Book Back Questions with Answers - Acoustics | Science
I. Choose the correct answer
1. When a sound wave travels through air, the air particles
2. Velocity of sound in a gaseous medium is 330 m s–1. If the pressure is increased by 4 times without causing a change in the temperature, the velocity of sound in the gas is
3. The frequency, which is audible to the human ear is
4. The velocity of sound in air at a particular temperature is 330 m s–1. What will be its value when temperature is doubled and the pressure is halved?
5. If a sound wave travels with a frequency of 1.25 × 104 Hz at 344 m s–1, the wavelength will be
6. The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?
7. Velocity of sound in the atmosphere of a planet is 500 m s–1. The minimum distance between the sources of sound and the obstacle to hear the echo, should be
II. Fill up the blanks
- Rapid back and forth motion of a particle about its mean position is called Vibration.
- If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating in both north and south.
- A whistle giving out a sound of frequency 450 Hz, approaches a stationary observer at a speed of 33 m s–1. The frequency heard by the observer is (speed of sound = 330 m s–1) 500 Hz.
- A source of sound is travelling with a velocity 40 km/h towards an observer and emits a sound of frequency 2000 Hz. If the velocity of sound is 1220 km/h, then the apparent frequency heard by the observer is 2068 Hz.
III. True or false (If false give the reason)
-
Sound can travel through solids, gases, liquids and even vacuum. - False
Reason: Sound waves cannot travel through vacuum.
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Waves created by Earth Quake are Infrasonic. - True
-
The velocity of sound is independent of temperature. - False
Reason: The velocity of sound is dependent on temperature.
-
The Velocity of sound is high in gases than liquids. - False
Reason: The velocity of sound is high in liquids than gases.
IV. Match the following
Answers:
V. Assertion and Reason Questions
Mark the correct choice as
a. If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
b. If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
c. Assertion is true, but the reason is false.
d. Assertion is false, but the reason is true.
1. Assertion: The change in air pressure affects the speed of sound.
Reason: The speed of sound in a gas is proportional to the square of the pressure.
2. Assertion: Sound travels faster in solids than in gases.
Reason: Solid posses a greater density than that of gases.
VI. Answer very briefly
1. What is a longitudinal wave?
It is a wave in which particles are vibrating along the direction of wave motion.
2. What is the audible range of frequency?
The audible range of frequency is 20 to 20,000 Hz.
3. What is the minimum distance needed for an echo?
The minimum distance needed for an echo 17.2 m.
4. What will be the frequency sound having 0.20 m as its wavelength, when it travels with a speed of 331 m s–1?
Frequency Sound = n
Wavelength λ = 0.20 m
Frequency of sound, $$n = \frac{V}{\lambda}$$
$$n = \frac{331}{0.20}$$
$$n = 1655 \text{ Hz}$$
5. Name three animals, which can hear ultrasonic vibrations.
Dogs, Bats and Mosquitoes.
VII. Answer briefly
1. Why does sound travel faster on a rainy day than on a dry day?
During rainy days, the humidity is more in the atmosphere. The speed of the sound generally increases with humidity and speed of sound in water is more than 4X then speed in air. So sound travels faster on a rainy day.
2. Why does an empty vessel produce more sound than a filled one?
In an empty vessel, only air is present inside it. Whenever sound is produced in an empty vessel, the vibration of air molecule will be more due to multiple reflections. But a filled vessel has very less number of air molecules than the contents in it, so it does not produce more sound.
3. Air temperature in the Rajasthan desert can reach 46°C. What is the velocity of sound in air at that temperature? (V0 = 331 m s–1)
Velocity of sound in gas at 0° C
V0 =331 ms-1
Air temperature in Rajasthan, T = 46° C
The formula for velocity at temperature T is:
$$V_T = (V_o + 0.61T)$$
$$V_T = 331 + (0.61 \times 46)$$
$$V_T = 331 + 28.06$$
$$V_T = 359.06 \text{ m/s}$$
4. Explain why, the ceilings of concert halls are curved.
They are made curved so that the sound after reflecting from the ceiling reaches every corner of the concert hall and the audience listen the sound clearly.
5. Mention two cases in which there is no Doppler effect in sound?
(i) When source(S) and listener (L) both are at rest.
(ii) When S and L move in such a way that distance between them remains constant.
(iii) When source S and L are moving in mutually perpendicular directions.
