NCERT Solutions for Class 9 Science Chapter 12 - Sound

Intext Exercise 1

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

Ans: When an object vibrates, it causes the neighbouring particles in the medium to vibrate successively. These vibrations are further passed to adjacent particles. Like ways these vibrations are passed from particle to particle and reach our ears.

Intext Exercise 2

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

Ans: When the school bell rings, the neighbouring particles in air are forced to vibrate simultaneously. This disturbance leads to the formation of a sound wave and when the bell moves forward, it pushes the air in front of it. This creates a region of high pressures known as compression. When the bell moves backwards, a region of low pressure known as rarefaction is created. In this way the bell continues to move forward and backward which produces a series of compressions and rarefactions. This leads to the sound of a bell when it propagates through air.

2. Why are sound waves called mechanical waves?

Ans: When sound waves propagate through a medium it causes the neighbouring particles to vibrate. Sound waves propagate due to interaction of particles in the medium due to a series of compressions and rarefactions. Hence, these 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?

Ans: Sound waves are mechanical waves and hence need a medium to propagate. As the moon is devoid of any atmosphere, we cannot hear any sound on the moon.

Intext Exercise 3

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

Ans: 

(a) The loudness of a sound depends on its amplitude. Higher the amplitude, the louder the sound.

(b) The pitch of a sound depends on its frequency. Higher the frequency, higher the pitch of sound.

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

Ans: In the case of guitar, the frequency of vibration produced is greater when compared to a car horn. Pitch of the sound depends on the frequency. Higher the frequency, the higher the pitch. So the guitar produces a higher pitched sound than a car horn.

Intext Exercise 4

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

Ans: Following are the definitions given below:

• Wavelength: Wavelength is defined as the distance between two consecutive compressions or two consecutive rarefactions. Its SI unit is metre (m).

• Frequency: Frequency of a sound wave is defined as the number of complete oscillations per second. It is measured in hertz (Hz).

• Time period: Time period of a sound wave is defined as the time taken by a sound wave to complete one cycle. Its SI unit is seconds (s).

• Amplitude: The amplitude of a sound wave is defined as the maximum extent of a vibration, measured from the position of equilibrium.

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

Ans: Speed, wavelength, and frequency of a sound wave are related by the equation given below:

\[Speed\text{ }\left( v \right)\text{ }=\text{ }Wavelength\text{ }\left( \lambda  \right)\text{ }\times \text{ }Frequency\text{ }\left( \upsilon  \right)\]

$\therefore v=\lambda \times \upsilon $

3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

Ans: In the above question it is given that:

Frequency of the sound wave is \[\upsilon =220\text{ }Hz\]

Speed of the sound wave is \[v=440\text{ }m/s\]

For a sound wave,

\[Speed\text{ }\left( v \right)\text{ }=\text{ }Wavelength\text{ }\left( \lambda  \right)\text{ }\times \text{ }Frequency\text{ }\left( \upsilon  \right)\]

$\therefore \lambda =\frac{440}{220}=2m$

Therefore, the wavelength of the sound wave is \[2\text{ }m\]. 

4. 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: In the above question it is given that:

$frequency=500Hz$

The time interval between two successive compressions is equal to the time period of the wave. We know that,

\[Time\text{ }period=\frac{1}{frequency}=\frac{1}{500}=0.002s\]

Intext Exercise 5

1. Distinguish between loudness and intensity of sound.

Ans: The intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. The loudness or softness of a sound is determined basically by its amplitude. The amplitude of the sound wave depends upon the force with which an object vibrates. Intensity helps to decide the amplitude of a sound wave, which in turn is recognized by the ear as loudness.

Intext Exercise 6

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

Ans: Sound travels the fastest in solids, i.e., iron followed by water and air at a particular temperature.

Intext Exercise 7

1. 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?

Ans: In the above question it is given that:

Speed of sound is \[v\text{ }=\text{ }342\text{ }m/s\].

Echo returns in time, \[t\text{ }=\text{ }3s\].

