5. Acoustics

I. Choose the best Answer

1. When a sound wave travels through air, the air particles _____.
(a) vibrate along the direction of the wave motion
(b) vibrate but not in any fixed direction
(c) vibrate perpendicular to the direction of the wave motion
(d) do not vibrate.

2. Velocity of sound in a gaseous medium is 330 ms^-1. If the pressure is increased by 4 times without causing a change in the temperature, the velocity of sound in the gas is:
(a) 330 ms^-1
(b) 660 ms¹
(c) 156 ms^-1
(d) 990 ms^-1

3. The frequency, which is audible to the human ear is _____.
(a) 50 kHz
(b) 20 kHz
(c) 15000 kHz
(d) 10000 kHz.

4. The velocity of sound in air at a particular temperature is 330 ms^-1. What will be its value when temperature is doubled and the pressure is halved?
(a) 330 ms^-1
(b) 165 ms^-1
(c) 330 × √2 ms^-1
(d) 320 × √2 ms^-1

5. If a sound wave travels with a frequency of 1.25 × 10⁴ Hz at 344 ms^-1, the wavelength will be _____.
(a) 27.52 m
(b) 275.2 m
(c) 0.02752 m
(d) 2.752 m.

6. The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?
(a) speed
(b) frequency
(c) wavelength
(d) none of these.

7. Velocity of sound in the atmosphere of a planet is 500 ms^-1. The minimum distance between the sources of sound and the obstacle to hear the echo, in point Is is:
(a) 17 m
(b) 20 m
(c) 25 m
(d) 50 m

II. Fill in the blanks

1. The rapid back and forth motion of a particle about its mean position is called vibrations.

2. 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.

3. A whistle giving out a sound of frequency 450 Hz, approaches a stationary observer at a speed of 33 ms^-1. The frequency heard by the observer is (speed of sound = 330 ms^-1) 500 Hz.

4. 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

1. Sound can travel through solids, gases, liquids and even vacuum.
Answer: False.
Reason: Sound can not travel in a vacuum. This is because sound waves are vibrating waves. In vacuum, where there are no atoms or molecules to vibrate.

2. Waves created by Earth Quake are Infrasonic.
Answer: True.

3. The velocity of sound is independent of temperature.
Answer: False.
Reason: For an ideal gas the velocity of sound depends on its temperature and is independent of gas pressure.

4. The velocity of sound is high in gases than liquids.
Answer: False.
Reason: Velocity of a sound wave is maximum in solids because they are more elastic in nature than liquids and gases. Since gases are least elastic in nature, the velocity of sound is the least in a gaseous medium.
So V_S > V_L > V_G.

IV. Match the following

1. Infrasonic	(c) 10 Hz
2. Echo	(d) Ultrasonography
3. Ultrasonic	(b) 22 kHz
4. High-pressure region	(a) Compressions

V. Assertion and Reason Type 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) The assertion is true, but the reason is false.
(d) The 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
Answer:
(d) The assertion is false, but the reason is true.

2. Assertion: Sound travels faster in solids than in gases.
Reason: Solid posses a greater density than that of gases.
Answer:
(b) If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.

VI. Answer very briefly

1. What is a longitudinal wave?

If the particles of the medium vibrate along the direction of propagation of the wave.

2. What is the audible range of frequency?

Audible range of frequency is from 20 Hz to 20000 Hz.

3. What is the minimum distance needed for an echo?

The minimum distance required to hear an echo is 1/20th part of the magnitude of the velocity of sound in air.

If you consider the velocity of sound as 344 ms^-1, the minimum distance required to hear an echo is 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?
Solution:

5. Name three animals, which can hear ultrasonic vibrations.
Mosquitos, bats and dogs are the three animals that can hear ultrasonic vibrations.

VII. Answer briefly

1. Why does sound travel faster on a rainy day than on a dry day?

When humidity increases, the speed of sound increases. That is why you can hear sound from long distances clearly during rainy seasons.

2. Why does an empty vessel produce more sound than a filled one?

In an empty vessel, multiple reflections of sound takes place. Hence more sound is produced in an empty vessel than tilled one.

3. Air temperature in the Rajasthan desert can reach 46°C. What is the velocity of sound in air at that temperature? (v₀ = 331 ms^-1)

Solution:

Velocity of sound, v₀ = 331 ms^-1
Air temperature, T = 46° C
Velocity of sound in air temperature v_T = (v₀ + 0.61T) ms^-1
= 331 + (0.61 × 46)
= 331 + 28.06
v_T = 359.06 ms^-1.

4. Explain why the ceilings of concert halls are curved.

The ceilings of concert halls are curved so that multiple reflections of sound waves can take place.
The parabolic surfaces are used to focus the sound at a particular point. Hence sound will be louder.

5. Mention two cases in which there is no Doppler effect in sound?

When source (S) and listener (L) both are at rest.

When S and L move in such a way that distance between them remains constant.
When source S and L are moving in mutually perpendicular directions.
If the source is situated at the centre of the circle along which the listener is moving.

