Chapter 7 – Sound | Chapter Solution Class 9

Publisher	: Santra publication pvt. ltd.
Book Name	: Madhyamik Physical Science And Environment
Class	: 9 (Madhyamik)
Subject	: Physical Science
Chapter Name	: Sound

Multiple Choice Questions (MCQ)

[Each of Mark-1]

Question 1

Sound can not travel through

1. air
2. water
3. steel
4. vacuum

Answer

vacuum

Explanation

Sound can travel through air, steel and water, but not through vacuum, as they lack the necessary particles for sound transmission.

Question 2

Which frequencies are not clearly audible to the human ear?

1. 30 Hz
2. 3000 Hz
3. 300 Hz
4. 30,000 Hz

Answer

30,000 Hz

Explanation

Frequencies below 20 Hz (infrasound) and above 20,000 Hz (ultrasound) are generally not clearly audible to the human ear.

Question 3

The frequency of sound waves can be expressed in

1. Hz
2. dB
3. m
4. m/s

Answer

Hertz (Hz)

Explanation

The frequency of sound waves can be expressed in Hertz (Hz), which represents the number of vibrations per second.

Question 4

Frequency (n) and time period (T) are related as

1. n\over t = 1
2. n × T = 1
3. n = T²
4. nT² = 1

Answer

n × T = 1

Explanation

Frequency (n) multiplied by Time period (T) equals 1, which can be written as n × T = 1.

Question 5

The audible range of frequency in Hz is

1. 20 to 20,000
2. 2000 to 20,000
3. 40 to 4000
4. 200 to 2000

Answer

20 to 20,000 Hz

Explanation

The audible range of frequency in Hz is generally considered to be from 20 to 20,000 Hz for the average human ear.

Question 6

Earthquake produces

1. audible sound
2. ultrasound
3. infrasonic sound
4. none of these

Answer

infrasonic sound

Explanation

Earthquakes can produce audible sound in the form of low-frequency rumbles known as infrasonic sound, which may be felt and heard by humans.

Question 7

In SONAR, we use

1. radio waves
2. ultrasonic waves
3. infrasonic waves
4. audible sound waves

Answer

ultrasonic waves

Explanation

In SONAR (Sound Navigation and Ranging), we use ultrasonic waves, which are high-frequency sound waves beyond the range of human hearing.

Question 8

Due to the increase in the frequency of the sounding source, its pitch

1. increases
2. decreases
3. remains the same
4. first increases and then decreases

Answer

increases

Explanation

Due to the increase in the frequency of the sounding source, its pitch increases. Higher frequency corresponds to higher pitch perception.

Question 9

Stethoscope works on the principle of

1. reflection of sound
2. refraction of sound
3. reverberation of sound
4. none of these

Answer

reflection of sound

Explanation

Stethoscope works on the principle of sound conduction through solid mediums, particularly the reflection of sound waves for auscultation (listening) purposes.

Question 10

The relation between velocity (v) of sound, wavelength (λ) and frequency (n) is

1. n = v\over λ
2. λ = vn
3. n = vλ
4. n = v^2\over λ

Answer

n = v/λ

Explanation

The correct relationship is n = v/λ, where n represents frequency, v represents velocity, and λ represents wavelength.

Answer in one word or in one sentence

[Each of Mark-1]

Question

1. What is the audible range of frequency?
2. Name the wave whose frequency is less than 20 Hz.
3. Name the wave which helps a bat to catch its prey.
4. What does SONAR stand for?
5. Between frequency and pitch of the sound, which one is a physical quantity?
6. On what factor does the quality of a musical sound depend?
7. On which property does the working principle of a stethoscope depend ?
8. Which characteristics of musical sound are responsible for noise pollution?
9. How long does transient sound persist in our brain? 
10. Unit of which physical quantity is a decibel ?

Answer

1. 20 to 20,000 Hz
2. Infrasound
3. Ultrasound
4. Sound Navigation and Ranging
5. Frequency
6. Timbre
7. A stethoscope utilizes the principle of multiple reflections of sound.
8. Loudness and intensity
9. The sensation of sound persists in our brain for about 0.1 s.
10. Sound level or intensity

Short answer type questions

[Each of Mark-2]

Question 1

What do you mean by longitudinal and transverse waves?

Answer

Longitudinal waves are waves in which the particles of the medium vibrate parallel to the direction of wave propagation. Examples of longitudinal waves include sound waves, pressure waves in fluids, and compressional waves in solids.

Transverse waves are waves in which the particles of the medium vibrate perpendicular to the direction of wave propagation. Examples of transverse waves include light waves, water waves, and seismic shear waves.

Question 2

Define frequency and wavelength of a wave.

Answer

Frequency refers to the number of oscillations or cycles of a wave that occur in one second, measured in Hertz (Hz).

Wavelength refers to the distance between two consecutive points in a wave that are in the same phase, such as crest to crest or trough to trough. It is typically measured in meters (m).

Question 3

Establish the relation between frequency, wavelength and wave velocity.

