Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 7 Sound: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
Summary
Sound is made when things shake. These shakes travel as waves through air or water to our ears. Sound needs a medium; it cannot travel in a vacuum. We hear sounds with frequencies from 20 Hertz to 20,000 Hertz. Higher frequencies are ultrasonic, lower are infrasonic. Wave amplitude is the size of the vibration. Other properties are time period, frequency (vibrations per second), and wavelength (wave distance). Wave speed is frequency times wavelength.
Sound waves reflect, or bounce, off surfaces. An echo is a reflected sound heard after the original. Clear echoes need a distant, large reflector and loud sound. Speed of sound can be found using echo time and distance. Bats and dolphins use echoes to navigate and hunt. SONAR uses ultrasonic echoes to find underwater objects. Ultrasonic waves are useful as they travel far, stay focused, and are not easily absorbed. Doctors use ultrasonic echoes for medical imaging.
Objects have natural vibrations, shaking on their own after a disturbance. Natural frequency depends on shape and size, like a pendulum or guitar string. Damped vibrations occur when friction makes vibrations fade. Forced vibrations happen when an external repeating force makes an object vibrate at the force’s frequency.
Resonance is a special forced vibration. It occurs when the external force’s frequency matches the object’s natural frequency. This causes very large amplitude vibrations, like pushing a swing at the right time to make it go higher. If you push a swing just as it starts to come back towards you, it will go higher with each push. This matching of the push timing with the swing’s own timing is like resonance. Examples include one tuning fork making another vibrate, or soldiers making a bridge shake if their marching matches its frequency. Radios and TVs use resonance to select specific signals.
Sound has characteristics. Loudness distinguishes loud from soft sounds and depends on wave amplitude; larger amplitude means louder sound. Pitch distinguishes high (shrill) from low (flat) sounds and depends on frequency; higher frequency means higher pitch. Quality (timbre) distinguishes sounds of the same loudness and pitch from different instruments, like a piano versus a flute. It depends on the sound’s waveform, or shape. Music is generally pleasant, with regular vibrations. Noise is often unpleasant, with irregular vibrations.
Workbook solutions (Concise/Selina)
Exercise (A)
MCQ
1. With respect to sound waves, which of the following statements are correct ?
(1) The maximum displacement of a particle on either side of its mean position is called the amplitude of the wave.
(2) Time taken by the particle of a medium to complete one vibration is called the time period.
(3) The number of vibrations made by the particle of a medium in one second is called the frequency.
(4) The frequency of a wave is different than the frequency of the source producing it.
(a) 1
(b) 1, 2
(c) 1, 2, 3
(d) 1, 2, 3 and 4
Answer: (c) 1, 2, 3
2. The speed V of a longitudinal wave is given by:
(a) V = √(γP/d)
(b) V = √γP
(c) V = √(T/m)
(d) V = √(γP/d)
Answer: (d) V = √(γP/d)
3. The speed V of a transverse wave is given by:
(a) V = √(T/m)
(b) V = √(γP/d)
(c) V = √Tm
(d) V = 1/√(γP)
Answer: (a) V = √(T/m)
4. Various types of mechanical waves are :
(a) electromagnetic, longitudinal, transverse
(b) electromagnetic, transverse
(c) longitudinal, transverse
(d) electromagnetic, transverse.
Answer: (c) longitudinal, transverse
5. Due to vibrations of medium particles, the energy transformation is from:
(a) heat energy to kinetic energy and vice versa
(b) kinetic energy to potential energy and vice versa
(c) heat energy to potential energy and vice versa
(d) potential energy to nuclear energy and vice versa
Answer: (b) kinetic energy to potential energy and vice versa
6. With the increase in temperature of a gas, the speed of sound:
(a) decreases
(b) increases
(c) remains the same
(d) cannot say
Answer: (b) increases
7. The condition for reflection of sound wave is:
(a) surface must be smooth
(b) surface must be polished
(c) reflecting surface must be bigger than the wavelength of the sound wave
(d) reflecting surface must be smaller than the wavelength of the sound wave
Answer: (c) reflecting surface must be bigger than the wavelength of the sound wave
8. For an echo, the reflected sound must reach the person at least ……………….. after the original sound is heard.
(a) 1 s
(b) 10 s
(c) 0.1 s
(d) 0.01 s
Answer: (c) 0.1 s
9. Time taken to hear the echo if distance between the listener and the obstacle d is given by:
(a) t = d x V
(b) t = 2d x V
(c) t = V/2d
(d) t = 2d/V
Answer: (d) t = 2d/V
10. To detect obstacles in their path, bats produce:
(a) infrasonic waves
(b) ultrasonic waves
(c) electromagnetic waves
(d) radio waves.
Answer: (b) ultrasonic waves
11. With respect to ultrasonic waves, which of the following statements is correct ?
(a) they travel undeviated through a long distance
(b) they have a speed greater than the speed of sound in a medium
(c) they are not easily absorbed in a medium
(d) they can be confined to a narrow beam
Answer: (b) they have a speed greater than the speed of sound in a medium
12. Infrasonic sound waves have a frequency:
(a) between 20 Hz to 10,000 Hz
(b) below 20 Hz
(c) above 20,000 Hz
(d) between 20 Hz to 20,000 Hz
Answer: (b) below 20 Hz
13. The diagram shown below depicts a displacement-time graph of particle a wave travelling with speed 20 m/s.
The frequency of the wave is:
(a) 50 Hz
(b) 33.3 Hz
(c) 100 Hz
(d) 1000 Hz
Answer: (a) 50 Hz
14. Assertion (A): Sound waves can travel in vacuum, but light waves cannot.
Reason (R): Light is an electromagnetic wave, but sound is a mechanical wave.
(a) both A and R are true and R is the correct explaination of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
15. Assertion (A) : The speed of sound in a gas increases with an increase in humidity.
Reason (R): Density of a gas decreases with increase in humidity.
(a) both A and R are true and R is the correct explaination of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explaination of A
16. Assertion (A): The flash of lightening is seen before the sound of thunder is heard.
Reason (R): The speed of sound is greater than the speed of light.
(a) both A and R are true and R is the correct explaination of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
Very Short Answer Type Questions
1. State two factors on which the speed of a wave travelling in a medium depends.
Answer: The speed of a wave travelling in a medium depends on two factors:
(i) Elasticity of the medium
(ii) Density of the medium.
The speed V is given by the relation V = √(E/d), where E is the elasticity and d is the density of the medium.
Short Answer Type Questions
1. What are mechanical waves ?
Answer: Mechanical waves are waves that require a medium for their propagation. For instance, sound is a mechanical wave.
2. Define the following terms in relation to a wave : (a) amplitude, (b) frequency, (c) wavelength, and (d) wave velocity.
Answer: In relation to a wave:
(a) Amplitude (a) is the maximum displacement of the particle of the medium on either side of its mean position.
(b) Frequency (f) is the number of vibrations made by the particle of the medium in one second.
(c) Wavelength (λ) is the distance travelled by a wave in one time period of vibration of the particle of the medium.
(d) Wave velocity (V) is the distance travelled by the wave in one second.
3. A wave passes from one medium to another medium. Mention one property of the wave out of speed, frequency or wavelength (i) which changes, (ii) which does not change.
Answer: When a wave passes from one medium to another medium:
(i) The property that changes is its wavelength (or speed).
(ii) The property that does not change is its frequency.
4. State two differences between light and sound waves.
Answer: Two differences between light and sound waves are:
(i) Sound waves are mechanical waves and require a material medium for their propagation, meaning they cannot travel in a vacuum. Light waves, on the other hand, are electromagnetic waves and can travel in a vacuum.