Distance travelled by sound is given by: \[v\text{ }\times \text{ }t\text{ }=\text{ }342\text{ }\times \text{ }3\text{ }=\text{ }1026\text{ }m\]

As the sound travels and gets reflected back, the distance of the reflecting surface from the source will be: \[\frac{1026}{2}\text{= 513 }m\].

Intext Exercise 8

1. Why are the ceilings of concert halls curved?

Ans: Ceilings of concert halls are curved so that the sound waves can spread uniformly in all directions after reflection.

Intext Exercise 9

1. What is the audible range of the average human ear?

Ans: The audible range of an average human ear is between \[20\text{ }Hz\] to \[20,000\text{ }Hz\]. Humans cannot hear sounds with frequency less than \[20\text{ }Hz\]and greater than \[20,000\text{ }Hz\].

2. What is the range of frequencies associated with

(a) Infrasound?

Ans: Infrasound has frequencies less than \[20\text{ }Hz\].

(b) Ultrasound?

Ans: Ultrasound has frequencies more than \[20,000\text{ }Hz\].

 Intext Exercise 10

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?

Ans: In the above question it is given that:

Time taken by the sonar pulse to return is \[t\text{ }=\text{ }1.02\text{ }s\].

Speed of soundwave is \[v\text{ }=\text{ }1531\text{ }m\text{ /}s\]. 

\[Distance\text{ }of\text{ }the\text{ }cliff\text{ }from\text{ }the\text{ }submarine\text{ }=\text{ }Speed\text{ }of\text{ }sound\text{ }\times \text{ }Time\text{ }taken\]

\[\therefore Distance\text{ }of\text{ }the\text{ }cliff\text{ }from\text{ }the\text{ }submarine=1.02\times 1531=1561.2m\]

As the sound travels and gets reflected back, actual distance will be:

$\frac{1561.62}{2}=780.31m$

NCERT Exercise

1. What is sound and how is it produced?

Ans: Sound is a form of energy which produces the sensation of hearing produced by vibration. When an object vibrates, it causes the neighbouring particles of the medium to vibrate which are further passed to adjacent particles. This creates a disturbance in the medium, which travels in the form of waves. Hence, when this disturbance reaches the ear, sound is produced.

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

Ans: Consider the figure given below:

The most common medium through which sound travels is air. When a vibrating object moves forward, it pushes and compresses the air forward creating a region of high pressure. This region is called a compression (C). This compression starts to move away from the vibrating object. 

When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R). As the object moves back and forth rapidly, a series of compressions and rarefactions are created in the air. These make the sound wave that propagates through the medium. Compression is the region of high pressure and rarefaction is the region of low pressure and the pressure is related to the number of particles of a medium in a given volume. 

More density of the particles in the medium gives more pressure and vice versa. Hence, propagation of sound can be visualised as propagation of density variations or pressure variations in the medium.

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

Ans: Consider the figure given below:

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Follow the procedure given below to show that sound needs a material medium for its propagation:

Take an electric bell and an airtight glass bell jar such that the electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump, as shown in Fig. 

If you press the switch you will be able to hear the bell. Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes fainter, although the same current is passing through the bell. After some time when less air is left inside the bell jar you will hear a very feeble sound. When there is no air present inside, vacuum is produced. Sound cannot travel through a vacuum. This indicates that sound needs a material medium for its propagation.

4. Why is a sound wave called a longitudinal wave?

Ans: In the case of sound waves, the particles do not move from one place to another but they simply oscillate back and forth about their position of rest. In sound waves the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. Hence, a longitudinal wave is called a sound wave.

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

Ans: The characteristic of the sound which helps you to identify your friend by his voice while sitting with others in a dark room is the quality or timber of sound which enables us to distinguish one sound from another having the same pitch and loudness.

6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?

Ans: Velocity of sound is \[344\text{ }m/s\] and that of light is \[3\text{ }\times \text{ }{{10}^{8}}m/s\] . As the speed of light is greater than that of sound, the sound of thunder requires longer time than light to reach Earth. Therefore, before we hear thunder, a flash is seen.