VIII. Problem Corner
1. A sound wave has a frequency of 200 Hz and a speed of 400 m s–1 in a medium. Find the wavelength of the sound wave.
Given
Frequency of a wave, n = 200 Hz
Speed of sound, V = 400 ms-1
To find
Wavelength, λ = ?
Solution
Velocity of light, $$V = n\lambda$$ $$\lambda = \frac{V}{n} = \frac{400}{200}$$ $$\lambda = 2 \text{ m}$$
2. The thunder of cloud is heard 9.8 seconds later than the flash of lightning. If the speed of sound in air is 330 m s–1, what will be the height of the cloud?
Given
Time, t = 9.8 s
Speed of sound, V = 330 ms-1
To find
Height of the cloud, d = ?
Solution
We know, $$V = \frac{d}{t}$$ Height of the cloud, $$d = V \times t$$ $$d = 330 \times 9.8$$ $$d = 3234 \text{ m}$$
3. A person who is sitting at a distance of 400 m from a source of sound is listening to a sound of 600 Hz. Find the time period between successive compressions from the source?
Given
Frequency of sound, n = 600 Hz
To find
Time period between successive compressions, T = ?
Solution
$$T = \frac{1}{n} = \frac{1}{600} = 0.00166 \text{ s}$$ Time period, T ≈ 0.0017 s
4. An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval between the transmission and reception of the wave is 1.6 seconds. What is the depth of the sea, if the velocity of sound in the seawater is 1400 m s–1?
Given
Time interval between sending and receiving of the wave, t = 1.6 s
Velocity of sound in sea wave, V = 1400 ms-1
To find
Depth of the sea, d = ?
Solution
$$2d = V \times t$$ Depth of the sea, $$d = \frac{1400 \times 1.6}{2}$$ $$d = \frac{2240}{2}$$ $$d = 1120 \text{ m}$$
5. A man is standing between two vertical walls 680 m apart. He claps his hands and hears two distinct echoes after 0.9 seconds and 1.1 second respectively. What is the speed of sound in the air?
Given
Time of first echo, t1 = 0.9 s
Time of second echo, t2 = 1.1 s
Total distance between walls = 680 m. The total distance travelled by sound for both echoes is 2d, where d is the distance between the walls.
To find
Speed of sound in air, V = ?
Solution
The total time for sound to travel to both walls and back to the man is t = t1 + t2. The total distance covered is 2d.
Speed of sound in air, $$V = \frac{2d}{t_1+t_2}$$
$$V = \frac{2 \times 680}{0.9 + 1.1}$$
$$V = \frac{1360}{2}$$
Speed of sound in air, V = 680 m/s
6. Two observers are stationed in two boats 4.5 km apart. A sound signal sent by one, under water, reaches the other after 3 seconds. What is the speed of sound in the water?
Given
Distance between two observers, d = 4.5 km = 4500 m
Time taken to reach underwater, t = 3 s
To find
Speed of sound, V = ?
Solution
Speed of sound in water, $$V = \frac{d}{t}$$ $$V = \frac{4.5 \text{ km}}{3 \text{ s}} = 1.5 \text{ km/s}$$ or $$V = \frac{4500 \text{ m}}{3 \text{ s}} = 1500 \text{ m/s}$$
7. A strong sound signal is sent from a ship towards the bottom of the sea. It is received back after 1s.What is the depth of sea given that the speed of sound in water 1450 m s–1?
Solution
Total time for signal to go and come back = 1 s.
Time taken by the signal to reach the bottom of the sea, t = 1/2 s = 0.5 s.
Speed of sound in water, V = 1450 m/s
Depth of the sea (distance travelled by signal), d = Speed × Time
$$d = V \times t$$
$$d = 1450 \times 0.5$$
Depth of the sea, d = 725 m
IX. Answer in Detail
1. What are the factors that affect the speed of sound in gases?
(i) Effect of density: Velocity of sound in gas is inversely proportional to the square root of density of the gas. Hence, the velocity decreases as the density of the gas increases.
$$v \propto \sqrt{\frac{1}{d}}$$
(ii) Effect of temperature: Velocity of sound in a gas is directly proportional to the square root of its temperature. Velocity of sound in a gas increases with increase in temperature, $$v \propto \sqrt{T}$$. Velocity at temperature T is given by the following equation.
$$v_T = (v_0 + 0.61 T) \text{ ms}^{-1}$$
Here, v0 is the velocity of sound in the gas at 0° C. For air, v0 = 331 ms-1. Hence, the velocity of sound changes by 0.61 ms-1 when temperature changes by each degree celcius.
