SOUND - Q&A

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 do not travel all the way from the vibrating object to the ear. A particle of the medium in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle. As a result of this, the adjacent particle gets displaced from its position of rest. After displacing the adjacent particle, the first particle comes back to its original position. This process continues in the medium till the sound reaches your ear.

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

Answer: When the school bell is struck with a hammer, it starts vibrating. These vibrations force the adjacent air particles to vibrate. The forward motion of the bell pushes the air in front of it, creating a region of high pressure called compression. The backward motion creates a region of low pressure called rarefaction. As the bell continues to vibrate forward and backward, a series of compressions and rarefactions is created in the air, which propagates as a sound wave.

3. Why are sound waves called mechanical waves?

Answer: Sound waves are called mechanical waves because they require a material medium (like air, water, or steel) for their propagation. They cannot travel through a vacuum.

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

Answer: No, you will not be able to hear any sound produced by your friend on the moon. This is because there is no atmosphere (medium) on the moon, and sound needs a material medium to propagate.

Questions on Page 132

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

Answer:
(a) Loudness: It is determined by the amplitude of the sound wave. Larger amplitude results in a louder sound.
(b) Pitch: It is determined by the frequency of the sound wave. Higher frequency results in a higher pitch (shriller sound).

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

Answer: The guitar has a higher pitch. This is because the frequency of vibrations produced by a guitar string is typically higher than that of a car horn.

Questions on Page 133

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

Answer:
- Wavelength (λ): The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength. Its SI unit is metre (m).
- Frequency (ν): The number of complete oscillations (or compressions and rarefactions) per unit time is called the frequency. Its SI unit is Hertz (Hz).
- Time Period (T): The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period. Its SI unit is second (s).
- Amplitude (A): The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave.

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

Answer: The relationship between speed (v), frequency (ν), and wavelength (λ) is given by:
Speed = Wavelength × Frequency
v = λ × 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:
Given:
Frequency (ν) = 220 Hz
Speed (v) = 440 m/s
Formula: v = λ × ν
λ = v / ν
λ = 440 / 220
λ = 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 equal to the time period (T) of the wave.
Given:
Frequency (ν) = 500 Hz
Formula: T = 1 / ν
T = 1 / 500
T = 0.002 s
(Note: The distance of 450 m is not needed to find the time period).

Questions on Page 134

1. Distinguish between loudness and intensity of sound.

Answer:
- Loudness: It is a physiological response of the ear to the intensity of sound. It depends on the sensitivity of the ear. Even if two sounds have equal intensity, one might sound louder than the other to a person because their ear detects it better.
- Intensity: It is the amount of sound energy passing each second through a unit area. It is an objective physical quantity and does not depend on the sensitivity of the ear.

Questions on Page 135

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

Answer: Sound travels fastest in iron (solids), slower in water (liquids), and slowest in air (gases).

Questions on Page 136

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:
Given:
Time for echo (t) = 3 s
Speed of sound (v) = 342 m s^-1
Distance traveled by sound = v × t = 342 × 3 = 1026 m
Since the sound travels to the reflecting surface and comes back, the total distance is 2d.
2d = 1026 m
d = 1026 / 2
d = 513 m
The distance of the reflecting surface is 513 m.

2. Why are the ceilings of concert halls curved?

Answer: The ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall. The curved surface acts like a concave mirror, reflecting sound waves towards the audience, ensuring uniform distribution of sound.

Exercises

1. What is sound and how is it produced?

Answer: Sound is a form of energy which produces a sensation of hearing in our ears. It is produced by vibrating objects. When an object vibrates, it sets the particles of the medium around it into vibration, creating sound waves.

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

Answer:
[Image of compressions and rarefactions in a sound wave]
When a vibrating object (like a tuning fork prong) moves forward, it pushes and compresses the air in front of it, creating a region of high pressure called compression (C). When the vibrating object moves backward, it creates a region of low pressure called rarefaction (R). As the object vibrates rapidly, a series of compressions and rarefactions is created in the air, which propagates as a sound wave.

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

Answer:
Bell Jar Experiment:
1. Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the bell jar.
2. Connect the bell jar to a vacuum pump.
3. Switch on the bell. You can hear its sound.
4. Now start the vacuum pump. As the air is pumped out, the sound becomes fainter, although the same current is passing through the bell.
5. When very little air is left, the sound becomes very feeble.
6. If the air is completely removed, no sound is heard.
This proves that sound needs a material medium for propagation.

4. Why is sound wave called a longitudinal wave?

Answer: A sound wave is called a longitudinal wave because the particles of the medium vibrate back and forth parallel to the direction of propagation of the wave.

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

Answer: The characteristic is Quality or Timbre. It 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?

Answer: This happens because the speed of light (3 × 10⁸ m/s) is much greater than the speed of sound (344 m/s in air). Light travels almost instantly to us, while sound takes some time to cover the distance.

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^-1.

Answer:
Given speed of sound (v) = 344 m s^-1.
Formula: λ = v / ν

(i) For frequency ν₁ = 20 Hz:
λ₁ = 344 / 20 = 17.2 m

(ii) For frequency ν₂ = 20 kHz = 20000 Hz:
λ₂ = 344 / 20000 = 0.0172 m

The typical wavelengths are 17.2 m and 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 air and in aluminium to reach the second child.