7. A person has a hearing range from \[20\text{ }Hz\] to \[20\text{ k}Hz\]. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as \[344\text{ }m/s\].

Ans: In the above question it is given that:

A person has a hearing range from \[20\text{ }Hz\] to \[20\text{ k}Hz\].

Speed of sound in air is \[344\text{ }m/s\].

We know that:

\[Speed\text{ }=\text{ }Wavelength\text{ }\times \text{ }Frequency\]

$\therefore v=\lambda \times \upsilon $

 For, ${{\upsilon }_{1}}=20\text{ }Hz$

$\therefore {{\lambda }_{1}}=\frac{v}{{{\upsilon }_{1}}}=\frac{344}{20}=17.2m$

 For, ${{\upsilon }_{2}}=20\text{ k}Hz$

$\therefore {{\lambda }_{2}}=\frac{v}{{{\upsilon }_{2}}}=\frac{344}{20000}=0.0172m$

Hence, humans have the wavelength range for hearing from \[0.0172\text{ }m\] to \[17.2\text{ }m\].

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 time taken by the sound wave in air and in aluminium to reach the second child.

Ans: Consider the length of the aluminium rod to be d.

Speed of sound waves in aluminium at \[25{}^\circ C\] is ${{v}_{Al}}=6420m/s$.

Therefore, time taken by the sound wave to reach the other end will be:

${{t}_{Al}}=\frac{d}{{{v}_{Al}}}=\frac{d}{6420}$

Speed of sound waves in air at \[25{}^\circ C\] is ${{v}_{Air}}=346m/s$.

Therefore, the time taken by a sound wave to reach the other end will be:

${{t}_{Air}}=\frac{d}{{{v}_{Air}}}=\frac{d}{346}$

Hence, the ratio of time taken by the sound wave in air and aluminium will be:

$\frac{{{t}_{Air}}}{{{t}_{Al}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55$.

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

Ans: In the above question it is given that:

Frequency of sound is \[100\text{ }Hz\].

\[Total\text{ }time\text{ }=\text{ }1\text{ }min\text{ }=\text{ }60\text{ }s\].

We know that:

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

\[Number\text{ }of\text{ }oscillations\text{ }=\text{ }Frequency\text{ }\times \text{ }Total\text{ }time\]

\[Number\text{ }of\text{ }oscillation=\text{ }100\text{ }\times \text{ }60\text{ }=\text{ }6000\]

Hence, the source vibrates \[6000\] times in a minute, producing a frequency of \[100\text{ }Hz\].

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

Ans: The incident and the reflected sound wave create the same angle at the point of incidence with the normal to the surface. In addition, the sound wave incident, the sound wave reflected, and the normal sound wave to the point of incidence are all in the same plane. Hence, sound follows the same laws of reflection as light does.

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 remain the same. Do you hear the echo sound on a hotter day?

Ans: An echo is heard when the time interval between the original sound and the sound reflected is at least 0.1s. Sound speed in a medium is directly proportional to the temperature. 

Hence, time interval will be inversely directly proportional to the temperature. Therefore, the time interval between the original sound and the sound reflected will decrease on a hotter day. 

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

Ans: Following are the two practical applications of reflection of sound waves:

(a) SONAR: SONAR is a technology where reflection of sound is used to measure the distance and speed of underwater objects.

(b) Stethoscope: A stethoscope is a device where the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection 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, \[g\text{ }=\text{ }10\text{ }m/{{s}^{2}}\] and \[speed\text{ }of\text{ }sound\text{ }=\text{ }340\text{ }m/{{s}^{2}}\].

Ans: In the above question it is given that:

Height of the tower is \[s\text{ }=\text{ }500\text{ }m\].

Velocity of sound is \[v\text{ }=\text{ }340\text{ }m/s\].

Acceleration due to gravity is \[g\text{ }=\text{ }10\text{ }m/{{s}^{2}}\].  

As the stone is initially at rest, initial velocity of the stone will be \[u\text{ }=\text{ 0 m/s}\].