Answer

The relation between frequency (f), wavelength (λ), and wave velocity (v) is given by the equation:

v = f * λ

where v represents the wave velocity, f represents the frequency, and λ represents the wavelength.

Question 4

Define time-period. What is its relation with frequency?

Answer

The time period refers to the time taken for one complete cycle or oscillation of a wave.

The relation between time period (T) and frequency (f) is:

T = 1/f

where T is the time period and f is the frequency. In other words, the time period is the reciprocal of the frequency.

Question 5

What is an echo? Mention one of its uses.

Answer

An echo is the reflection of sound waves of a surface back to the listener. One of its uses is determining the distance or location of objects using echolocation.

Question 6

What are the characteristics of sound? Of them which one depends on frequency?

Answer

The main characteristics of sound are:

• Pitch: The characteristic of a musical sound by which a sharp or a shrill sound is distinguished from a dull, flat sound of the same intensity is known as its pitch.
• Loudness:  Loudness is the property by which a loud sound can be distinguished from a faint one, both having the same pitch.
• Timbre: The quality of a sound that distinguishes it from other sounds of the same pitch and loudness. Timbre is determined by the waveform of the sound wave.

Of these characteristics, pitch is the only one that depends on frequency. The higher the frequency of the sound wave, the higher the pitch of the sound.

Question 7

What is reverberation? Give an example.

Answer

Reverberation refers to the persistence of sound in an enclosed space due to multiple reflections. An example is the echo heard in a large empty hall after clapping or shouting.

Question 8

What is ultrasonic waves? What is ultrasonography? 

Answer

Ultrasonic waves are sound waves with frequencies higher than the upper limit of human hearing (typically above 20,000 Hz). They are used in various applications such as imaging, cleaning, and measuring distances.

Ultrasonography, also known as ultrasound imaging, is a medical imaging technique that uses ultrasonic waves to visualize internal body structures. It is commonly used for diagnostic purposes, such as examining organs, tissues, and pregnancies.

Question 9

Mention the use of ultrasonic waves for the echolocation of bats and whales.

Answer

Bats: Bats emit ultrasonic pulses from their mouths or noses. The pulses bounce off objects in the environment and the echoes return to the bat’s ears. The bat’s brain then analyzes the echoes to determine the distance, size, and location of the objects.

Whales: Whales emit ultrasonic pulses from their nasal passages. The pulses bounce off objects in the water and the echoes return to the whale’s ears. The whale’s brain then analyzes the echoes to determine the distance, size, and location of the objects.

Question 10

How ultrasonic waves are used to kill microbes and bacteria?

Answer

Ultrasonic waves can be used to kill microbes and bacteria through a process called ultrasonic sterilization. The high-frequency vibrations disrupt their cellular structures, leading to their destruction or inactivation.

Question 11

Mention the factors on which the intensity of sound depends.

Answer

The intensity of sound depends on factors such as the amplitude of the sound wave, the distance from the sound source, and the medium through which the sound propagates.

Question 12

What is sound pollution? Mention one of its adverse effects.

Answer

The presence of loud, unwanted and disturbing sound in our environment is called noise pollution.

One of its adverse effects is the potential to cause hearing damage or hearing loss over time, especially when exposed to loud or prolonged noise.

Question 13

Mention three sources of sound pollution.

Answer

Three sources of sound pollution are:

1. Traffic noise
2. Industrial machinery and equipment
3. Construction activities

Question 14

Mention two ways of controlling sound pollution.

Answer

Two ways of controlling sound pollution are:

1. Implementing noise barriers or soundproofing measures.
2. Avoid Bursting of cracker.

Long answer type questions

[Each of Mark-2/3]

Question 1

What is sound? Prove by an experiment that the vibration of a source is required to produce sound.

Answer

Question 2

Prove by an experiment that a material medium is necessary for the propagation of sound.

Answer

To prove that a material medium is necessary for the propagation of sound, we can use the following experiment with a bell jar:

Question 3

What is a wave? Define the amplitude of a wave. How sound wave is generated in our vocal cords?

Answer

A wave is a disturbance that travels through space and time. It is characterized by a periodic oscillation of a quantity, such as pressure, electric field strength, or displacement of a medium.

The amplitude of a wave is the maximum displacement of the medium from its equilibrium position. It is a measure of the strength of the wave.

Sound waves are generated in our vocal cords when air from our lungs is pushed through the narrow gap between the vocal cords. The tension and shape of the vocal cords can be adjusted by muscles to produce different pitches and tones, creating sound waves that travel through the air as speech or singing.

Question 4

What do you mean by an echo? What should be the minimum distance of the reflector to hear the echo of a transient sound? Mention some practical applications of echo.

Answer

An echo is a reflection of a sound wave that is perceived as a distinct repetition of the original sound.

The minimum distance of the reflector to hear the echo of a transient sound is approximately 17 meters in air.

Some practical applications of echo include:

• Echolocation: Bats and whales use echolocation to navigate and find prey in darkness or murky water. They emit ultrasonic pulses and listen to the echoes to create a mental map of their surroundings.
• Sonar: Sonar is a system that uses sound waves to detect and locate objects underwater. It is used by ships and submarines to navigate and avoid obstacles.