Answer:
Let the length of the rod be d.
Speed of sound in air (v_air) ≈ 346 m/s (at 25°C).
Speed of sound in aluminium (v_Al) ≈ 6420 m/s (at 25°C).

Time taken in air (t_air) = d / v_air
Time taken in aluminium (t_Al) = d / v_Al

Ratio = t_air / t_Al = (d / v_air) / (d / v_Al) = v_Al / v_air
Ratio = 6420 / 346
Ratio ≈ 18.55 : 1

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

Answer:
Frequency = 100 Hz = 100 vibrations per second.
Time = 1 minute = 60 seconds.
Total vibrations = Frequency × Time
Total vibrations = 100 × 60 = 6000 times.

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

Answer: Yes, sound follows the same laws of reflection as light does.
1. The angle of incidence of sound is equal to the angle of reflection.
2. The incident sound wave, the reflected sound wave, and the normal at the point of incidence 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 remains the same. Do you hear echo sound on a hotter day?

Answer: The sensation of sound persists in our brain for about 0.1 s. To hear a distinct echo, the time interval between the original sound and the reflected sound must be at least 0.1 s.
Speed of sound increases with temperature. On a hotter day, the speed of sound is higher. Therefore, the sound will travel faster and return sooner. If the time taken is less than 0.1 s, we will not hear a distinct echo. If it is still more than 0.1 s, we will hear it. Generally, if the distance is just enough for an echo on a cold day, the echo might disappear on a hotter day.

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

Answer:
1. Megaphones or Loudhailers: These are designed to send sound in a particular direction without spreading it in all directions, using multiple reflections.
2. Stethoscope: In a stethoscope, the sound of the patient's heartbeat reaches the doctor's ears by multiple reflections 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 = 10 m s^-2 and speed of sound = 340 m s^-1.

Answer:
Step 1: Time taken by stone to reach the pond (t₁)
s = ut + ½gt²
500 = 0 × t₁ + ½ × 10 × t₁²
500 = 5 t₁²
t₁² = 100
t₁ = 10 s

Step 2: Time taken by sound to travel up (t₂)
t₂ = Distance / Speed of sound
t₂ = 500 / 340
t₂ ≈ 1.47 s

Total Time:
Total time = t₁ + t₂ = 10 + 1.47 = 11.47 s

14. 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:
Given:
Speed (v) = 339 m s^-1
Wavelength (λ) = 1.5 cm = 0.015 m
Frequency (ν) = v / λ
ν = 339 / 0.015
ν = 22600 Hz

Audibility: The audible range for humans is 20 Hz to 20,000 Hz. Since 22,600 Hz is greater than 20,000 Hz, it will not be audible (it is ultrasonic).

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

Answer:
Reverberation: The persistence of sound in a big hall due to repeated reflection from the walls, ceiling, and floor is called reverberation.
How to reduce it: It can be reduced by covering the roof and walls of the hall with sound-absorbent materials like compressed fibreboard, rough plaster, or draperies. The seat materials are also selected for their sound-absorbing properties.

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

Answer:
Loudness: It is the measure of the response of the ear to the sound. It distinguishes a loud sound from a faint one.
Factors: It depends on the amplitude of vibrations. If the amplitude is large, the sound is loud. If the amplitude is small, the sound is feeble.

17. Explain how bats use ultrasound to catch a prey.

Answer: Bats emit high-pitched ultrasonic squeaks while flying. These squeaks get reflected by objects or prey (like insects) in their path and return to the bat's ear. By detecting these reflected waves, the bat can determine the distance, direction, and nature of the prey, allowing it to catch the prey even in the dark.

18. How is ultrasound used for cleaning?

Answer: Objects to be cleaned are placed in a cleaning solution, and ultrasonic waves are passed through the solution. The high frequency of ultrasound causes the dust, grease, and dirt particles to detach and drop out. Thus, the object gets thoroughly cleaned. This is used for spiral tubes, odd-shaped parts, electronic components, etc.

19. Explain the working and application of a sonar.

Answer:
Working: SONAR (Sound Navigation And Ranging) consists of a transmitter and a detector installed in a ship. The transmitter produces and transmits ultrasonic waves. These waves travel through water, strike underwater objects, and get reflected back. The detector senses the reflected waves and converts them into electrical signals.
The distance (d) of the object is calculated using the formula 2d = v × t, where v is speed of sound in water and t is the time interval between transmission and reception.
Application: It is used to measure the depth of the sea and to locate underwater hills, valleys, submarines, icebergs, and sunken ships.

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.

Answer:
Given:
Total time (t) = 5 s
Distance (d) = 3625 m
Total distance traveled by sound = 2d = 2 × 3625 = 7250 m
Speed (v) = Total Distance / Time
v = 7250 / 5
v = 1450 m s^-1

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

Answer: Ultrasound 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 or crack, the ultrasound gets reflected back indicating the presence of the flaw or defect.

22. Explain how the human ear works.

Answer:
- The outer ear (pinna) collects sound from the surroundings.
- The collected sound passes through the auditory canal.
- At the end of the auditory canal is a thin membrane called the eardrum or tympanic membrane. The compressions and rarefactions of sound waves make the eardrum vibrate.
- These vibrations are amplified several times by three bones (hammer, anvil, and stirrup) in the middle ear.
- The middle ear transmits these vibrations to the inner ear (cochlea).
- In the inner ear, the vibrations are converted into electrical signals, which are sent to the brain via the auditory nerve. The brain interprets them as sound.