(ii) The speed of light (approximately 3 × 10⁸ m s⁻¹ in vacuum) is significantly greater than the speed of sound (e.g., approximately 340 m s⁻¹ in air at room temperature). This is why, for example, the flash of lightning is seen before the sound of thunder is heard.
5. What do you mean by reflection of sound ? State one condition for the reflection of a sound wave. Name a device in which reflection of sound wave is used.
Answer: Reflection of sound refers to the bouncing back of sound waves when they strike a surface or an obstacle.
One condition for the reflection of a sound wave is that the size of the reflecting surface (reflector) must be large enough as compared to the wavelength of the sound wave.
A device in which reflection of sound waves is used is SONAR (Sound Navigation and Ranging), which uses reflected ultrasonic waves to detect underwater objects or determine the depth of the sea.
6. What is meant by an echo ? State two conditions necessary for an echo to be heard distinctly.
Answer: An echo is the sound heard after reflection from a rigid obstacle. Two conditions necessary for an echo to be heard distinctly are:
(i) The minimum distance between the source of sound and the reflector must be such that the reflected sound reaches the ear at least 0.1 seconds after the original sound is heard. This is because the sensation of sound persists in our ears for about 0.1 seconds.
(ii) The size of the reflector must be large enough as compared to the wavelength of the sound wave.
(iii) Additionally, the intensity of the sound should be such that the reflected sound reaching the ear is sufficiently loud to be audible.
7. A man is standing at a distance of 12 m from a cliff. Will he be able to hear a clear echo of his sound ? Give a reason for your answer.
Answer: No, the man will likely not be able to hear a clear echo of his sound.
Reason: To hear a distinct echo, the reflected sound must reach the ear at least 0.1 seconds after the original sound is produced. If the speed of sound in air is taken as approximately 340 m s⁻¹, the minimum distance (d) to the reflector for a distinct echo is calculated as d = (V × t) / 2 = (340 m s⁻¹ × 0.1 s) / 2 = 17 m. Since the man is standing only 12 m away from the cliff, which is less than the required minimum distance of 17 m, the reflected sound will return in less than 0.1 seconds and will overlap with the original sound, making it difficult to distinguish the echo clearly.
8. State two applications of echo.
Answer: Two applications of echo are:
(i) Bats and dolphins use echoes for navigation and hunting. Bats emit ultrasonic waves and listen to the echoes to detect obstacles in their path (sound ranging) and locate prey.
(ii) SONAR (Sound Navigation and Ranging) systems use echoes of ultrasonic waves to determine the depth of the sea, locate underwater objects like submarines, shipwrecks, or shoals of fish.
9. Explain how the speed of sound can be determined by the method of echo.
Answer: The speed of sound in air can be determined by the method of echo as follows:
- Produce a sound from a place at a known distance, say ‘d’, from a reflecting surface. This distance ‘d’ should be sufficiently large, for example, at least 50 m, to ensure a clear echo.
- Measure the time interval ‘t’ between the instant the sound is produced and the instant the echo is heard. This can be done using a stopwatch with a small least count, like 0.01 s.
- The sound travels a total distance of 2d (to the reflector and back).
- The speed of sound ‘V’ is then calculated using the formula:
V = Total distance travelled / Time interval = 2d / t. - To improve accuracy, the experiment should be repeated several times, and the average value of the speed of sound V is then determined.
10. State the use of echo by a bat, dolphin and fisherman.
Answer: The use of echo by a bat, dolphin and fisherman are:
- Bat: Bats use echoes for sound ranging. When the sounds produced by flying bats get reflected back from an obstacle in front of them, by hearing the echo, bats come to know, even in the dark, the location of the obstacle. This allows them to turn away from their path and fly safely without colliding with the obstacle.
- Dolphin: Dolphins detect their enemy and obstacles by emitting ultrasonic waves and hearing their echo. They also use ultrasonic waves for hunting their prey.
- Fisherman: A trawlerman or fisherman sends an ultrasonic pulse into the sea and receives the pulse reflected from a shoal of fish in a detector. By recording the total time ‘t’ of the to-and-fro journey of the pulse, the distance ‘d’ of the fish can be calculated using the relation d = Vt/2, where V is the speed of ultrasonic waves in sea water.
11. How do bats avoid obstacles in their way, when in flight?
Answer: Bats avoid obstacles in their way when in flight by using sound ranging. They emit sounds, often ultrasonic, which get reflected back from any obstacle in front of them. By hearing these echoes, bats can determine the location of the obstacle, even in complete darkness. This allows them to change their path and fly safely without colliding with the obstacles.
12. What is meant by sound ranging ? Give one use of sound ranging.
Answer: Sound ranging is the process of detecting the presence and determining the distance of an object by using the reflection of sound waves (echoes).
One use of sound ranging is by bats to detect obstacles in their path and navigate safely, especially in the dark.
13. Name the waves used for sound ranging. State one reason for their use. Why are the waves mentioned by you not audible to us ?
Answer: The waves used for sound ranging are ultrasonic waves.
One reason for their use is that they can travel undeviated through a long distance. Other properties include being able to be confined to a narrow beam and not being easily absorbed in a medium.
These ultrasonic waves are not audible to us (human beings) because their frequency is above 20,000 Hz, which is beyond the upper limit of the audible range of frequency for human ears (20 Hz to 20,000 Hz).
14. What is ‘SONAR’ ? State the principle on which it is based.
Answer: ‘SONAR’ stands for Sound Navigation And Ranging.
The principle on which SONAR is based is the reflection of sound waves, specifically ultrasonic waves. Ultrasonic waves are sent out from a transmitter on a ship. These waves travel through the water and are reflected back when they strike an obstacle, such as an enemy submarine, an iceberg, a sunken ship, or the seabed. The reflected waves (echoes) are received by a detector on the ship. By measuring the time interval ‘t’ between the transmission and reception of the ultrasonic wave, and knowing the speed ‘V’ of sound in water, the distance ‘d’ of the obstacle can be calculated using the formula d = (V × t) / 2.
15. State the use of echo in medical field.
Answer: In the medical field, the echo method using ultrasonic waves is employed for imaging human organs. This technique is known as ultrasonography and is used to get images of organs such as the liver, gall bladder, uterus, womb, etc. Similarly, echocardiography, which also uses the principle of echoes, is used to obtain the image of the human heart.
Numerical Questions
1. The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 m s⁻¹, calculate : (i) the number of waves produced in one second, and (ii) the time in which one wave is produced.
Answer:
Given:
Wavelength (λ) = 20 cm
Wave velocity (v) = 24 m s⁻¹
To find:
(i) Number of waves produced in one second (Frequency, f) = ?
(ii) Time in which one wave is produced (Time period, T) = ?
Solution:
First, we need to ensure all units are consistent. We will convert the wavelength from centimeters (cm) to meters (m).
Wavelength (λ) = 20 cm
=> λ = 20 / 100 m
=> λ = 0.2 m
(i) Calculation for the number of waves produced in one second (Frequency, f):
We know the relationship between wave velocity (v), frequency (f), and wavelength (λ) is:
v = f × λ
To find the frequency (f), we rearrange the formula:
f = v / λ
=> f = 24 / 0.2
=> f = 120 s⁻¹ or 120 Hz.
Therefore, the number of waves produced in one second is 120.
(ii) Calculation for the time in which one wave is produced (Time period, T):
The time in which one wave is produced is the time period (T), which is the reciprocal of the frequency (f).
T = 1 / f
=> T = 1 / 120
=> T = 0.00833 s.