VIII. Problem Corner

1. A sound wave has a frequency of 200 Hz and a speed of 400 ms^-1 in a medium. Find the wavelength of the sound wave.
Solution:

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 ms^-1, what will be the height of the cloud?

Solution:

Speed of sound in air v = 330 m/s
Time to hear thunder t = 9.8 s
∴ Height of the cloud = v × t
= 330 × 9.8
= 3234 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?
Solution:

The time period between successive compressions from the source is 0.00167 seconds.

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 ms^-1?
Solution:

The depth of the sea is 1120 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 seconds respectively. What is the speed of sound in the air?
Solution:

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?
Answer:
Distance of boats d = 4.5 × 10³ m
= 4500 m.
Time t = 3 s
Speed of sound in water = distance / time

= 4500 m / 3 s

= 1500 m/s.

7. A strong sound signal is sent from a ship towards the bottom of the sea. It is received back after 1 s What is the depth of sea given that the speed of sound in water 1450 ms^-1?
Solution:
This question is based on echo, the formula for echo is

Velocity × time/2 = distance

Velocity is 1450 time is 1 s

Just multiply them we got,

1450 / 2 = D

725 = D

The distance is 725 m.

IX. Answer in Detail

1. What are the factors that affect the speed of sound in gases?

In the case of gases, the following factors affect the velocity of sound waves.
Effect of density : The velocity of sound in a gas is inversely proportional to the square root of the density of the gas. Hence, the velocity decreases as the density of the gas increases. v ∝ 1/√d

Effect of temperature : The velocity of sound in a gas is directly proportional to the square root of its temperature. The velocity of sound in a gas increases with the increase in temperature v ∝ √T. Velocity at temperature T is given by the following equation:
V_r = (v_o + 0.61 T)ms^-1.
Here, v_o is the velocity of sound in the gas at 0° C. For air, v_o = 331 ms^-1. Hence, the velocity of sound changes by 0.61 ms^-1 when the temperature changes by one degree Celsius.

Effect of relative humidity : When humidity increases, the speed of sound increases. That is why you can hear sound from long distances clearly during rainy seasons.

2. What is mean by the reflection of sound? Explain.

Reflection of Sound:
The bouncing of sound waves from the interface between two media is termed as the reflection of sound.
(a) Reflection at the boundary of a rarer medium

Consider a wave travelling in a solid medium striking on the interface between the solid and the air. The compression exerts a force F on the surface of the rarer medium.
As a rarer medium has smaller resistance for any deformation, the surface of separation is pushed backwards.
As the particles of the rarer medium are free to move, a rarefaction is produced at the interface. Thus, compression is reflected as rarefaction and a rarefaction travels from right to left.

(b) Reflection at the boundary of a denser medium

A longitudinal wave travels in a medium in the form of compressions and rarefactions. Suppose a compression travelling in the air from left to right reaches a rigid wall.
The compression exerts a force F on the rigid wall. 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.
Thus, a compression travelling towards the rigid wall is reflected back as a compression. That is the direction of compression is reversed.

When the sound waves are reflected from the curved surfaces, the intensity of the reflected waves is changed.
When reflected from a convex surface, the reflected waves are diverged out and the intensity is decreased.
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.
Parabolic surfaces are used when it is required to focus the sound at a particular point. Hence, many halls are designed with parabolic reflecting surfaces.
In elliptical surfaces, sound from one focus will always be reflected 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 are the vibration with frequency greater than 20 KHz. Human ear cannot detect the ultrasonic vibration.

(b) (i) Ultrasonic waves are used in ultrasonography.
(ii) It is used to get signal images of a developing embryo in the mother’s uterus.
(in) They are used to forecast about tsunami and earthquake.

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?
Answer:
Echo: An echo is the sound reproduced due to the reflection of the original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc.

(a) (i) The persistence of hearing for human ears is 0.1 second. This means that we can hear two sound waves clearly if the time interval between the two sounds is at least 0.1 s. Thus, the minimum time gap between the original sound and an echo must be 0.1 s.

(ii) The above criterion can be satisfied only when the distance between the source of the sound and the reflecting surface would satisfy the following equation:

Velocity = distance travelled by sound/ time taken

V = 2d / t

d = vt /2

Since, t = 0.1 second,

then d = V × 0.1/2 = V/20

(b) (i) 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.
(ii) This is a safe testing tool, as it does not use any harmful radiations.

(c) Apparatus required:
A source of sound pulses, a measuring tape, a sound receiver and a stopwatch.
Calculation of speed of sound:
The sound pulse emitted by the source travels a total distance of 2nd while travelling from the source to the wall and then back to the receiver.
The time taken for this has been observed to be ‘t’. Hence, the speed of the sound wave is given by:
Speed of Sound = distance travelled /time taken= 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?
(a) Sound
(b) Light
(c) both (a) and (b)
(d) data not sufficient.
Answer: (b) light
Explanation: The light wave has a longer wavelength because it has much greater speed.

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.

Echo of sound is heard with less velocity and intensity on a hotter day. Because the velocity of sound is directly proportional to temperature of air. On a hotter day temperature will be higher so velocity of sound will be changed by higher values.