Class 10 Physics – Chapter 13: Sound

Board: Punjab Board
Subject: Physics
Class: 10th

A. Multiple Choice Questions (MCQs)

13.1 What is required for the production of sound?

(a) A vacuum
(b) Vibrations of a medium
(c) Only air
(d) Amplification of sound

Correct Answer: (b) Vibrations of a medium

13.2 An echo is produced due to:

(a) Refraction of sound waves
(b) Reflection of sound waves
(c) Diffraction of sound waves
(d) Absorption of sound waves

Correct Answer: (b) Reflection of sound waves

13.3 An example of a longitudinal wave is:

(a) Light wave
(b) Water wave
(c) Sound wave
(d) Radio wave

Correct Answer: (c) Sound wave

13.4 In which medium does sound travel the fastest?

(a) Gases
(b) Liquids
(c) Solids
(d) Vacuum

Correct Answer: (c) Solids

13.5 What is the approximate speed of sound in air?

(a) \(150\ \text{m s}^{-1}\)
(b) \(330\text{–}350\ \text{m s}^{-1}\)
(c) \(500\ \text{m s}^{-1}\)
(d) \(1{,}000\ \text{m s}^{-1}\)

Correct Answer: (b) \(330\text{–}350\ \text{m s}^{-1}\)

13.6 Why is acoustic protection important in building design?

(a) To increase noise pollution
(b) To reduce echo and improve sound quality
(c) To increase the speed of sound
(d) To increase the frequency of sound

Correct Answer: (b) To reduce echo and improve sound quality

13.7 For a normal person, the audible frequency range for sound waves lies between:

(a) 10 Hz and 10 kHz
(b) 20 Hz and 20 kHz
(c) 25 Hz and 25 kHz
(d) 30 Hz and 30 kHz

Correct Answer: (b) 20 Hz and 20 kHz

B. Short Answer Questions

13.1 How does the medium affect the speed of sound?

Answer:

The speed of sound depends on the density and elasticity of the medium. Sound generally travels fastest in solids, slower in liquids, and slowest in gases because the particles and elastic properties of the medium determine how efficiently vibrations are transmitted.

13.2 Do sound waves show reflection?

Answer:

Yes, sound waves reflect when they strike a surface, obstacle, or wall. The bouncing back of sound waves into the same medium is called reflection of sound. This phenomenon gives rise to echoes.

13.3 What factors influence the loudness of sound?

Answer:

The main factors influencing the loudness of sound are:

Amplitude of vibration: A larger amplitude produces a louder sound.
Area of the vibrating body: A larger vibrating surface can produce a louder sound.
Distance from the sound source: Loudness generally decreases as the distance from the source increases.

13.4 Why do we hear an echo when we clap near a building?

Answer:

When we clap near a tall building, the sound waves travel towards the building, strike its hard surface, and reflect back to our ears. If the reflected sound arrives at least \(0.1\) seconds after the original sound, our brain perceives it as a separate sound called an echo.

13.5 What is ultrasound, and how is it used in medicine?

Answer:

Ultrasound: Ultrasound consists of sound waves with frequencies higher than \(20{,}000\ \text{Hz}\) or \(20\ \text{kHz}\), which are beyond the normal human audible range.

Medical uses:

Ultrasonography is used to produce images of internal organs.
It is used to monitor fetal development during pregnancy.
It helps detect conditions such as gallstones and kidney stones.
High-intensity ultrasound can be used to break up kidney stones non-invasively.

C. Constructed Response Questions

13.1 Why does a tuning fork placed on a wooden table produce a louder sound than when held in the air? What does this tell us about the role of surface area in sound propagation?

Answer:

Why the sound is louder: When the stem of a vibrating tuning fork is pressed against a wooden tabletop, it causes the table to vibrate at the same frequency.

Role of surface area: The wooden table has a much larger surface area than the prongs of the tuning fork. It sets a larger volume of surrounding air into vibration, producing a louder sound. This demonstrates that increasing the vibrating surface area can increase the loudness of sound.