Therefore, the time in which one wave is produced is 0.00833 seconds.
2. Calculate the minimum distance in air required between the source of sound and the obstacle to hear an echo. Take speed of sound in air = 350 m s⁻¹.
Answer:
Given:
Speed of sound in air (v) = 350 m s⁻¹
To find:
Minimum distance between the source and the obstacle (d) = ?
Solution:
For a human ear to distinguish an echo from the original sound, the time interval between the original sound and the reflected sound must be at least 0.1 seconds. This is due to the persistence of hearing.
Therefore, the minimum time taken for the echo to be heard (t) = 0.1 s.
Let ‘d’ be the minimum distance between the source of sound and the obstacle.
The sound travels from the source to the obstacle and then reflects back to the source.
So, the total distance traveled by the sound = d + d = 2d.
We know the relationship between speed, distance, and time is:
Speed = Total Distance / Time
=> v = 2d / t
To find the distance ‘d’, we rearrange the formula:
=> 2d = v × t
=> d = (v × t) / 2
Substituting the known values into the formula:
=> d = (350 × 0.1) / 2
=> d = 35 / 2
=> d = 17.5 m.
Thus, the minimum distance in air required between the source of sound and the obstacle to hear an echo is 17.5 m.
3. What should be the minimum distance between the source and reflector in water so that the echo is heard distinctly ? (The speed of sound in water = 1400 m s⁻¹).
Answer:
Given:
Speed of sound in water (v) = 1400 m s⁻¹
Time interval for a distinct echo (persistence of hearing) (t) = 0.1 s
To find:
Minimum distance between the source and reflector (d) = ?
Formula:
We know that for an echo, the sound travels from the source to the reflector and back to the source.
Total distance travelled by sound = 2 × distance between source and reflector (d)
The relationship between speed, distance, and time is:
Speed of sound (v) = Total distance / Time taken (t)
=> v = 2d / t
Solution:
Using the formula,
v = 2d / t
To find the distance (d), we can rearrange the formula:
2d = v × t
=> d = (v × t) / 2
Now, substituting the given values into the formula:
=> d = (1400 × 0.1) / 2
=> d = 140 / 2
=> d = 70 m.
Therefore, the minimum distance between the source and the reflector in water for the echo to be heard distinctly is 70 m.
4. A man standing 25 m away from a wall produces a sound and receives the reflected sound. (a) Calculate the time after which he receives the reflected sound if the speed of sound in air is 350 m s⁻¹. (b) Will the man be able to hear a distinct echo ? Explain the answer.
(a) Calculate the time after which he receives the reflected sound if the speed of sound in air is 350 m s⁻¹.
Answer:
Given:
Distance of the man from the wall (d) = 25 m
Speed of sound in air (v) = 350 m s⁻¹
To find:
Time after which the reflected sound is received (t) = ?
Solution:
For the man to hear the reflected sound (echo), the sound must travel from the man to the wall and then back to the man.
Therefore, the total distance traveled by the sound = Distance to the wall + Distance from the wall
Total distance = d + d = 2d
Now,
Total distance = 2 × 25 m
=> Total distance = 50 m
We know the formula for time is:
Time = Total Distance / Speed
=> t = 2d / v
=> t = 50 / 350
=> t = 5 / 35
=> t = 1 / 7
=> t ≈ 0.143 s
Thus, the man will receive the reflected sound after approximately 0.143 seconds.
(b) Will the man be able to hear a distinct echo ? Explain the answer.
Answer:
Yes, the man will be able to hear a distinct echo.
Explanation:
The sensation of sound persists in the human brain for about 0.1 seconds. This phenomenon is known as the persistence of hearing. For a person to hear a distinct echo, the time interval between the original sound and the reflected sound must be at least 0.1 seconds.
From the calculation in part (a), the time interval is t ≈ 0.143 s.
Comparing the calculated time interval with the minimum time required:
0.143 s > 0.1 s
Since the time taken for the reflected sound to reach the man is greater than the persistence of hearing (0.1 s), the original sound and the reflected sound will be perceived as separate sounds. Therefore, the man will be able to hear a distinct echo.
5. A RADAR sends a signal with a speed of 3 × 10⁸ ms⁻¹ to an aeroplane at a distance 300 km from it. After how much time is the signal received back after reflection from the aeroplane ?
Answer:
Given:
Speed of the signal (v) = 3 × 10⁸ ms⁻¹
Distance of the aeroplane from the RADAR (d) = 300 km
To find:
Total time taken for the signal to be received back (t) = ?
Solution:
First, we need to convert the distance from kilometers (km) to meters (m) to have consistent units with the speed.
We know that 1 km = 1000 m = 10³ m.
=> d = 300 km
=> d = 300 × 10³ m
=> d = 3 × 10² × 10³ m
=> d = 3 × 10⁵ m
The signal travels from the RADAR to the aeroplane and then reflects back to the RADAR. Therefore, the total distance travelled by the signal is twice the distance between the RADAR and the aeroplane.
Total distance (D) = 2 × d
=> D = 2 × (3 × 10⁵ m)
=> D = 6 × 10⁵ m
Now, we can calculate the total time taken for the signal’s journey using the formula:
Time = Total Distance / Speed
t = D / v
=> t = (6 × 10⁵ m) / (3 × 10⁸ ms⁻¹)
=> t = (6/3) × 10⁵⁻⁸ s
=> t = 2 × 10⁻³ s
Therefore, the signal is received back after 2 × 10⁻³ seconds.
6. A man standing 48 m away from a wall fires a gun. Calculate the time after which an echo is heard. (The speed of sound in air is 320 m s⁻¹).
Answer:
Given:
Distance of the man from the wall (d) = 48 m
Speed of sound in air (v) = 320 m s⁻¹
To find:
Time after which an echo is heard (t) = ?
Solution:
For an echo to be heard, the sound has to travel from the man to the wall and then reflect back to the man.
Therefore, the total distance traveled by the sound is twice the distance between the man and the wall.
Total distance traveled by sound (D) = 2 × Distance from the wall
=> D = 2 × d
=> D = 2 × 48 m
=> D = 96 m
We know the relationship between speed, distance, and time is:
Speed = Total Distance / Time
v = D / t
To find the time (t), we rearrange the formula:
Time = Total Distance / Speed
=> t = D / v
Substituting the given values into the formula:
=> t = 96 m / 320 m s⁻¹
=> t = 0.3 s
Thus, the man will hear the echo after 0.3 seconds.
7. A ship on the surface of water sends a signal and receives it back from a submarine inside water after 4 s. Calculate the distance of the submarine from the ship. (The speed of sound in water is 1450 m s⁻¹).
Answer:
Given:
Speed of sound in water (v) = 1450 m s⁻¹
Total time taken for the signal to travel to the submarine and back (t) = 4 s
To find:
The distance of the submarine from the ship (d) = ?
Solution:
The formula for distance is:
Distance = Speed × Time
The time given (4 s) is the total time for the sound signal to travel from the ship to the submarine and reflect back to the ship.
Let ‘d’ be the distance of the submarine from the ship.
The total distance traveled by the sound signal is from the ship to the submarine and back, which is d + d = 2d.
Now, we can write the formula as:
Total Distance = Speed × Total Time
=> 2d = v × t
Substituting the given values into the formula:
=> 2d = 1450 × 4
=> 2d = 5800 m
This is the total distance traveled by the signal. To find the distance of the submarine from the ship (d), we need to divide this by 2.
=> d = 5800 / 2
=> d = 2900 m
Therefore, the distance of the submarine from the ship is 2900 m.
8. A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340 m s⁻¹, find the distance between the cliff and the observer.
