NCERT Solutions Class 9 Science Chapter 12 Sound

Given:

Frequency, \( \nu = 220 \text{ Hz} \)
Speed of sound,\( v = 440 \text{ m/s} \)

We know:

\( ⇒ \ v = \lambda \cdot \nu \)
\( ⇒ 440 = \lambda \cdot 220 \)
\( ⇒ \lambda = \frac{440}{220} = 2 \text{ m} \)

So, the wavelength of the sound wave is 2 metres.

6. 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 equal to the time period \( T \) of the sound wave. Therefore, it can be calculated as follows:

Given \( f = 500 \, \text{Hz} \)

\( T = \frac{1}{500} = 0.002 \text{seconds} \)

So, the time interval between successive compressions is 0.002 s.

7. Distinguish between loudness and intensity of sound.

Answer:

NCERT Textbook for Class 9 Science – Page 167

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 iron as compared to water and air at a particular temperature.

NCERT Textbook for Class 9 Science – Page 168

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?

Solution:

Given:

Speed of sound, \( v = 342 \, \text{m/s} \)
Time, \( t = 3 \, \text{s} \)

Distance traveled by sound:

\( \text{Distance} = v \times t = 342 \times 3 = 1026 \, \text{m}\)

Since the echo travels to the reflecting surface and back. So, sound must travel twice the distance between the reflecting surface and the source.

\( \text{Distance to surface} = \frac{1026}{2} = 513 \, \text{m} \)

So, the reflecting surface is 513 metres away from the source.

NCERT Textbook for Class 9 Science – Page 169

1. Why are the ceilings of concert halls curved?

Answer:

The ceilings of concert halls are curved so that sound waves reflect and spread evenly in all directions. This helps the sound reach every part of the hall clearly, allowing the audience to hear properly from all seats.

NCERT Textbook for Class 9 Science – Page 170

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

Answer:

The audible range of the average human ear is from 20 Hz to 20,000 Hz. Sounds with frequencies below 20 Hz or above 20,000 Hz cannot be heard by humans.

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

Answer:

(a) Infrasound are frequencies less than 20 Hz.
(b) Ultrasonic sounds have frequencies greater than 20,000 Hertz.

NCERT Textbook for Class 9 Science – Page 172

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?

Solution:

Given:

Speed of sound,\( v = 1531 \,\text{m/s}, \)
Time taken, \( t = 1.02 \,\text{s} \)

Total distance travelled by the sonar pulse:

\( \text{Distance} = v \times t = 1531 \times 1.02 = 1561.62 \,\text{m} \)

Since the pulse travels to the cliff and back, the actual distance of the cliff from the submarine is:  \( \frac{\text{Total distance travelled by sonar pulse}}{2} \)

\(⇒ \frac{1561.62}{2} = 780.81 \,\text{m} \)

So, the underwater cliff is 780.81 metres away from the submarine.

NCERT Textbook for Class 9 Science – Page 174

1. What is sound, and how is it produced?

Answer:

Sound is a form of energy that produces the sensation of hearing. It is produced when an object vibrates. These vibrations cause the adjacent particles of the medium to vibrate. This creates a disturbance that travels through the medium in the form of sound waves. When these waves reach our ears, we hear sound.

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

Answer:

Sound is produced when objects vibrate. When a vibrating object moves forward, it pushes the air particles ahead, creating an area of high pressure called compression (C), as shown in the above Figure. This compression travels away from the source. When the vibrating object moves backwards, it leaves behind a region of low pressure called rarefaction (R), as shown in the above Figure. Continuous back-and-forth vibration produces a series of compressions and rarefactions in the air. These make the sound wave that propagates through the medium. Compressions are high-pressure regions, while rarefactions are low-pressure regions. The pressure depends on the number of particles in a given volume. More density of the particles in the medium gives more pressure and lower density produces lower pressure. Thus, propagation of sound can be seen as the movement of pressure or density variations through the medium.

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

Answer:

To show that sound needs a material medium to travel, an electric bell experiment can be performed. An electric bell is suspended inside an airtight glass bell jar connected to a vacuum pump (as shown in the figure below).

When the switch is turned on, the bell rings and the sound is clearly heard as air is present inside the jar. Now, air is gradually removed using the vacuum pump. As the air is pumped out, the sound of the bell becomes weaker, although the bell can still be seen vibrating. When almost all the air is removed and a vacuum is created inside the jar, no sound is heard at all. This experiment proves that sound cannot travel through a vacuum and requires a material medium for its propagation.

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

Answer:

In sound waves, the individual particles of the medium move in a direction parallel to the direction in which the sound wave travels. The particles do not move from one place to another but they simply oscillate about their mean positions. Hence, sound waves are longitudinal waves.