13.2 If sound needs a medium to travel, how does the bell jar experiment prove this requirement? What happens when air is removed, and why?

Answer:

Bell jar experiment: An electric bell is suspended inside a sealed glass bell jar connected to a vacuum pump. When the bell is switched on, its ringing sound is clearly audible.

When air is removed: As the vacuum pump gradually removes air from the jar, the sound becomes fainter until it is almost inaudible, even though the hammer can still be seen striking the gong.

Reason: Sound is a mechanical wave that requires a material medium, such as air, to propagate. When air is removed, fewer particles are available to transmit sound vibrations. In an ideal vacuum, sound cannot travel because there are no material particles to carry the vibrations.

13.3 A person hears an echo when shouting near a tall building. Why must the building be at least 17 metres away for the echo to be heard clearly?

Answer:

The human ear retains the sensation of a sound for about \(0.1\) seconds. For a distinct echo, the reflected sound must reach the listener at least \(0.1\) seconds after the original sound.

Using the formula:

\[ d = v \times t \]

Where:

Speed of sound, \(v \approx 340\ \text{m s}^{-1}\)
Time, \(t = 0.1\ \text{s}\)

The total distance travelled by the sound to the building and back is:

\[ 2d = v \times t \]

\[ 2d = 340 \times 0.1 = 34\ \text{m} \]

Therefore, the distance to the building is:

\[ d = \frac{34}{2} = 17\ \text{m} \]

Hence, the reflecting surface must be approximately \(17\) metres away for a distinct echo to be heard under these conditions.

13.4 Why do voices of men sound deeper than those of women or children, even when they are speaking at the same intensity? Explain using the relationship between frequency and pitch.

Answer:

Pitch and frequency: Pitch is the characteristic of sound that allows us to distinguish between high-pitched or shrill sounds and low-pitched or deep sounds. Pitch depends on frequency.

Higher frequency produces a higher-pitched sound.
Lower frequency produces a lower-pitched sound.

Comparison: Adult men generally have longer and thicker vocal cords that vibrate at lower frequencies, producing deeper voices. Women and children generally have shorter, thinner vocal cords that vibrate at higher frequencies, producing higher-pitched voices.

13.5 In what ways do soft materials like carpets or curtains improve the acoustics of a room? How do they affect echoes and sound quality?

Answer:

Soft, porous materials such as carpets, curtains, and foam panels absorb some sound energy instead of reflecting it back into the room.

Effect on echoes: They reduce unwanted sound reflections, reverberation, and persistent echoes.

Effect on sound quality: By reducing overlapping reflections, these materials make speech and music clearer and easier to understand, thereby improving room acoustics.

13.6 When a person claps near a canyon and hears the echo after a few seconds, how can they calculate the distance to the canyon? What principle is used in this method?

Answer:

Principle used: Reflection of sound waves, using the echo-location principle.

Method and calculation:

Measure the time interval (\(t\)) between clapping and hearing the echo.
Use the speed of sound in air, approximately \(343\ \text{m s}^{-1}\).
Remember that the sound travels to the canyon wall and back, covering twice the distance.

The formula is:

\[ 2d = v \times t \]

Therefore:

\[ \boxed{d = \frac{v \times t}{2}} \]

Substitute the measured time and the speed of sound into the formula to calculate the distance to the canyon.

D. Comprehensive Questions

13.1 What is the approximate speed of sound in air, and what factors affect it?

Answer:

Approximate speed: At room temperature (around \(20^\circ\text{C}\)), the speed of sound in air is approximately \(343\ \text{m s}^{-1}\).

Factors affecting the speed of sound:

Temperature: The speed of sound increases as the temperature of air increases because the molecules move faster and transmit vibrations more quickly.
Humidity: Sound generally travels faster in humid air than in dry air because humid air is less dense under comparable conditions.
Density and elasticity of the medium: The speed of sound depends on both the density and the elastic properties of the medium. Greater elasticity generally increases the speed, while density also affects it.

13.2 Why does sound travel faster in solids than in liquids and gases? Explain.

Answer:

Sound is a mechanical wave transmitted through the vibrations and interactions of neighbouring particles.

