Hello students and welcome to the important chapter on Sound! Mastering the concepts in this chapter early on is key, and being able to tackle numerical problems on your own is crucial. With solutions to all the in-text questions, exercises, and activities in one place, we’ve got you covered. Plus, we’ve included an extra problem set to help you hone your problem-solving skills and further enhance your understanding. Engaging illustrations make learning even more enjoyable. So, let’s dive in!
Solutions to In Text Questions of NCERT Class 9 Science Chapter 11 Sound –
Page 129:
1. How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it sets the particles of the medium around it vibrating. These particles then make the particles adjacent to them vibrate. This process continues in the medium until the disturbance created by a source of sound reaches your ear.
2. Explain how sound is produced by your school bell.
Answer:
The school bell vibrates back and forth rapidly and creates a series of compressions and rarefactions in the air. These start to move away from the vibrating object and make the sound wave that propagates through the medium.
3. Why are sound waves called mechanical waves?
Answer:
Sound waves are called mechanical waves because they require a medium (such as air, water, or solids) to travel through. These waves propagate by causing particles in the medium to oscillate back and forth in the direction of wave motion. Since sound waves are characterised by the motion of particles in the medium, they are called mechanical waves.
4. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
You will not be able to hear any sound produced by your friend because sound needs a medium to propagate. The moon has no atmosphere and hence sound cannot travel through vacuum.
Page 132:
1. Which wave property determines (a) loudness, (b) pitch?
Answer:
(a) The amplitude of a sound wave determines the loudness. Higher the amplitude, louder the sound.
(b) The frequency of a sound wave determines the pitch. Higher the pitch, higher the frequency of the sound.
2. Guess which sound has a higher pitch: guitar or car horn?
Answer:
The sound of the guitar has a higher frequency and hence has a higher pitch.
Page 132:
1. What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
Wavelength: The distance between two compressions and rarefactions is known as the wavelength. The wavelength is usually represented by λ and its SI unit is metre (m).
Frequency: The number of oscillations per unit time is the frequency of the sound wave. It is usually represented by ν. Its SI unit is hertz.
Time period: The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave. In other words, we can say that the time taken for one complete oscillation is called the time period of the sound wave. It is represented by the symbol T. Its SI unit is second (s).
Amplitude: The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave. It is usually represented by the letter A.
2. How are the wavelength and frequency of a sound wave related to its speed?
Answer:
The relation is:
Speed (v) = Wavelength (λ) × Frequency (ν)
3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Speed (v) = Wavelength (λ) × Frequency (ν)
or, v = λν
or, 440 = λ × 220
or, λ = 440/220
or, λ = 2 m
Hence, the wavelength of the sound wave = 2 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?
Answer:
The time interval between successive compressions is the time period (T).
T = 1/ν
or, T = 1/500
or, T = 0.002 s
Page 133:
1. Distinguish between loudness and intensity of sound.
Answer:
The amount of sound energy passing each second through unit area is called the intensity of sound. We sometimes use the terms “loudness” and “intensity” interchangeably, but they are not the same. Loudness is a measure of the response of the ear to the sound. Even when two sounds are of equal intensity, we may hear one as louder than the other simply because our ear detects it better.
Page 133:
1. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
Sound travels the fastest in solids. Therefore, sound traves the fastest at a particular temperature in iron.
Page 134:
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 m s–1?
Answer:
Speed of sound (v) = 342 ms-1
Sound of echo returns in time (t) = 3 s
Therefore, distance travelled by sound = v × t = 342 × 3 = 1026 m.
The sound covers the distance between the source and the reflecting surface twice.
Therefore, the distance of the reflecting surface from the source = 1026/2 = 513 m.
Page 135:
1. Why are the ceilings of concert halls curved?
Answer:
Ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall and everyone can hear the sound.
Page 136:
1. What is the audible range of the average human ear?
Answer:
The audible range of the average human ear is 20 Hz – 20,000 Hz.