Answer:
Given:
Frequency of the pendulum (f) = 5 vibrations per second = 5 Hz
Number of vibrations (n) = 8
Velocity of sound in air (v) = 340 m s⁻¹
To find:
Distance between the cliff and the observer (d) = ?
Solution:
First, we need to find the time taken for the pendulum to complete 8 vibrations.
Time period of the pendulum (T) is the time taken for one vibration.
T = 1 / Frequency (f)
=> T = 1 / 5
=> T = 0.2 seconds.
Now, we find the total time taken for the echo to be heard, which is the time for 8 vibrations.
Total time (t) = Number of vibrations (n) × Time period (T)
=> t = 8 × 0.2
=> t = 1.6 seconds.
This total time (t) is the time taken for the sound to travel from the observer to the cliff and back to the observer.
Let ‘d’ be the distance between the observer and the cliff.
The total distance traveled by the sound = d (to the cliff) + d (back from the cliff) = 2d.
Using the formula, Distance = Velocity × Time:
Total distance traveled by sound = Velocity of sound × Total time
=> 2d = v × t
=> 2d = 340 × 1.6
=> 2d = 544 m
=> d = 544 / 2
=> d = 272 m.
Therefore, the distance between the cliff and the observer is 272 m.
9. A person standing between two vertical cliffs produces a sound. Two successive echoes are heard at 4 s and 6 s. Calculate the distance between the cliffs. (Speed of sound in air = 320 m s⁻¹).
Answer:
Given:
Time for the first echo, t₁ = 4 s
Time for the second echo, t₂ = 6 s
Speed of sound in air, v = 320 m s⁻¹
To find:
The distance between the cliffs, D = ?
Solution:
Let the person be standing at a distance d₁ from the nearer cliff and d₂ from the farther cliff. The first echo heard will be from the nearer cliff.
For the first echo, the sound travels from the person to the nearer cliff and back.
The total distance travelled by the sound = 2d₁
The time taken for the first echo = t₁ = 4 s
Using the formula, Distance = Speed × Time
=> 2d₁ = v × t₁
=> 2d₁ = 320 × 4
=> 2d₁ = 1280 m
=> d₁ = 1280 / 2
=> d₁ = 640 m
Now, for the second echo, the sound travels from the person to the farther cliff and back.
The total distance travelled by the sound = 2d₂
The time taken for the second echo = t₂ = 6 s
Using the formula, Distance = Speed × Time
=> 2d₂ = v × t₂
=> 2d₂ = 320 × 6
=> 2d₂ = 1920 m
=> d₂ = 1920 / 2
=> d₂ = 960 m
The total distance between the two cliffs is the sum of the distances of the person from each cliff.
Distance between the cliffs, D = d₁ + d₂
=> D = 640 + 960
=> D = 1600 m
Therefore, the distance between the cliffs is 1600 m.
10. A person standing at a distance x in front of a cliff fires a gun. Another person B standing behind the person A at distance y from the cliff hears two sounds of the fired shot after 2s and 3s respectively. Calculate x and y (take speed of sound 320 ms⁻¹).
Answer:
Given:
Speed of sound in air, V = 320 ms⁻¹
Time for the first sound, t₁ = 2 s
Time for the second sound, t₂ = 3 s
Distance of person A from the cliff = x
Distance of person B from the cliff = y
To find:
The distances x and y.
Solution:
Person B hears two sounds after the gun is fired by person A.
- The first sound is the direct sound that travels from person A to person B.
- The second sound is the reflected sound (echo) that travels from person A to the cliff and then to person B.
For the first (direct) sound:
The sound travels a distance equal to the separation between person A and person B. Since person B is behind person A, this distance is (y – x).
Time taken = t₁ = 2 s
Using the formula, Distance = Speed × Time:
y – x = V × t₁
=> y – x = 320 × 2
=> y – x = 640 m …..(i)
For the second (reflected) sound:
The sound travels from person A to the cliff (distance x) and after reflection travels from the cliff to person B (distance y).
Total distance travelled = x + y
Time taken = t₂ = 3 s
Using the formula, Distance = Speed × Time:
x + y = V × t₂
=> x + y = 320 × 3
=> x + y = 960 m …..(ii)
Now, we solve the two simultaneous equations:
- y – x = 640
- y + x = 960
Adding equation (i) and equation (ii):
(y – x) + (y + x) = 640 + 960
2y = 1600
y = 1600 / 2
y = 800 m
Substituting the value of y = 800 m into equation (ii):
x + 800 = 960
x = 960 – 800
x = 160 m
Therefore, the distance of person A from the cliff is x = 160 m, and the distance of person B from the cliff is y = 800 m.
11. On sending an ultrasonic wave from a ship towards the bottom of a sea, the time interval between sending the wave and receiving it back is found to be 1.5 s. If the velocity of wave in sea water is 1400 m s⁻¹, find the depth of sea.
Answer:
Given:
Time interval between sending and receiving the wave (t) = 1.5 s
Velocity of the wave in sea water (v) = 1400 m s⁻¹
To find:
Depth of the sea (d) = ?
Solution:
The total distance travelled by the ultrasonic wave is from the ship to the seabed and back to the ship. This is twice the depth of the sea.
Let the depth of the sea be ‘d’.
Total distance travelled by the wave = 2d
We know the formula:
Distance = Velocity × Time
Using the given values:
Total distance = v × t
=> 2d = 1400 m s⁻¹ × 1.5 s
=> 2d = 2100 m
=> d = 2100 / 2
=> d = 1050 m
Therefore, the depth of the sea is 1050 m.
12. The figure below shows the distance-displacement graph of two waves A and B. Compare (i) the amplitude, (ii) the wavelength of the two waves.
Answer:
Given:
A distance-displacement graph for two waves, A (magenta wave) and B (purple wave).
To find:
(i) A comparison of the amplitudes of wave A and wave B.
(ii) A comparison of the wavelengths of wave A and wave B.
Solution:
We will determine the amplitude and wavelength for each wave by reading the values from the provided graph.
(i) Comparison of Amplitude
The amplitude of a wave is its maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position.
From the graph, for wave A (magenta), the maximum displacement (peak) from the equilibrium position is 7 cm.
=> Amplitude of A = 7 cm.
From the graph, for wave B (purple), the maximum displacement (peak) from the equilibrium position is 4 cm.
=> Amplitude of B = 4 cm.
Now, comparing the two amplitudes:
Since 7 cm > 4 cm, the amplitude of wave A is greater than the amplitude of wave B.
(ii) Comparison of Wavelength
The wavelength (λ) of a wave is the distance over which the wave’s shape repeats. It is the distance between consecutive corresponding points of the same phase, such as two adjacent crests or troughs.
From the graph, for wave A (magenta), one complete cycle starts at a distance of 0 cm and ends at a distance of 8 cm.
=> Wavelength of A (λ_A) = 8 cm.
From the graph, for wave B (purple), one complete cycle starts at a distance of 0 cm and ends at a distance of 12 cm.
=> Wavelength of B (λ_B) = 12 cm.
Now, comparing the two wavelengths:
Since 12 cm > 8 cm, the wavelength of wave B is greater than the wavelength of wave A.