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

Answer:

The characteristic of sound that helps us identify a person by their voice is its quality or timber. Even if two sounds have the same pitch and loudness, they differ in quality, which allows us to distinguish one person’s voice from another in a dark room.

6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Answer:
The speed of light is about \( 3 \times 10^8 \) m/s, while the speed of sound in air is only about 344 m/s. So, light travels much faster than the sound. That is the reason why we hear the thunder a few seconds after the flash of thunder is seen, though flash and thunder occur simultaneously.

7. 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.

Solution:

Given:
Speed of sound in air, v= 344 m/s

For sound waves, Speed = Wavelength × frequency

\( \ v = \lambda \cdot \nu \)

(a) For frequency \( \nu_1 = 20 \,\text{Hz} \)

\( ∴\lambda_1 = \frac{v}{\nu_1} = \frac{344}{20} = 17.2 \,\text{m} \)

(b) For frequency \( f_2 = 20{,}000 \,\text{Hz} \)

\( ∴\lambda_2 = \frac{v}{\nu_2} = \frac{344}{20{,}000} = 0.0172 \,\text{m} \)​=0.0172m

Hence, humans have the wavelength range for hearing from 17.2 m to 0.0172 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 times taken by the sound wave in the air and in aluminium to reach the second child.

Solution:

Let the length of the aluminium rod be, \(d\)

Speed of sound in aluminium, \(v_{\text{Al}} = 6420 \,\text{m/s}\)

Speed of sound in air, \(v_{\text{air}} = 346 \,\text{m/s}\)

Time taken by sound to travel through aluminium:

\( t_{\text{Al}} = \frac{d}{v_{\text{Al}}} = \frac{d}{6420} \)​

Time taken by sound to travel through air:

\( t_{\text{air}} = \frac{d}{346} \)​

Required ratio of times (air : aluminium):

\( \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{d/346}{d/6420} = \frac{6420}{346} \approx 18.55 \)

So, the ratio of time taken by sound in air to that in aluminium is 18.55 : 1.

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

Solution:

Given:
Frequency of sound, \(\nu = 100\,\text{Hz}\)
Time in one minute, \(t = 60\,\text{s}\)

We know that: Frequency is defined as the number of oscillations per second. It is given by the relation: Number of oscillations = Frequency × Total time

\( \text{Number of oscillations} = \nu \times t = 100 \times 60 = 6000 \)

So, the source vibrates 6000 times in one minute.

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

Answer:

Yes, sound follows the same laws of reflection as light.

1. The angle of incidence of a sound wave is equal to the angle of reflection with respect to the normal at the point of incidence.
2. The incident sound wave, the reflected sound wave, and the normal all lie in the same plane.

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?

Answer:

An echo is heard only if the time interval between the original sound and its reflection is at least 0.1 seconds. The speed of sound increases with temperature, so on a hotter day, sound travels faster. As a result, the time interval between the original sound and the reflected sound decreases. If this interval becomes greater than 0.1 seconds, the echo can be heard.

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

Answer:

Two practical applications of the reflection of sound waves are:

1. SONAR: Reflection of sound is used to measure the speed and distance of underwater objects. This technique is called SONAR.
2. Stethoscope: In a stethoscope, the sound of a patient’s heartbeat reaches the doctor’s ear through multiple reflections of sound.

13. A stone 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? Giving, g = 10 m/s² and speed of sound = 340 m/s.

Answer:

Given:

• Height of tower, \(s = 500 \,\text{m}\)
• Acceleration due to gravity, \(g = 10 \,\text{m/s²}\)
• Speed of sound, \(v = 340 \,\text{m/s}\)
• Initial velocity of stone, \(u = 0\)

Let the time taken by the stone to reach the water be, \((t_1)\)

According to the second equation of motion:

\(s = ut_1 + \frac{1}{2} g t_1^2\)

\(⇒500 = 0 + \frac{1}{2} \times 10 \times t_1^2\)
\(⇒500 = 5 t_1^2\)
\(⇒t_1^2 = 100 \implies t_1 = 10 \,\text{s}\)

Now, the time taken by sound to travel back to the top will be, \((t_2)\)

\(t_2 = \frac{\text{Distance}}{\text{Speed of sound}} = \frac{500}{340} \approx 1.47 \,\text{s}\)

Total time to hear the splash:

\(t = t_1 + t_2 = 10 + 1.47 \approx 11.47 \,\text{s}\)

Hence, the splash is heard at the top after 11.47 seconds.