2. What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer:
(a) The frequency of infrasound is below 20 Hz.
(b) The frequency of ultrasound is above 20,000 Hz.
Solutions to Exercises (Page No 138) of NCERT Class 9 Science Chapter 11 Sound –
1. What is sound and how is it produced?
Answer:
Sound is a type of energy caused by vibrations. When an object vibrates, it sets the particles of the medium around it vibrating. These particles then make the particles adjacent to them vibrate. This process continues in the medium until the disturbance created by a source of sound reaches your ear and hence, sound is produced.
2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When a vibrating object such as a tuning fork moves forward, it pushes and compresses the air particles in front of it creating a region of high pressure. This region is called a compression (C), as shown in the figure. 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 shown in the figure. As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air.
3. Why is sound wave called a longitudinal wave?
Answer:
In sound waves, individual particles of the medium vibrate in a direction parallel to the direction of propagation of the disturbance. Hence sound waves are longitudinal waves.
4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality of a sound helps you to identify a particular voice while sitting with others in a dark room. Even though the pitch and loudness are same, quality of a sound is unique.
5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
The speed of sound is much less (344 m/s) as compared to the speed of light (3 × 108 m/s). Hence, sound of thunder takes more time to reach the earth as compared to light from the lightning flash. Hence, thunder is heard after lightning.
6. 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 (v) = Wavelength (λ) × Frequency (ν)
or, v = λν
For 20 Hz:
or, 344 = λ × 20
or, λ = 344/20
or, λ = 17.2 m
For 20 kHz:
or, 344 = λ × 20000
or, λ = 344/20000
or, λ = 0.0172 m
Therefore, humans can hear sounds between wavelengths 0.0172 m and 17.2 m.
7. 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.
Answer:
From Table 12.1:
Speed of sound in aluminium (vAl) = 6420 m/s.
Speed of sound in air (vAir) = 346 m/s.
Let length of the aluminium rod be l.
Time taken for sound to reach the other end via aluminium rod (TAl) = l/6420.
Time taken for sound to reach the other end via air (TAir) = l/346.
TAir/TAl = l/346 ÷ l/6420
or, TAir/TAl = 6420/346
or, TAir/TAl = 18.55
The ratio of times taken by the sound wave in air and in aluminium to reach the second child is 18:55:1.
8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
The frequency of a source of sound is 100 Hz.
Therefore, in 1 second it vibrates 100 times.
Therefore, in 1 minute or 60 seconds, it vibrates (100 × 60) = 6000 times.
9. Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, sound follows the same laws of reflection as light does. The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence. The incident ray, reflected ray and normal all lie in the same plane.
10. 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?
Answer:
An echo occurs when the time gap between the original sound and its reflection is at least 0.1 seconds. With rising temperatures, the speed of sound also increases. Consequently, on warmer days, the time gap between the reflected and original sound diminishes and can become less than 0.1 seconds. Therefore, the echo will not be heard on a hotter day.
11. Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves are:
(a) Stethoscope: In stethoscopes the sound of the patient’s heartbeat reaches the doctor’s ears by multiple reflection of sound.
(b) Detection of Defects in Metal: Ultrasonic waves are passed through the metal and get reflected by any defects present in the interior of the metal. Thus, defects can be detected.
12. 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 of the tower (s) = 500 m.
Acceleration due to gravity (g) = 10 m s−1.
Initial of the stone (u) = 0.
Let time taken by the stone to fall into the pond be t1.
We know,
s= ut1 + (½) a(t1)2
500 = 0 x t1 + (½) 10 (t1)2 (Here a = g)
(t1)2 = 100
t1 = 10 s
Now, speed of sound (v) = 340 m s−1.
Let (t2) taken by sound to reach the top be t2.
t2 = 500/340 = 1.47 s
Total time (t) = t1 + t2
t = 10 s + 1.47 s
t = 11.47 s
Therefore, the splash heard at the top after 11.47 s.