Exercise (B)
MCQ
1. The period (or frequency) of natural vibrations depends on:
(a) shape
(b) size
(c) external force
(d) both (a) and (b)
Answer: (d) both (a) and (b)
2. The natural frequency of a simple pendulum of length 1.0 m on earth’s surface is: [Take g = 9.8 m/s²]
(a) 0.5 Hz
(b) 100 Hz
(c) 50 Hz
(d) 5 Hz
Answer: (a) 0.5 Hz
3. A wire stretched between two fixed supports is plucked exactly in the middle and then released. It executes (neglect the resistance of the medium) :
(a) resonant vibrations
(b) natural vibrations
(c) damped vibrations
(d) forced vibrations
Answer: (b) natural vibrations
4. In an organ pipe with one closed end, the frequencies of different modes are in the ratio:
(a) 1:2:3 …..
(b) 1:3:5 …..
(c) 1:2:4 …..
(d) 2:4:6 …..
Answer: (b) 1:3:5 …..
5. The frequency f of a vibration in a stretched string depends on:
(a) length l
(b) radius r
(c) tension T
(d) all of these
Answer: (d) all of these
6. The frequency f of the note produced by a string can be increased by:
(a) decreasing the length l of the string
(b) decreasing the radius r of the string
(c) increasing the Tension T of the string
(d) All of these
Answer: (d) All of these
7. If l is the length of the string stretched between its ends, the wavelength of different modes in Figure (1), (2) and (3) will be :
(a) 2l/2, 2l, 2l/3
(b) 2l/3, 2l/2, 2l
(c) 2l, 2l/2, 2l/3
(d) 2l, 2l/3, 2l/2
Answer: (c) 2l, 2l/2, 2l/3
8. When a body vibrates under a periodic force, the vibrations of a body are:
(a) forced vibrations
(b) damped vibrations
(c) resonant vibrations
(d) natural vibrations
Answer: (a) forced vibrations
9. A slim branch of a tree is pulled and released. It makes ………… vibrations.
(a) damped
(b) natural
(c) forced
(d) resonant
Answer: (a) damped
10. A body is vibrating under the influence of an external periodic force of frequency exactly equal to the natural frequency of the vibrations of the body. It executes:
(a) damped vibrations
(b) forced vibrations
(c) resonant vibrations
(d) natural vibrations
Answer: (c) resonant vibrations
11. At resonance, a loud sound is heard because the body vibrating with a large ………… sends forth a large amount of energy in the medium.
(a) amplitude
(b) frequency
(c) wavelength
(d) time period
Answer: (a) amplitude
12. The sound box of a musical instrument is so constructed that the air column inside it has a natural frequency ………… that of the string stretched in it.
(a) greater than
(b) smaller than
(c) equal to
(d) cannot say
Answer: (c) equal to
13. Column X shows the kind of vibrations and column Y shows their examples.
(A) Natural vibrations (1) A tuning fork when stroked on a rubber pad.
(B) Damped vibrations (2) When a guitarist plucks or strums the guitar strings with his/her fingers.
(C) Forced vibrations (3) When a troop crosses a suspension bridge.
(D) Resonant vibrations (4) A load suspended from a spring.
Choose the correct pairing:
(a) A—(4) B—(2) C—(3) D—(1)
(b) A—(4) B—(1) C—(2) D—(3)
(c) A—(4) B—(3) C—(1) D—(2)
(d) A—(2) B—(3) C—(4) D—(1)
Answer: (b) A—(4) B—(1) C—(2) D—(3)
Very Short Answer Type Questions
1. (a) Name one factor on which the frequency of sound emitted due to vibrations in an air column depends.
Answer: The frequency of sound emitted due to vibrations in an air column depends on the length of the air column.
(b) How does the frequency depend on the factor stated in part (a).
Answer: The frequency is inversely proportional to the length of the air column, so f ∝ 1/l.
2. State one way of increasing the frequency of a note produced by an air column.
Answer: One way of increasing the frequency of a note produced by an air column is by decreasing the length of the air column.
3. How does the frequency of sound given by a stretched string depend on its (a) length, (b) tension ?
Answer: The frequency of sound given by a stretched string depends on its length as f ∝ 1/l, and on its tension as f ∝ √T.
4. A tuning fork is vibrating in air. State whether the vibrations are natural or damped ?
Answer: If a tuning fork is vibrating in air, the vibrations are damped.
5. State the condition for the resonance to occur.
Answer: Resonance occurs only when the applied force causes forced vibration in the body and the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.
6. Complete the following sentence : Resonance is a special case of …….. vibrations, when frequency of the driving force is …….. natural frequency of the driven body.
Answer: Resonance is a special case of forced vibrations, when frequency of the driving force is equal to the natural frequency of the driven body.
Short Answer Type Questions
1. What do you understand by natural vibrations of a body? Give one example.
Answer: The periodic vibrations of a body in the absence of any external force on it, are called natural (or free) vibrations. An example is if the bob of a simple pendulum is displaced slightly from its mean (or rest) position, it starts vibrating with its natural frequency.
2. What is meant by the natural frequency of vibrations of a body ? Name one factor on which it depends.
Answer: The natural frequency of vibrations of a body is the constant frequency at which a body, capable of vibrating, vibrates. The period (or frequency) of vibration depends on the shape and size (or structure) of the body.
3. State one condition for a body to execute natural vibrations.
Answer: A body executes natural vibrations only when restoring forces are present. Natural vibrations persist only in vacuum.
4. State two ways of increasing the frequency of vibrations of a stretched string.
Answer: Two ways of increasing the frequency of vibrations of a stretched string are (a) by increasing the tension in the string and (b) by decreasing the length of the string.
5. What adjustments would you make for tuning a stringed instrument for it to emit a note of a desired frequency ?
Answer: To tune a stringed instrument to emit a note of a desired frequency, adjustments can be made by changing the tension T in the string, or by changing the length l of the string. For example, the frequency f of the note can be increased by increasing the tension T of the string, or by decreasing the length l of the string.
6. A blade, fixed at one end, is made to vibrate by pressing its other end and then releasing it. State one way in which the frequency of vibrations of the blade can be lowered.
Answer: One way in which the frequency of vibrations of the blade can be lowered is by increasing the length of the blade or by sticking a small weight on the blade at its free end so as to increase its mass.
7. How does the medium affect the amplitude of the natural vibrations of a body ?
Answer: The surrounding medium offers resistance (or friction) to the motion, so the energy of the vibrating body continuously decreases due to which the amplitude of vibration gradually decreases.
8. What are forced vibrations ? Give one example to illustrate your answer.
Answer: The vibrations of a body which take place under the influence of an external periodic force acting on it, are called forced vibrations. An example is when the stem of a vibrating tuning fork is pressed against the top of a table, the tuning fork produces forced vibrations in the table top.
9. On keeping the stem of a vibrating tuning fork on the surface of a table, a loud sound is heard. Give reason.
Answer: On keeping the stem of a vibrating tuning fork on the surface of a table, a loud sound is heard because forced vibrations are produced on the surface of the table. The table top has a much larger vibrating area than the tuning fork, so the forced vibrations of the table top send forth a greater energy and produces a louder sound.
10. State two differences between natural and forced vibrations.
Answer: Two differences between natural and forced vibrations are:
- Natural vibrations occur in the absence of any resistive or external force, while forced vibrations occur in a medium in the presence of an external periodic force.
- The frequency of natural vibration depends on the shape and size of the body and remains constant, while the frequency of forced vibration is equal to the frequency of the applied force and changes with change in the frequency of the applied force.
11. State two differences between forced and resonant vibrations.
Answer: Two differences between forced and resonant vibrations are:
- Forced vibrations occur when a body vibrates under an external periodic force of frequency different from its natural frequency, while resonant vibrations occur when the frequency of the external periodic force is exactly equal to the natural frequency of the body.
- The amplitude of forced vibration is small, while the amplitude of resonant vibration is very large.