Question 5

How would you determine the depth of a sea with the help of SONAR technology?

Answer

SONAR (Sound Navigation and Ranging) technology is used to determine the depth of a sea or water body by measuring the time it takes for sound waves to travel from a transmitter to the seafloor and back to a receiver. Here’s a simplified explanation of how this works:

1. Transmitting Sound Waves: A SONAR system sends out a pulse of sound, usually in the form of a high-frequency sound wave, into the water.
2. Sound Wave Propagation: The sound wave travels through the water at a constant speed, which is roughly 1,500 meters per second (about 1.5 kilometres per second) in seawater. The speed of sound in water can vary slightly depending on factors like water temperature and salinity.
3. Reflection from Seafloor: When the sound wave reaches the seafloor, it reflects off it because of the density difference between water and the seafloor.
4. Receiving the Echo: A receiver in the SONAR system detects the reflected sound wave or echo when it returns.
5. Calculating Depth: The time it takes for the sound pulse to travel to the seafloor and back to the receiver is recorded. Since you know the speed of sound in water and the time it took for the echo to return, you can use a simple formula: Depth (in meters) = (Speed of Sound in Water × Time) / 2.

Question 6

What are the characteristics of sound? On what factors do they depend?

Answer

The main characteristics of sound are:

• Pitch: The characteristic of a musical sound by which a sharp or a shrill sound is distinguished from a dull, flat sound of the same intensity is known as its pitch.
• Loudness:  Loudness is the property by which a loud sound can be distinguished from a faint one, both having the same pitch.
• Timbre: The quality of a sound that distinguishes it from other sounds of the same pitch and loudness. Timbre is determined by the waveform of the sound wave.

Question 7

What is sound pollution? Discuss the hazardous effect of it on public health. State possible remedial measures.

Answer

Numerical problems

Question 1

If the frequency of a sound wave is 250 Hz and the wavelength is 1.32 m, then what is the velocity of the wave? [Ans. 330 m/s]

Answer

The velocity of a wave can be calculated by multiplying the frequency and the wavelength. In this case, the frequency is 250 Hz and the wavelength is 1.32 m.

Velocity = Frequency × Wavelength

Velocity = 250 Hz × 1.32 m

Velocity = 330 m/s

Therefore, the velocity of the wave is 330 m/s.

Question 2

The velocity of a wave having wavelength 1.7 m is 340 m/s. Find the frequency of the wave. [Ans. 200 Hz]

Answer

The frequency of a wave can be calculated by dividing the velocity by the wavelength. In this case, the wavelength is 1.7 m and the velocity is 340 m/s.

Frequency = Velocity \over Wavelength

Frequency = 340 × 1.7

Frequency = 200 Hz

Therefore, the frequency of the wave is 200 Hz.

Question 3

The velocity of sound in air is 330 m/s and the frequency of a tuning fork is 660 Hz. What will be the distance covered by the sound is 200 complete vibrations of the fork? [Ans. 100 m]

Answer

To find the distance covered by sound in a given number of complete vibrations, we need to calculate the wavelength first. The wavelength can be determined by dividing the velocity of sound by the frequency.

Wavelength = Velocity \over Wavelength

Wavelength = 330 \over 660

Wavelength = 0.5 m

Since one complete vibration corresponds to one wavelength, the distance covered by sound in 200 complete vibrations is:

Distance = Wavelength × Number of Vibrations

Distance = 0.5 m × 200

Distance = 100 m

Therefore, the distance covered by the sound in 200 complete vibrations of the tuning fork is 100 m.

Question 4

A source of sound produces 20 compressions and 20 rarefactions in 0.2 seconds. The distance between compression and the consecutive rarefaction is 50 cm. Find the wavelength, frequency and time period of the wave. [Ans. 100 cm, 100 Hz, 0.01 sec.]

Answer

Given:

Number of compressions = 20

Number of rarefactions = 20

Time taken = 0.2 seconds

Distance between compression and rarefaction = 50 cm

Wavelength = Distance between two consecutive compression

= 50 cm × 2 = 100 cm

Frequency = Total\ No.\ of\ oscillation  \over Time

= 20 \over 0.2

= 100 Hz

Time period = 1 \over Frequency

= 1 \over 100

= 0.01 seconds

Therefore, the wavelength is 100 cm (0.5 m), the frequency is 100 Hz, and the time period is 0.01 seconds.

Question 5

The audible range of frequencies is 20 Hz to 20,000 Hz. Find the range of wavelengths corresponding to this frequency. Given, the velocity of sound 340 m/s. [Ans. Range: 0.017 m to 17 m]

Answer

To find the range of wavelengths corresponding to the audible frequency range, we can use the formula:

Wavelength = Velocity \over Frequency

For the lower frequency limit of 20 Hz:

Wavelength = 340 \over 20 = 17 m

For the upper frequency limit of 20,000 Hz:

Wavelength = 340 m \over 20,000 = 0.017 m (17 mm)

Therefore, the range of wavelengths corresponding to the audible frequency range is from 0.017 m (17 mm) to 17 m.