In solids: Particles are closely packed and connected by strong elastic forces. Vibrations are transmitted rapidly from one particle to the next.
In liquids: Particles are less tightly constrained than in solids, and sound generally travels more slowly.
In gases: Particles are much farther apart, and sound generally travels more slowly than in solids and liquids.

Approximate speeds in different media are:

Steel: \(5{,}000\ \text{m s}^{-1}\)
Water: \(1{,}500\ \text{m s}^{-1}\)
Air: \(340\ \text{m s}^{-1}\)

These values illustrate that sound generally travels fastest in solids, followed by liquids and gases. The exact speed depends on the properties of the material.

13.3 How do amplitude and frequency affect the quality of sound? Explain.

Answer:

1. Amplitude

Amplitude determines the loudness of a sound wave. A larger amplitude generally produces a louder sound, while a smaller amplitude produces a fainter sound.

2. Frequency

Frequency determines the pitch of a sound.

Higher frequency produces a higher-pitched or shriller sound.
Lower frequency produces a lower-pitched or deeper sound.

3. Quality (Timbre)

Timbre is the characteristic that allows us to distinguish between sounds produced by different instruments or sources, even when they have the same loudness and pitch. It depends on the waveform and the combination of fundamental and overtone frequencies.

13.4 How does an echo occur? Explain.

Answer:

An echo is a reflected sound heard separately from the original sound.

Occurrence: When a sound wave travels through air and strikes a broad, hard surface, such as a wall, mountain, or cliff, it reflects back towards the listener.

Condition: For a distinct echo, the reflected sound should arrive at least \(0.1\) seconds after the original sound. At a speed of sound of approximately \(340\ \text{m s}^{-1}\), this generally requires a reflecting surface about \(17\) metres away.

13.5 How do reflection, refraction, and diffraction affect the behaviour of sound waves?

Answer:

1. Reflection

Sound waves bounce back when they strike hard surfaces. Reflection can produce echoes and reverberation.

Example: Hearing an echo after shouting near a mountain.

2. Refraction

Sound waves bend when they pass through regions of air with different temperatures or other properties that affect sound speed. Changes in atmospheric conditions can influence how far sound travels.

Example: Sound may travel farther at night because temperature variations can bend sound waves towards the ground.

3. Diffraction

Sound waves bend around corners and spread through openings. This occurs because sound wavelengths can be comparable to the sizes of everyday openings and obstacles.

Example: Hearing people talking from another room even when they are not directly visible.

13.6 What are infrasound and ultrasound, and how are they used in different applications?

Answer:

1. Infrasound

Infrasound consists of sound waves with frequencies below \(20\ \text{Hz}\), which are below the normal human audible range.

Applications:

Monitoring certain signals associated with earthquakes and volcanic eruptions.
Studying long-range communication in animals, such as elephants.

2. Ultrasound

Ultrasound consists of sound waves with frequencies above \(20{,}000\ \text{Hz}\) or \(20\ \text{kHz}\), which are above the normal human audible range.

Applications:

Medical diagnostic imaging (sonography).
Cleaning delicate machinery parts.
Measuring ocean depth and detecting underwater objects using SONAR.
Non-destructive testing of materials.

E. Numerical Problems

13.1 At a particular temperature, the speed of sound in air is 333 m s⁻¹. If the wavelength of a sound wave is 3 cm, calculate the frequency of the sound wave. Is this frequency in the audible range of the human ear?

Given Data:

Speed of sound: \(v = 333\ \text{m s}^{-1}\)
Wavelength: \(\lambda = 3\ \text{cm} = 0.03\ \text{m}\)

To Find:

Frequency, \(f = ?\)
Whether the frequency is audible.

Formula:

\[ v = f\lambda \]

Rearranging:

\[ f = \frac{v}{\lambda} \]

Calculation:

\[ f = \frac{333}{0.03} \]

\[ f = 11{,}100\ \text{Hz} \]

Converting into kilohertz:

\[ f = 11.1\ \text{kHz} \]

Audibility Check:

The normal audible frequency range for humans is approximately \(20\ \text{Hz}\) to \(20{,}000\ \text{Hz}\).