12. Why is a loud sound heard at resonance ?
Answer: At resonance, a loud sound is heard because the body vibrating with a very large amplitude sends forth a large amount of energy in the medium. The body vibrates with a large amplitude thus conveying more energy to the ears, so a loud sound is heard.
13. When a troop crosses a suspension bridge, the soldiers are asked to break their steps. Why ?
Answer: When a troop crosses a suspension bridge, the soldiers are asked to break their steps because when soldiers march in steps, each soldier exerts a periodic force in the same phase and therefore the bridge executes the forced vibrations of frequency equal to the frequency of their steps. Incidentally if the bridge happens to have its natural frequency equal to the frequency of the steps of the soldiers, the bridge will vibrate with a large amplitude due to resonance and the suspension bridge may collapse.
14. Why are stringed instruments like guitar provided with a hollow sound box ?
Answer: Stringed instruments like guitar are provided with a hollow sound box because a vibrating string by itself produces a very weak sound which cannot be heard at a distance. The box is so constructed that the column of air inside it, has a natural frequency which is same as that of the string stretched on it, so that when the string is made to vibrate, the air column inside the box is set into forced vibrations, the frequency of which is the same as that of the string. Since the sound box has a large area, it sets a large volume of air into vibrations, so due to resonance, a loud sound is produced.
15. How do you tune your radio set to a particular station ? Name the phenomenon involved in doing so and define it.
Answer: To tune a radio set to a particular station, we adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive. The phenomenon involved is resonance. Resonance is a special case of forced vibrations. When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
Long Answer Type Questions
1. (a) Draw a graph between displacement and time for a body executing natural vibrations.
Answer: (a) A graph between displacement and time for a body executing natural vibrations shows displacement on the y-axis and time on the x-axis. The graph is a sine wave with constant amplitude +a and -a, and a constant period.
1. (b) Where can a body execute natural vibrations?
Answer: (b) A body can execute natural vibrations of constant amplitude only in vacuum.
2. The diagram ishows three ways in which a string of length l in an instrument can vibrate.
(a) Which of the diagram shows the principal note?
Answer: (a) Diagram (i) shows the principal note.
(b) Which vibration has a frequency four times that of the first?
Answer: (iii) has the frequency four times that of the first.
(c) Which vibration is of the longest wavelength?
Answer: (c) Vibration in diagram (i) is of the longest wavelength.
(d) What is the ratio of frequency of vibrations in diagrams (i) and (ii)?
Answer: (d) The ratio of frequency of vibrations in diagrams (i) and (ii) is 1 : 2.
3. Explain why strings of different thickness are provided on a stringed instrument.
Answer: Strings of different thickness are provided on a stringed instrument because the natural frequency of vibration of a stretched string is inversely proportional to the radius (or thickness) of the string (f ∝ 1/r). So notes of different frequencies can be obtained by producing vibrations in different strings of different thicknesses.
4. What are damped vibrations ? How do they differ from free vibrations ? Give one example of each.
Answer: Damped vibrations are the periodic vibrations of a body of decreasing amplitude in presence of a resistive force.
They differ from free (or natural) vibrations in several ways:
- The amplitude of natural vibrations remains constant (in vacuum) and the vibrations continue forever, whereas the amplitude of damped vibrations gradually decreases with time and ultimately the vibrations cease.
- There is no loss of energy in natural vibrations, while in each vibration of damped vibrations, there is some loss of energy in the form of heat.
- In natural vibrations, no external force acts on the vibrating body other than the restoring force, whereas in damped vibrations, in addition to the restoring force, a frictional or damping force acts on the body to oppose its motion.
- The frequency of natural vibrations depends on the size and shape of the body and it remains constant, while the frequency of damped vibrations is less than the natural frequency, and the decrease in frequency depends on the damping force.
An example of damped vibrations is when a slim branch of a tree is pulled and then released, it makes damped vibrations.
5. The diagram shows the displacement-time graph of a vibrating body.
(i) Name the kind of vibrations.
Answer: (i) The kind of vibrations shown are damped vibrations.
(ii) Give one example of such vibrations.
Answer: (ii) An example of such vibrations is a simple pendulum oscillating in air.
(iii) Why is the amplitude of vibrations gradually decreasing?
Answer: (iii) The amplitude of vibrations is gradually decreasing because the energy of the vibrating body continuously dissipates in doing work against the force of friction (resistive force) due to the surrounding medium.
(iv) What happens to the vibrations of the body after some time?
Answer: (iv) After some time, when the body has lost all its energy, it stops vibrating.
6. Draw a sketch showing displacement against time for a body executing damped vibrations.
Answer: A sketch showing displacement against time for a body executing damped vibrations shows displacement on the y-axis and time on the x-axis. The graph is a wave whose amplitude (+a and -a initially) gradually decreases with time until it becomes zero.
7. What is meant by resonance ? Describe a simple experiment to illustrate the phenomenon of resonance and explain it.
Answer: Resonance is a special case of forced vibrations. When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
A simple experiment to illustrate resonance involves two identical tuning forks A and B of the same frequency, mounted on two separate sound boxes with their open ends facing each other. When the prong of tuning fork A is struck on a rubber pad, it starts vibrating. On putting tuning fork A on its sound box, it is found that the other tuning fork B also starts vibrating and a loud sound is heard.
Explanation: The vibrating tuning fork A produces forced vibrations in the air column of its sound box. These vibrations are of large amplitude because of the large surface area of air in the sound box. They are communicated to the sound box of fork B. The air column of B starts vibrating with the frequency of fork A. Since the frequency of these vibrations is the same as the natural frequency of fork B, fork B readily picks up these vibrations and starts vibrating under resonance. Thus the two sound boxes help in communicating the vibrations and in increasing the amplitude of vibrations.
8. The figure shows two tuning forks A and B of the same frequency mounted on two separate sound boxes with their open ends facing each other. The fork A is set into vibration. (a) Describe your observation. (b) State the principle illustrated by this experiment.
Answer: (a) When fork A is set into vibration, it is observed that the other tuning fork B also starts vibrating and a loud sound is heard.
(b) The principle illustrated by this experiment is resonance. The vibrating tuning fork A produces forced vibrations in the air column of its sound box. These vibrations are communicated to the sound box of fork B. The air column of B starts vibrating with the frequency of fork A. Since the frequency of these vibrations is the same as the natural frequency of fork B, the fork B readily picks up these vibrations and starts vibrating under resonance.
9. In the figure, A, B, C and D are four pendulums suspended from the same elastic string XY. The lengths of pendulum A and D are equal, while the length of pendulum B is shorter and of the pendulum C is longer. Pendulum A is set into vibrations.
(a) What is your observation about the vibrations of pendulum D ?
Answer: (a) Pendulum D also starts vibrating initially with a small amplitude but gradually it acquires the same amplitude as pendulum A initially had. When the amplitude of pendulum D becomes maximum, the amplitude of pendulum A becomes minimum. This process continues.
(b) Give reason for your observation in part (a).
Answer: (b) The vibrations produced in pendulum A are communicated as forced vibrations to pendulum D through the rubber string XY. Pendulum D comes in the state of resonance, because the natural frequency of pendulum D is equal to that of A (being of the same length), and therefore there is an exchange of energy between the pendulums A and D.
(c) What type of vibrations take place in pendulums B and C ?
Answer: (c) Pendulums B and C vibrate with a very small amplitude. They remain in the state of forced vibrations.
(d) Give reason for the answer in part (c).
Answer: (d) The natural frequencies of pendulums B (shorter) and C (longer) are different from the natural frequency of pendulum A. Therefore, they do not come into resonance with pendulum A and only execute forced vibrations of small amplitude.