(iii) Effect of relative humidity: When humidity increases, the speed of sound increases. That is why we can hear sound from long distances clearly during rainy seasons.
2. What is mean by reflection of sound? Explain:
a) reflection at the boundary of a rarer medium
b) reflection at the boundary of a denser medium
c) Reflection at curved surfaces
When sound waves travel in a given medium and strike the surface of another medium, it can be bounced back into the first medium is called as reflection.
a) Reflection in rarer medium:
(i) Consider a wave travelling in a solid medium striking on the interface between the solid and the air.
(ii) The compression exerts a force F on the surface of the rarer medium.
(iii) As a rarer medium has smaller resistance for any deformation, the surface of separation is pushed backwards.
(iv) As the particles of the rarer medium are free to move, a rarefaction is produced at the interface.
(v) Thus, a compression is reflected as a rarefaction and a rarefaction travels from right to left.
b) Reflection in denser medium:
(i) A longitudinal wave travels in a medium in the form of compressions and rarefactions.
(ii) Suppose a compression travelling in air from left to right reaches a rigid wall. The compression exerts a force F on the rigid wall.
(iii) In turn, the wall exerts an equal and opposite reaction R = - F on the air molecules. This results in a compression near the rigid wall.
(iv) Thus, a compression travelling towards the rigid wall is reflected back as a compression. That is, the direction of compression is reversed.
c) Reflection in curved surfaces:
(i) When sound waves are reflected from curved surfaces, the intensity of reflected waves is changed.
(ii) When reflected from a convex surface, the reflected waves are diverged out and the intensity is decreased.
(iii) When sound is reflected from a concave surface, the reflected waves are converged and focused at a point. So the intensity of reflected waves is concentrated at a point.
(iv) Parabolic surfaces are used when it is required to focus the sound at a particular point.
(v) Hence, many halls are designed with parabolic reflecting surfaces.
(vi) In elliptical surfaces, sound from one focus will always be reflected to the other focus, no matter where it strikes the wall.
3. a) What do you understand by the term ‘ultrasonic vibration’?
b) State three uses of ultrasonic vibrations.
c) Name three animals which can hear ultrasonic vibrations.
a) Ultrasonic vibrations: These are high- frequency sound waves beyond the range of human hearing. The frequency, for these sound waves, is greater than 20 kHz.
b) Uses:
(i) To kill micro organisms.
(ii) To find direction and range of submarines.
(iii) To clean dental plates, jewellery and coins.
(iv) For welding.
c) Animals which can hear Ultrasonic vibrations:
(i) Dogs,
(ii) Bats,
(iii) Dolphins.
4. What is an echo?
a) State two conditions necessary for hearing an echo.
b) What are the medical applications of echo?
c) How can you calculate the speed of sound using echo?
An echo is the sound reproduced due to the reflection of original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc.
a) Conditions necessary for hearing an echo:
(i) The minimum time gap between the original sound and echo must be 0.1s.
(ii) The minimum distance required to hear an echo is 17.2 m.
b) The medical applications of echo:
The principle of echo is used in obstetric ultrasonography, which is used to create real-time visual images of the developing embryo or fetus in the mother’s uterus.
c) Calculate the speed of sound using echo:
(i) The sound pulse emitted by the source travels a total distance of 2d while traveling from the source to the wall and then back to the receiver.
(ii) The time taken for this has been observed to be “t”. Hence, the speed of sound wave is given by:
$$\text{Speed of sound} = \frac{\text{Distance travelled}}{\text{Time taken}} = \frac{2d}{t}$$
X. HOT Questions
1. Suppose that a sound wave and a light wave have the same frequency, then which one has a longer wavelength?
2. When sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound remain the same. Do you hear an echo sound on a hotter day? Justify your answer.
(i) An echo can only be heard if it reaches the ear after 0.1s. Time taken = Total distance / Velocity.
(ii) If the temperature rises (i.e. on a hotter day), the velocity of sound will increase. This in turn will decrease the time required for the sound to travel the same distance. If this time becomes less than 0.1s, an echo will not be heard.
Concept Map