Let the time taken by the stone to fall to the base of the tower be ${{t}_{1}}$

According to the second equation of motion:

\[s=u{{t}_{1}}+\frac{1}{2}g{{t}_{1}}^{2}\]

$\Rightarrow 500=\frac{1}{2}\left( 10 \right){{\left( {{t}_{1}} \right)}^{2}}$

$\Rightarrow {{\left( {{t}_{1}} \right)}^{2}}=100$

$\Rightarrow {{t}_{1}}=10s$.

Now, time taken by the sound to reach the top from the base of the tower will be

$\Rightarrow {{t}_{2}}=\frac{500}{340}=1.47s$

Hence, the splash is heard at the top after time $t={{t}_{1}}+{{t}_{2}}=10+1.47=11.47s$.

14. A sound wave travels at a speed of \[339\text{ }m/s\]. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Ans: In the above question it is given that:

Speed of sound is \[339\text{ }m/s\].

Wavelength of sound is $\lambda =\text{ }1.5\text{ }cm\text{ }=\text{ }0.015\text{ }m$.

We know that:

\[Speed\text{ }of\text{ }sound\text{ }=\text{ }Wavelength\text{ }\times \text{ }Frequency\]

$\Rightarrow v=\lambda \times \upsilon $

Therefore frequency $\upsilon $ will be:

$\Rightarrow \upsilon =\frac{v}{\lambda }=\frac{339}{0.015}=22600Hz$

As the frequency range of audible sound for humans lies between \[20\text{ }Hz\] to \[20,000\text{ }Hz\]. Since the frequency of the given sound is more than \[20,000\text{ }Hz\], it won’t be audible.

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

Ans: Reverberation is defined as persistence of sound (after the source stops producing sound) due to repeated reflection. When a sound is created in a big hall, it persists by repeated reflection from the walls until it is reduced to a value where it is no longer audible. 

To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are selected based on their sound absorbing properties.

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

Ans: The measure of the response of the ear to the sound is defined as the loudness of sound. 

The loudness or softness of sound is determined basically by its amplitude which depends upon the force with which an object is made to vibrate. A loud sound has high energy. 

Loudness depends on the amplitude of vibrations such that loudness is proportional to the square of the amplitude of vibrations.

17. Explain how bats use ultrasound to catch prey.

Ans: Bats produce high-pitched, ultrasonic squeaks. These high-pitched squeaks are reflected by objects like prey and brought back to the ear of the bat. This helps the bat to understand his prey's distance.

18. How is ultrasound used for cleaning?

Ans: Ultrasound waves are used for cleaning by passing through the objects kept in the cleaning solution. Their high frequency removes dirt from the objects.

19. Explain the working and application of a sonar.

Ans: SONAR is an abbreviation for Sound Navigation And Ranging. 

It is an acoustic device used with the help of ultrasounds to measure the depth, direction and speed of underwater objects like submarines. It is also used to measure sea and ocean depth. 

The SONAR produces and transmits an ultrasonic sound beam. 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 using the time (t) taken by the echo to return with speed (v) is given by

\[2d\text{ }=\text{ }v\text{ }\times \text{ }t\]

This method of measuring distance is termed as ‘echo-ranging’.

20. A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Ans: In the above question it is given that:

Time taken to hear the echo is \[t\text{ }=\text{ }5\text{ }s\].

Distance of the object from the submarine is \[d\text{ }=\text{ }3625\text{ }m\].

Total distance travelled by the sonar waves during the transmission and reception in water is 2d

Therefore, the velocity of sound in water will be:

$v=\frac{2d}{t}=\frac{2\times 3625}{5}=1450m/s$

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

Ans: Metallic components are generally used in construction of big structures like buildings, bridges, machines and also scientific equipment. Ultrasounds are used to detect cracks and flaws in metal blocks. 

The cracks or holes inside the metal blocks that are invisible from outside reduce the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. 

If there is even a small defect, the ultrasound gets reflected back. This indicates the presence of the flaw or any defect. 

Finding defects using Ultrasound:

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22. Explain how the human ear works.

Ans: Human ear has three sections, the outer ear, the middle ear and the inner ear. Pinna of the external ear performs the function of collecting the sound vibrations. They enter into the ear canal. 