10. A vibrating tuning fork, held over an air column of a given length with its one end closed, produces a loud audible sound. Name the phenomenon responsible for it and explain the observation.
Answer: The phenomenon responsible is resonance.
Explanation: The vibrating tuning fork held just above the mouth of the tube A (containing the air column) produces forced vibrations in the air column of tube A. When the natural frequency of the air column becomes equal to the frequency of the tuning fork, the vibrations of the air column are in resonance with those of the tuning fork. At this point, a loud sound is heard because the air column vibrates with a large amplitude.
11. In the figure, A, B, C and D represent the test tubes each of height 20 cm which are filled with water up to heights of 12 cm, 14 cm, 16 cm and 18 cm respectively. If a vibrating tuning fork is placed over the mouth of test tube D, a loud sound is heard.
(a) Describe the observations with the tubes A, B and C when the vibrating tuning fork is placed over the mouth of each of these tubes.
Answer: (a) No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.
(b) Give the reason for your observation in each tube.
Answer: (b) Frequency of air column in tube D is equal to the frequency of the tuning fork. Resonance occurs with the air column in tube B whereas no resonance occurs with the air column of tubes A and C. The frequency of vibrations of air column in tube B is same as the frequency of vibrations of air column in tube D because the length of air column in tube D is 20 – 18 = 2 cm and that in tube B is 20 – 14 = 6 cm (i.e., three times). On the other hand, the frequency of vibrations of air column in tubes A and C are not equal to the frequency of vibrations of air column in tube D.
(c) State the principle illustrated by the above experiment.
Answer: (c) The principle illustrated is resonance. When the frequency of vibrations of air column is equal to the frequency of the vibrating tuning fork, resonance occurs.
Exercise (C)
MCQ
1. The intensity of a sound wave in air is proportional to the :
(a) square of amplitude of vibrations
(b) square of frequency of vibrations
(c) density of air
(d) all of the above
Answer: (d) all of the above
2. For normal ears, sensitivity is maximum at the frequency:
(a) 5 kHz
(b) 10 kHz
(c) 1 kHz
(d) 500 Hz
Answer: (c) 1 kHz
3. By reducing the amplitude of a sound wave, its :
(a) pitch decreases
(b) loudness decreases
(c) loudness increases
(d) pitch increases
Answer: (b) loudness decreases
4. Sounds above ……………….. are termed as noise pollution.
(a) 75 dB
(b) 120 dB
(c) 100 dB
(d) 80 dB
Answer: (b) 120 dB
5. Two sounds of the same loudness and same pitch produced by two different instruments differ in their:
(a) amplitudes
(b) frequencies
(c) waveforms
(d) all of the above
Answer: (c) waveforms
6. Instruments A and B are producing sounds of the same amplitude, same waveform but frequencies F and 2F respectively. Which of the following statements is correct?
(a) both A and B differ in quality
(b) B is grave but A is shrill
(c) A is grave but B is shrill
(d) B is louder than A
Answer: (c) A is grave but B is shrill
7. Which of the following statements is incorrect with respect to music ?
(a) It is pleasant to the ears.
(b) Its waveform is irregular.
(c) The sound level is low.
(d) It is produced by periodic vibrations.
Answer: (b) Its waveform is irregular.
8. Out of the following, which instrument emits a monotone ?
(a) guitar
(b) flute
(c) piano
(d) tuning fork
Answer: (d) tuning fork
9. The pitch of a note depends on:
(a) frequency
(b) amplitude
(c) waveform
(d) loudness
Answer: (a) frequency
10. The voice of a woman is of …………. pitch than that of a man.
(a) lower
(b) higher
(c) equal
(d) cannot say
Answer: (b) higher
11. The quality of a musical sound depends on :
(a) number of subsidiary notes
(b) relative amplitudes
(c) both (a) and (b)
(d) none of the above
Answer: (c) both (a) and (b)
12. Which statement about sound waves is incorrect?
(a) Sound waves can be reflected, refracted and transmitted.
(b) High pitch sound waves have a high frequency.
(c) Quiet sound waves have a large amplitude.
(d) Sound waves are longitudinal waves.
Answer: (c) Quiet sound waves have a large amplitude.
13. Assertion (A): When we start filling an empty pitcher with water, the pitch of the sound produced goes on decreasing.
Reason (R): The pitch of sound increases with an increase in frequency.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
Very Short Answer Type Questions
1. (a) Which of the following quantity determines the loudness of a sound wave?
(i) wavelength, (ii) frequency, and (iii) amplitude.
Answer: (iii) amplitude.
(b) How is loudness related to the quantity mentioned above in part (a)?
Answer: Loudness is proportional to the square of the amplitude (loudness ∝ (amplitude)²).
2. If the amplitude of a wave is doubled, what will be the effect on its loudness?
Answer: Its loudness will become four times.
3. Two waves of the same pitch have amplitudes in the ratio 1:3. What will be the ratio of their (i) loudness, and (ii) frequencies?
Answer: (i) The ratio of their loudness will be 1:9.
(ii) The ratio of their frequencies will be 1:1.
4. Name the unit in which loudness of sound is measured.
Answer: The unit in which loudness of sound is measured is phon.
5. Name the unit used to measure the sound level.
Answer: The unit used to measure the sound level is decibel (dB).
6. What is the safe limit of sound level in dB for our ears?
Answer: The safe limit of sound level in dB for our ears is up to 80 dB.
7. Name the subjective property of sound related to its frequency.
Answer: The subjective property of sound related to its frequency is pitch.
8. The frequencies of notes given by flute, guitar and trumpet are respectively 400 Hz, 200 Hz and 500 Hz. Which one of these has the highest pitch?
Answer: The trumpet, with a frequency of 500 Hz, has the highest pitch.
9. Complete the following sentences:
(a) The pitch of sound increases if its frequency ……………..
Answer: The pitch of sound increases if its frequency increases.
(b) If the amplitude of a sound is halved, its loudness becomes ……………..
Answer: If the amplitude of a sound is halved, its loudness becomes one-fourth.
10. Name the characteristic which enables one to distinguish the sound of two musical instruments even if they are of the same pitch and same loudness.
Answer: The characteristic is quality (or timbre).
11. Which characteristic of sound makes it possible to recognise a person by his voice without seeing him?
Answer: The quality of sound makes it possible to recognise a person by his voice without seeing him.
12. State the factor that determines
(i) the pitch of a note,
(ii) the loudness of the sound heard, and
(iii) the quality of the note.
Answer: (i) The factor that determines the pitch of a note is its frequency.
(ii) The factor that determines the loudness of the sound heard is its amplitude.
(iii) The factor that determines the quality of the note is its wave form.
13. Name the characteristic of sound affected due to a change in its (i) amplitude, (ii) wave form, and (iii) frequency.
Answer: (i) A change in amplitude affects the loudness of sound.
(ii) A change in wave form affects the quality (or timbre) of sound.
(iii) A change in frequency affects the pitch of sound.
14. Sketches I to IV show sound waves, all formed in the same time interval. Which diagram shows
(i) a note from a musical instrument,
(ii) a soft (or feeble) note,
(iii) a bass (or low frequency) note.
Answer: (i) Diagram IV shows a note from a musical instrument.
(ii) Diagram I shows a soft (or feeble) note.
(iii) Diagram II shows a bass (or low frequency) note.
15. The figure shows wave patterns of three sounds A, B and C. Name the characteristic of sound which is the same between (i) A and B, (ii) B and C, and (iii) C and A.
Answer: (i) Between A and B, loudness and quality are the same.