13. 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) = Wavelength (λ) × Frequency (ν)
or, v = λν
v = 339 m s–1
λ = 1.5 cm = 0.015
Therefore,
339 = 1.5 × ν
or, ν = 339/0.015
or, ν = 22600 Hz
The audible range of sound for human beings extends from about 20 Hz to 20000 Hz. The frequency 22600 Hz falls beyond that range and is not audible.
14. What is reverberation? How can it be reduced?
Answer:
The repeated reflection of sound in a closed enclosed space is called reverberation. To reduce reverberation, the roof and walls of the enclosed spaces can be covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies.
15. What is loudness of sound? What factors does it depend on?
Answer:
Loudness is a measure of the response of the ear to the sound. Even when two sounds are of equal intensity, we may hear one as louder than the other simply because our ear detects it better. Loudness depends on the amplitude of the sound wave. The larger the amplitude, the louder the sound.
16. How is ultrasound used for cleaning?
Answer:
Objects requiring cleaning are submerged in a cleaning solution, where they are subjected to ultrasonic sound waves. The high frequency of these waves aids in dislodging dirt and impurities from the objects, facilitating effective cleaning.
17. Explain how defects in a metal block can be detected using ultrasound.
Answer:
When ultrasounds are passed through a metal block, even small defects reflect the ultrasound back. This indicates the presence of the flaw or defect.
Solutions to All Activities of NCERT Class 9 Science Chapter 11 Sound
1. Complete Activity 11.1 (Page 127).
• Take a tuning fork and set it vibrating by striking its prong on a rubber pad. Bring it near your ear.
• Do you hear any sound?
• Touch one of the prongs of the vibrating tuning fork with your finger and share your experience with your friends.
• Now, suspend a table tennis ball or a small plastic ball by a thread from a support [Take a big needle and a thread, put a knot at one end of the thread, and then with the help of the needle pass the thread through the ball]. Touch the ball gently with the prong of a vibrating tuning fork (Fig. 11.1).
• Observe what happens and discuss with your friends.
Answer:
2. Complete Activity 11.2 (Page 127).
• Fill water in a beaker or a glass up to the brim. Gently touch the water surface with one of the prongs of the vibrating tuning fork, as shown in Fig. 11.2.
• Next dip the prongs of the vibrating tuning fork in water, as shown in Fig. 11.3.
• Observe what happens in both the cases.• Discuss with your friends why this happens.
Answer:
3. Complete Activity 11.3 (Page 128)• Make a list of different types of musical instruments and discuss with your friends which part of the instrument vibrates to produce sound.
Answer:
4. Complete Activity 11.4 (Page 129).
• Take a slinky. Ask your friend to hold one end. You hold the other end. Now stretch the slinky as shown in Fig. 11.5(a). Then give it a sharp push towards your friend.
•What do you notice? If you move your hand pushing and pulling the slinky alternatively, what will you observe? • If you mark a dot on the slinky, you will observe that the dot on the slinky will move back and forth parallel to the direction of the propagation of the disturbance.
Answer:
5. Complete Activity 11.5 (Page 133).
• Take two identical pipes, as shown in Fig. 11.9. You can make the pipes using chart paper. The length of the pipes should be sufficiently long as shown.
• Arrange them on a table near a wall.
• Keep a clock near the open end of one of the pipes and try to hear the sound of the clock through the other pipe.
• Adjust the position of the pipes so that you can best hear the sound of the clock.
• Now, measure the angles of incidence and reflection and see the relationship between the angles.
• Lift the pipe on the right vertically to a small height and observe what happens. (In place of a clock, a mobile phone on vibrating mode may also be used.)
Answer:
Extra Questions to Complement Solutions to NCERT Class 9 Science Chapter 11 Sound
Very Short Answer Type:
1. What kind of body produces sound?
Answer:
Vibrating body.
2. Sound will travel faster through iron or water?
Answer:
Iron. Speed of sound in solids is more than in liquids.