These vibrations strike the eardrum, making it vibrate. The vibrations from the eardrum reach the middle ear which contains three small bones: malleus (hammer-shaped), incus (anvil shaped) and stapes (stirrup-shaped). They are meant for magnifying sound vibrations. 

The vibrations are further transmitted to the cochlea which is the inner part of the ear. The cochlea is filled with fluid which transmits the vibrations. The motion of the vibrations in the cochlea is detected by tiny hairs connected to nerves at this point. 

The vibrations are converted into electrical signals and carried by the auditory nerves to the brain where the sensation of the sound is realized.

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NCERT Solutions for Class 9th Science Chapter 12 Sound - Free PDF Download

Even if you are planning to have a Commerce or Arts background in the future, it is necessary to have basic knowledge of science which can help you in future. NCERT Solutions for Class 9 Chapter 12 ensures that the student gains a basic knowledge of Sound chapter Class 9. The Class 9th Science Chapter 12 Sound is easy to understand and can help secure good marks in the exam.

You can opt for Chapter 12 - Sound NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science

• Chapter 1 - Matter in Our Surroundings

• Chapter 2 - Is Matter Around us Pure

• Chapter 3 - Atoms and Molecules

• Chapter 4 - Structure of Atom

• Chapter 5 - The Fundamental Unit of Life

• Chapter 6 - Tissues

• Chapter 7 - Diversity in Living Organisms

• Chapter 8 - Motion

• Chapter 9 - Force and Laws of Motion

• Chapter 10 - Gravitation

• Chapter 11 - Work and Energy

• Chapter 12 - Sound

• Chapter 13 - Why do We Fall ill

• Chapter 14 - Natural Resources 

• Chapter 15 - Improvement in Food Resources

NCERT Solution Class 9 Science Chapter 12 - Sound

Ch Sound Class 9 comes under unit III of the syllabus. This unit is based upon Motion, Force and Work. The Class 9 Science Chapter 12 includes topics like propagation and production of sound.

The topic propagation of sound is meant for educating students on how sound needs a medium to travel, how the waves of sound are considered to be longitudinal, what are the characteristics of sound waves and the speed of the sound in different media. The topics mentioned here are explained with the help of practical activities, diagrams and in between exercises. The solution for Class 9 Science Chapter 12 will help you to understand the two main topics of sound, i.e., propagation and production of sound very easily.

This ch 12 science class 9 will help the students to learn about the reflections of sound ECHO, reverberation and uses of multiple reflections of sound. All these aspects related to sound are explained with proper laboratory examples, thus keeping the readers engaged with the chapter. Students can check their level of learning by comparing their answers with the solutions given in the chapter for different problems. The concept of many topics such as the range of hearing, applications of ultrasound, etc. will get studied with the help of proper practical laboratory examples. With the NCERT Solutions for class 9th science chapter 12 sound, you can strengthen your basics that will be helpful in the future.

In this chapter sound, class 9 students will learn about the human ear and its structures. Students are advised to go through all the problems and solutions given in this chapter for making sure that they will score good marks in the exam. The NCERT solutions for class 9 science ch 12 has been prepared by some of the best teachers who are experts in this field.

Class 9 Science Chapter 12 Sound Weightage Marks

The Sound chapter Class 9 comes under unit III that is based on Motion, Force and Work. In the annual examination, this unit holds a weightage of 27 marks. The Sound chapter Class 9 NCERT Solution includes the solutions to the problems given in this chapter of the textbook. The Chapters are listed below:

Ex 12.1 - Production of Sound

Ex 12.2 - Propagation of Sound

Ex 12.3 - Reflection of Sound

Ex 12. 4 - Range of Hearing

Ex 12.5 - Applications of Ultrasound

Ex 12.6 - Structure of Human Ear

Benefits of Sound Chapter Class 9 NCERT Solutions

The chapter sound class 9 is an easy chapter and can help students to secure a higher percentage in their exams. Some of the benefits of NCERT Solutions for Class 9 chapter sound are: