Therefore, an echo can be heard only if the time interval between the original sound and the reflected sound is greater than 0.1 s.

Q12. Give two practical applications of reflection of sound waves.

Solution:

(i) Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.

(ii) Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.

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.

Solution:

Height of the tower, s = 500 m

Velocity of sound, v = 340 m s-1

Acceleration due to gravity, g = 10 m sˆ’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 + ½ gt12

500 = 0 x t1 + ½ x 10 x t12

t12 = 100

t1 = 10s

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.

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?

Solution: Speed  ⇒ 339 m/s

Wavelength ⇒ 1.5 cm

⇒ 1.5 / 100

⇒ 0.015 m

Speed = Wavelength x Frequency

⇒ 339 m/s  = 0.015m x Frequency

⇒ Frequency = 339 m/s / 0.015m

= 22600 Hz.

It is more than 20000 Hz. Hence, it will not be audible to the human ear.

Q15. What is reverberation? How can it be reduced?

Solution:

Persistence of sound due to repeated reflection is known as reverberation. As the source produces sound, it starts traveling in all directions. Once it reaches the wall of a room, it is partly reflected back from the wall. 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.

To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fiberboard, rough plastic, heavy curtains, and cushioned seats can be used to reduce reverberation.

Q16. What is loudness of sound? What factors does it depend on?

Solution:

A loud sound has high energy. 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.

Solution:

Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are reflected by objects such as preys and returned to the bat’s ear. This allows a bat to know the distance of his prey.

Q18. How is ultrasound used for cleaning?

Solution:

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.

Q19. Explain the working and application of a sonar.

Solution:

SONAR stands 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’.

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.

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 = 2X3625 /5 = 1450 ms-1

Q21. Explain how defects in a metal block can be detected using ultrasound.

Solution:

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.

Q22. Explain how the human ear works.

Solution:

Different sounds produced in our surroundings are collected by pinna that sends these sounds to the ear drum via the ear canal. The ear drum starts vibrating back and forth rapidly when the sound waves fall on that. The vibrating eardrum sets the tiny bone hammer into vibration. The vibrations are passed from the hammer to the second bone anvil, and eventually to the third bone stirrup. The vibrating stirrup strikes on the membrane of the fenestra and passes its vibration to the liquid within the cochlea. This produces electrical impulses in nerve cells. The cranial nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and that we get a sensation of hearing.

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