(ii) Between B and C, none of the characteristics (loudness, pitch, quality) are the same.
(iii) Between C and A, pitch is the same.
Short Answer Type Questions
1. Name three characteristics of a musical sound.
Answer: Three characteristics of a musical sound are (1) loudness, (2) pitch (or shrillness), and (3) quality (or timbre).
2. Why is the loudness of the sound heard by a plucked wire increased when it is mounted on a sound board?
Answer: When a plucked wire is mounted on a sound board, the sound board is set into forced vibrations. Since the sound board has a much larger vibrating area than the wire, it sets a larger volume of air into vibration. This results in a greater amount of sound energy being radiated, and hence the loudness of the sound increases.
3. Define the term intensity of a sound wave. State the unit in which it is measured.
Answer: The intensity of a sound wave at a point of the medium is the amount of sound energy passing per second normally through a unit area at that point. Its unit is watt per metre² (W m⁻²).
4. How is loudness of sound related to the intensity of wave producing it?
Answer: Loudness of sound depends on its intensity. Experimentally, the relationship between loudness L and intensity I is given by L = K log₁₀ I, where K is a constant of proportionality. This means loudness increases with the increase in intensity, but not in the same proportion.
5. Comment on the statement ‘loudness of sound is of subjective nature, while intensity is of objective nature.’
Answer: Loudness of sound is subjective because it depends not only on the intensity of the sound wave but also on the sensitivity of the listener’s ears. Sounds of the same intensity might appear to have different loudness to different people, or even to the same person if the frequencies are different, due to varying ear sensitivity. Intensity, on the other hand, is an objective, measurable quantity representing the sound energy passing per unit area per second.
6. State three factors which affect the loudness of a sound heard by a listener.
Answer: Three factors that affect the loudness of a sound heard by a listener are:
(i) The amplitude of vibrations (loudness is proportional to the square of the amplitude).
(ii) The distance from the source (loudness varies inversely as the square of the distance).
(iii) The surface area of the vibrating body (a larger surface area produces a louder sound).
7. The bells of a temple are big in size. Why?
Answer: The bells of a temple are made big in size because loudness depends on the surface area of the vibrating body. A larger vibrating area, as in a big bell, sends forth a greater amount of energy, and hence the sound heard is louder.
8. What is meant by noise pollution? Name one source of sound causing noise pollution.
Answer: Noise pollution is the disturbance produced in the environment due to undesirable loud and harsh sound, typically of a level above 120 dB. One source of sound causing noise pollution is a loudspeaker (or siren, moving vehicles).
9. What determines the pitch of a sound?
Answer: The pitch of a sound or note depends on its frequency.
10. Name and define the characteristic which enables one to distinguish two sounds of the same loudness, but of different frequencies, given by the same instrument.
Answer: The characteristic is pitch. Pitch is that characteristic of sound by which an acute (or shrill) note can be distinguished from a grave (or flat) note of the same loudness and quality. Since the frequencies are different, their pitches will be different.
11. How is it possible to detect the filling of a bottle under a water tap by hearing the sound at a distance?
Answer: As the water level in a bottle under a water tap rises, the length of the air column above the water level decreases. This causes the frequency of the sound produced by the vibrating air column to increase, making the sound become shriller. By hearing this change in shrillness from a distance, one can get an idea of the water level in the bottle.
12. Two identical guitars are played by two persons to give notes of the same loudness and pitch. Will they differ in quality? Give a reason for your answer.
Answer: Yes, they can differ in quality. The reason is that the quality of a musical sound depends on the number of subsidiary notes (overtones) and their relative amplitudes present along with the principal (fundamental) vibration. Different persons might pluck the strings of identical guitars in slightly different ways, exciting different combinations or relative amplitudes of these subsidiary vibrations, even if the fundamental pitch and overall loudness are the same. This results in different wave forms and hence different quality.
13. State one difference between a musical note and noise.
Answer: A musical note is produced by regular, periodic vibrations and has a regular wave form, whereas noise is produced by an irregular succession of disturbances and has an irregular wave form.
Long Answer Type Questions
1. How does the wave pattern of a loud note differ from that of a soft note? Draw a diagram.
Answer: The wave pattern of a loud note differs from that of a soft note primarily in its amplitude. A loud note corresponds to a wave of large amplitude, while a soft note corresponds to a wave of small amplitude. If both notes have the same pitch and quality, their frequency and basic wave form will be the same.
2. Draw a diagram to show the wave pattern of a high pitch note and a low pitch note, but of the same loudness.
Answer: A high pitch note has a higher frequency (more waves per unit time or shorter wavelength) compared to a low pitch note, which has a lower frequency (fewer waves per unit time or longer wavelength). If they have the same loudness, their amplitudes will be the same.
3. The diagram below shows three different modes of vibration P, Q and R of the same string of a given length.
(a) Which vibration will produce a louder sound and why?
Answer: Vibration R will produce a louder sound because its amplitude is maximum, and loudness is proportional to the square of the amplitude.
(b) Which vibration will produce a sound of maximum shrillness (or pitch) and why?
Answer: Vibration P will produce a sound of maximum shrillness (or pitch) because its frequency is maximum (it has the most number of loops, indicating a higher harmonic).
(c) What is the ratio of the wavelength of vibrations P and R?
Answer: The ratio of the wavelength of vibration P to vibration R (λₚ : λᵣ) is 1 : 3.
4. How do the two sounds of same loudness and same pitch produced by different instruments differ? Draw diagrams to illustrate your answer.
Answer: Two sounds of the same loudness (same amplitude) and same pitch (same fundamental frequency) produced by different instruments differ in their quality or timbre. This difference is due to their different wave forms, which arise because different instruments produce different combinations of subsidiary notes (overtones or harmonics) along with the principal (fundamental) note. The number and relative amplitudes of these subsidiary notes determine the complexity and shape of the wave form.
5. Two musical notes of the same pitch and same loudness are played on two different instruments. Their wave patterns are shown in the figure. Explain why are the wave patterns different.
Answer: The wave patterns are different because, although the notes have the same pitch (same fundamental frequency) and same loudness (same overall amplitude), the two different instruments (Violin and Piano) produce different combinations of subsidiary vibrations or overtones along with the fundamental frequency. The number and relative intensities of these overtones vary from instrument to instrument, leading to a unique mixture of frequencies. This mixture determines the overall shape of the sound wave, which is its wave form. Thus, the different wave forms reflect the different quality or timbre of the sound produced by each instrument.
6. A microphone is connected to the Y-input of a C.R.O. Three different sounds are made in turn in front of the microphone. Their traces (a), (b) and (c) produced on the screen are shown in the figure.
(i) Which trace is due to the loudest sound? Give reason for your answer.
Answer: Trace (b) is due to the loudest sound because its amplitude is the largest, and loudness of a sound is directly related to the amplitude of the sound wave.
(ii) Which trace is due to the sound with the lowest pitch? Explain your answer.
Answer: Trace (a) is due to the sound with the lowest pitch because its frequency is the lowest. This is evident as it shows the fewest number of complete waves in the given time interval compared to traces (b) and (c).
7. In what respect does the wave pattern of noise and music differ? Draw diagrams to explain your answer.
Answer: The wave pattern of music differs from that of noise primarily in its regularity and periodicity. Music is produced by regular, periodic vibrations, resulting in a smooth and regular wave form. The component waves in music are generally similar without sudden changes in their wavelength and amplitude.
Noise, on the other hand, is produced by an irregular succession of disturbances, resulting in a harsh, discordant, and irregular wave form. The component waves in noise often change their character suddenly and are of short duration.
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