3. What kind of sound is used to clean electronic components?
Answer:
Ultrasound (frequency greater than 20,000 Hz)
4. What kind of sound is used to image internal organs of the human body?
Answer:
Ultrasound.
5. How is the distance between successive compressions and rarefactions of a sound wave related to the wavelength?
Answer:
Distance = Half the wavelength.
6. You are given a curved surface and a flat surface. Which will give you a better reflection of sound?
Answer:
Curved surface.
7. Can sound travel through space?
Answer:
No, there is no medium for sound to travel, only vacuum is present.
8. You cannot hear a sound, but a moth can hear it. What kind of sound is it?
Answer:
Ultrasound.
9. Speed of sound is lower in a medium than the speed in liquid at the same temperature. What kind of medium is it?
Answer:
Gas.
10. Sound waves emanating from a source travel in ________ lines right after they leave the source.
Answer:
Straight.
Multiple Choice Questions (MCQ):
1. The time period of a sound wave is 0.0003 seconds. The sound is
(A) Infrasound
(B) Ultrasound
(C) In between audible range
(D) Frequency cannot be determined
Answer: (C) In between audible range
We know, ν = 1/T.
Therefore ν = 1/0.0003 = 3333.3 Hz.
Audible range is from 20 Hz to 20,000 Hz
Therefore, 3333.3 Hz falls within the audible range.
2. If we want to increase the loudness of a sound we should increase the
(A) Amplitude
(B) Pitch
(C) Wavelength
(D) Frequency
Answer: (A) Amplitude
3. The velocity of sound in a medium is 200 m/s and the distance between two compressions is 0.5 m. What is the frequency of the sound?
(A) 100 Hz
(B) 200 Hz
(C) 250 Hz
(D) 400 Hz
Answer: (D) 400 Hz
Wavelength = distance between two compressions = 0.5 m
Velocity = 200 m/s.
Speed (v) = Wavelength (λ) × Frequency (ν)
or, v = λν
200 = 0.5(ν)
or, ν = 400 Hz.
4. The velocity of a sound in air at 250 C is 346 m/s. What will happen to the velocity of sound at the same conditions if the frequency is doubled?
(A) Remains the same
(B) Increases
(C) Decreases
(D) First increases then decreases
Answer: (A) Remains the same
The speed of sound remains constant in a particular medium at a particular temperature.
5. Which of the following animals do not produce ultrasound?
(A) Dolphins
(B) Rhinoceros
(C) Bats
(D) Porpoises
Answer: (B) Rhinoceros
Rhinoceroses produced infrasound.
Short Answer Type:
1. How does the frequency of a sound wave affect its pitch?
Answer:
The frequency of a sound wave directly affects its pitch. Higher the frequency of the sound, higher is its pitch.
2. Which of the following sound waves has a higher pitch?
Answer: The number of oscillations per unit time is known as the frequency. The frequency of vibration of wave (B) is higher, hence the pitch is higher.
3. The time period of a wave is 0.2 s and the time period of another wave is 0.4 s. What is the ratio of the frequencies of the waves?
Answer:
We know, ν = 1/T.
Therefore ν1 = 1/0.2 = 5 Hz.
ν2 = 1/0.4 = 2.5 Hz.
ν1/ν2 = 5/2.5 = 2.
Therefore, ratio of the frequencies of the waves is 2:1.
4. Provide a real-life example of how compressions and rarefactions are utilized in sound transmission or communication.
Answer:
Inside a microphone, there is a diaphragm that vibrates when sound waves from the surrounding environment hit it. When sound waves reach the diaphragm, they cause compressions and rarefactions in the air. The diaphragm responds to these changes in pressure by moving back and forth, converting the mechanical energy of sound waves into electrical signals. The electrical signals are then amplified and converted back into sound waves.
5. Why does a curved surface show better reflection of sound than a straight surface?
Answer:
In case of a curved surface the direction of the normal is different at every point. Hence, the direction of the reflected ray will also be different at each point. Hence, the reflected sound will reach in all directions.