14. A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer:

Given:
Speed of sound, \(v = 339 \,\text{m/s}\)
Wavelength of sound, \(\lambda = 1.5 \,\text{cm} = 0.015\,\text{m}\)

We know that, Speed of sound = Wavelength × Frequency

\(v = \lambda\times\nu\)
\(⇒\nu = \frac{v}{\lambda} = \frac{339}{0.015} = 22{,}600 \,\text{Hz}\)

The audible range for the human ear is from 20 Hz to 20,000 Hz. Since the frequency of the given sound wave is 22,600 Hz, therefore, it is not audible.

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

Answer:

Reverberation is the persistence of sound in an enclosed space due to repeated reflections from walls, ceilings, and floors of a big hall or room.

Reverberation can be reduced by using sound-absorbing materials. Walls and ceilings are covered with materials like compressed fibreboard, rough plaster or draperies. Seat materials are selected based on their sound absorbing properties.

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

Answer:

Loudness of sound is the measure of how loud or soft a sound is as perceived by the human ear. It is a physiological response and helps us distinguish between faint and loud sounds.

Loudness mainly depends on the amplitude of vibrations. A sound with greater amplitude has more energy and is heard as louder. Loudness is proportional to the square of the amplitude of vibration.

17. Explain how bats use ultrasound to catch prey.

Answer:

Bats search out their prey by by emitting and detecting reflections of ultrasonic waves. These sounds travel through the air and strike the obstacles or prey, then bounce back to the bat’s ears. By analyzing the reflected sound waves, the bat can identify the position, distance, and nature of the prey or obstacle.

18. How is ultrasound used for cleaning?

Answer:

The objects to be cleaned are first placed in a cleaning solution and then the ultrasonic
waves are passed through that solution. The high-frequency ultrasonic waves detach and remove dust, grease, and dirt from the objects. As a result, the objects get thoroughly cleaned, even in areas that are difficult to reach.

19. Explain the working and application of a sonar.

Answer:

Working of SONAR:
SONAR stands for Sound Navigation and Ranging. SONAR consists of a transmitter and a detector and is installed in a boat or a ship. The transmitter produces and transmits (sends) ultrasonic waves into the water. These waves travel through seawater and hit an object on the seabed. After striking the object, the waves are reflected back and received by the detector. The detector converts these ultrasonic waves into electrical signals, which are then analysed. The distance of the object can be calculated by knowing the speed of sound in water and the time taken for the sound to go and return. If the time taken is t and the speed of sound in seawater is v, then the total distance travelled by the sound waves is 2d = v × t.

This method is called echo-ranging.

Applications of SONAR:

The sonar technique is used to measure the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.

20. 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:

Given:

• Time taken to hear the echo, \(t = 5 \, \text{s}\)
• Distance of object from submarine, \(d = 3625 \, \text{m}\)

We know that the total distance travelled by the sound wave is twice the distance to the object: $$2d$$

Speed of sound in water: \(v = \frac{\text{Total distance}}{\text{Time}} = \frac{2d}{t}\)​

\( ⇒ v = \frac{2 \times 3625}{5} = \frac{7250}{5} = 1450 \, \text{m/s}\)

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

Ultrasound can be used to detect cracks or flaws in metal blocks. These metal blocks are often part of structures like buildings, bridges, machines, or scientific instruments. Any internal cracks or holes, which cannot be seen from the outside, weaken the structure. When ultrasonic waves are sent through the metallic block, a detector on the other side monitors the waves that pass through. If there is even a small defect, the ultrasound get reflected back instead of passing through, indicating the presence of the flaw or defect, as shown in the following Figure.

22. Explain how the human ear works.
Solution:

The outer part of the ear is called the ‘pinna’. It collects sounds from the surroundings and passes through the auditory canal. At the end of the auditory canal, there is a thin membrane called the eardrum. When a compression of sound reaches the eardrum, it moves the eardrum inward, and when a rarefaction arrives, it moves outward. In this way the eardrum vibrates. These vibrations are increased in strength by three bones in the middle ear—the hammer, anvil, and stirrup. The middle ear then passes these stronger vibrations to the inner ear. In the inner ear, the vibrations are converted into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve. The brain interprets these signals as sound.

NCERT Solutions for Class 9 Science Chapter 1 Matter In Our Surroundings

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure?

NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules

NCERT Solutions for Class 9 Science Chapter 4 Structure Of The Atom

NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit Of Life

NCERT Solutions for Class 9 Science Chapter 6 Tissues

NCERT Solutions for Class 9 Science Diversity In Living Organism

NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science Chapter 9 Force And Laws Of Motion

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

NCERT Solutions for Class 9 Science Chapter 11 Work And Energy

NCERT Solutions for Class 9 Science Chapter 12 Sound

NCERT Solutions for Class 9 Science Chapter 13 Why Do We Fall ill

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

NCERT Solutions for Class 9 Science Chapter 15 Improvement In Food Resources