Since \(11{,}100\ \text{Hz}\) lies within this range, the sound is audible.

Result: The frequency is \(11.1\ \text{kHz}\), and it is within the human audible range.

13.2 A clock chimes 48 times in 1 minute. Calculate the frequency and period of the chimes.

Given Data:

Number of chimes: \(N = 48\)
Total time: \(t = 1\ \text{minute} = 60\ \text{s}\)

To Find:

Frequency, \(f = ?\)
Time period, \(T = ?\)

Formulae:

\[ f = \frac{N}{t} \]

\[ T = \frac{1}{f} \]

Calculation:

1. Frequency

\[ f = \frac{48}{60} \]

\[ f = 0.8\ \text{Hz} \]

2. Time Period

\[ T = \frac{1}{0.8} \]

\[ T = 1.25\ \text{s} \]

Result: Frequency \(= 0.8\ \text{Hz}\), and time period \(= 1.25\ \text{s}\).

13.3 A car horn emits a sound with a wavelength of 0.5 m. If the speed of sound in air is 340 m s⁻¹, calculate the frequency of the sound produced by the car’s horn.

Given Data:

Wavelength: \(\lambda = 0.5\ \text{m}\)
Speed of sound: \(v = 340\ \text{m s}^{-1}\)

To Find:

Frequency, \(f = ?\)

Formula:

\[ v = f\lambda \]

Rearranging:

\[ f = \frac{v}{\lambda} \]

Calculation:

\[ f = \frac{340}{0.5} \]

\[ f = 680\ \text{Hz} \]

Result: The frequency of the car horn is \(680\ \text{Hz}\).

13.4 A wave travels with a speed of 500 m s⁻¹ and has a wavelength of 2 metres. Calculate the time period of the wave.

Given Data:

Speed: \(v = 500\ \text{m s}^{-1}\)
Wavelength: \(\lambda = 2\ \text{m}\)

To Find:

Time period, \(T = ?\)

Formulae:

\[ v = f\lambda \]

Rearranging:

\[ f = \frac{v}{\lambda} \]

The time period is:

\[ T = \frac{1}{f} \]

Calculation:

1. Calculate frequency

\[ f = \frac{500}{2} \]

\[ f = 250\ \text{Hz} \]

2. Calculate time period

\[ T = \frac{1}{250} \]

\[ T = 0.004\ \text{s} \]

Result: The time period of the wave is \(0.004\ \text{s}\).

13.5 A research boat sends a sound wave straight to the seabed and receives the echo 2 seconds later. The speed of sound in seawater is 1600 m s⁻¹. Find the depth of the sea at this position.

Given Data:

Total round-trip time: \(t = 2\ \text{s}\)
Speed of sound in seawater: \(v = 1600\ \text{m s}^{-1}\)

To Find:

Depth of the seabed, \(d = ?\)

Formula:

The sound travels to the seabed and back, so the total distance is \(2d\).

\[ 2d = vt \]

Rearranging:

\[ d = \frac{vt}{2} \]

Calculation:

\[ d = \frac{1600 \times 2}{2} \]

\[ d = 1600\ \text{m} \]

Result: The depth of the sea is \(1600\ \text{m}\).

13.6 A person claps his hands near a mountain and hears the echo after 6 seconds. If the speed of sound is 343 m s⁻¹, what is the distance of the mountain from the person?

Given Data:

Total time taken for the echo: \(t = 6\ \text{s}\)
Speed of sound: \(v = 343\ \text{m s}^{-1}\)

To Find:

Distance to the mountain, \(d = ?\)

Formula:

Since the sound travels to the mountain and back:

\[ 2d = vt \]

Rearranging:

\[ d = \frac{vt}{2} \]

Calculation:

\[ d = \frac{343 \times 6}{2} \]

\[ d = \frac{2058}{2} \]

\[ d = 1029\ \text{m} \]

Result: The distance of the mountain from the person is \(1029\ \text{m}\).

Notes by Sir Dilawar

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