6. How do longitudinal waves propagate compared to transverse waves?
Answer:
Longitudinal waves propagate by causing particles of the medium to oscillate back and forth in the direction parallel to the wave’s motion. In contrast, transverse waves cause particles to oscillate perpendicular to the direction of the wave’s motion.
7. The source of a sound is 10 m away from a wall and a person stands halfway in between. If the speed of sound in air is 346 m/s, will the person hear the echo?
Answer:
Distance to be covered by sound = 10 m + 5 m = 15 m.
Speed of sound = 346 m/s.
Time taken = Distance/ Speed = 15/346 = 0.14 seconds.
To hear a distinct echo the time interval between the original and the reflected one must be 0.1 seconds.
Since 0.14 > 0.1, the person will hear the echo.
8. Why do musical instruments such as trumpets and shehanais have distinctive shapes?
Answer:
In these instruments, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience. Thus, sound is not spread in all directions and the target audience can clearly hear the sound.
9. How can you detect a tumour inside a patient’s body using ultrasound?
Answer:
Ultrasonic sound waves are sent through the tissues of the body and get reflected from a region where there is a change of tissue density which happens when a tumour is present. These waves are then converted into electrical signals that are used to generate images of the organ. These images are then displayed on a monitor or printed on a film. This technique is called ‘ultrasonography’.
10. Why do animals behave strangely before an earthquake?
Answer:
Certain animals can hear the low-frequency infrasound produced by earthquakes before the main shockwaves begin. Thus, they get disturbed.
Fill in the Blanks:
(a) Sound waves are ____________ waves.
(b) Earthquakes produce __________ before the main shockwaves begin.
(c) Sound of long ___________ cannot be used in the detection of cracks.
(d) In a sound wave rarefaction has _________ density and compression has _________ density.
(e) Rhinoceroses communicate using infrasound of frequency as low as _________.
Answers:
(a) Sound waves are longitudinal waves.
(b) Earthquakes produce infrasound before the main shockwaves begin.
(c) Sound of long wavelengths cannot be used in the detection of cracks.
(d) In a sound wave rarefaction has low density and compression has high density.
(e) Rhinoceroses communicate using infrasound of frequency as low as 5 Hz.
Match and Pair:
| Column A | Column B |
| (i) Ultrasound | (a) Single frequency |
| (ii) Loudness | (b) Low frequency |
| (iii) Tone | (c) Crack detection |
| (iv) Note | (d) Amplitude |
| (v) Infrasound | (e) Multiple frequencies |
Answer:
| Column A | Column B |
| (i) Ultrasound | (c) Crack detection |
| (ii) Loudness | (d) Amplitude |
| (iii) Tone | (a) Single frequency |
| (iv) Note | (e) Multiple frequencies |
| (v) Infrasound | (b) Low frequency |
++++++++++++++
Frequently Asked Questions (FAQs) on NCERT Solutions to Class 9 Science Chapter 11 Sound
We have provided you with the ultimate solution hub for all your study needs. By answering all in-text questions, exercise questions and activities, we’ve compiled everything you need in one convenient place. Plus, we’ve sweetened the deal with a bonus set of extra practice problems to supercharge your learning. Also, feel free to download our PDF solutions for free, anytime you need them!
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The following topics are covered:
11.1 Production of Sound
11.2 Propagation of Sound
11.3 Reflection of Sound
11.4 Range of Hearing
11.5 Applications of Ultrasound
Here are the number of questions for the chapter:
(i) 12 Theoretical Questions (Questions 1 – 5, 9 – 11, 14 – 17)
(ii) 5 Numerical-based Questions (Questions 6 – 8, 12, 13)
It is vital that you understand the fundamental concepts of production and propagation of sound via alternate compressions and rarefactions. Also, learn what the frequency, amplitude and speed of a sound wave are because you will see both conceptual questions and numericals from these parts in your exams. Reflection of sound is another important concept and you will see theoretical questions and numericals from this part in your exams.
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