Acoustics

Class 10 | Science — Physics | Unit 5

Class 10 Unit 5 Science

UNIT-5 ACOUSTICS

Learning Objectives

By the end of this section, the students will be able to:

• Understand how sound is produced and transmitted.
• Relate the speed of sound, its frequency, and its wavelength.
• Know the speed of sound in various media.
• Explain the factors affecting the speed of sound in a gaseous medium.
• Demonstrate the phenomenon of reflection of sound.
• Determine the speed of sound using the method of echo.
• Understand Doppler Effect.
• Solve numerical problems related to the above topics.

INTRODUCTION

Sound plays a major role in our lives. We communicate with each other mainly through sound. In our daily life, we hear a variety of sounds produced by different sources like humans, animals, vehicle horns, etc etera . Hence, it becomes inevitable to understand how sound is produced, how it is propagated and how you hear the sound from various sources. It is sometimes misinterpreted that acoustics only deals with musical instruments and design of auditoria and concert halls. But acoustics is a branch of physics that deals with production, transmission, reception, control, and effects of sound. You have studied about propagation and properties of sound waves in ninth standard. In this lesson we will study about reflection of sound waves, Echo and Doppler effect.

5.1 SOUND WAVES

When you think about sound, the questions that arise in your minds are: How is sound produced? How does sound reach our ears from various sources? What is sound? Is it a force or energy? Let us answer all these questions.

By touching a ringing bell or a musical instrument while it is producing music, you can conclude that sound is produced by vibrations. The vibrating bodies produce energy in the form of waves, which are nothing but sound waves (Figure 5.1). Suppose you and your friend are on the Moon. Will you be able to hear any sound produced by your friend? As the Moon does not have air, you will not be able to hear any sound produced by your friend. Hence, you understand that the sound produced due to the vibration of different bodies needs a material medium like air, water, steel, etc etera, for its propagation. Hence, sound can propagate through a gaseous medium or a liquid medium or a solid medium.

ACTIVITY 1

Take a squeaky toy or old mobile phone and put it inside a plastic bag. Seal the bag with the help of a candle or with a thread. Fill a bucket with water and place the bag in the water bucket and squeeze the toy or ring the mobile. You will hear a low sound. Now place your ear against the side of the bucket and squeeze the toy or ring the mobile phone again. You will hear a louder sound.

5.1.1 Longitudinal Waves

Sound waves are longitudinal waves that can travel through any medium (solids, liquids, gases) with a speed that depends on the properties of the medium. As sound travels through a medium, the particles of the medium vibrate along the direction of propagation of the wave. This displacement involves the longitudinal displacements of the individual molecules from their mean positions. This results in a series of high- and low-pressure regions called compressions and rarefactions as shown in figure 5.2.

5.1.2 Categories of sound waves based on their frequencies

(1) Audible waves – These are sound waves with frequency ranging between 20 Hz and 20,000 Hz. These are generated by vibrating bodies such as vocal cords, stretched strings etc etera.

(2) Infrasonic waves – These are sound waves with a frequency below 20 Hz that cannot be heard by the human ear. example waves produced during earthquake, ocean waves, sound produced by whales, etcetera.

(3) Ultrasonic waves – These are sound waves with a frequency greater than 20 kHz, Human ear cannot detect these waves, but certain creatures like mosquito, dogs, bats, dolphins can detect these waves. example waves produced by bats.

5.1.3 Difference between the sound and light waves

Serial number.	SOUND	LIGHT
1	Medium is required for the propagation.	Medium is not required for the propagation.
2	Sound waves are longitudinal.	Light waves are transverse.
3	Wavelength ranges from 1.65 cm to 1.65 meter.	Wave length ranges from 4 times 10 power minus 7 meter to 7 times 10 power minus 7 meter.
4	Sound waves travel in air with a speed of about 340 meter per second at NTP.	Light waves travel in air with a speed of 3 times 10 power 8 meter per second.

5.1.4 Velocity of sound waves

When you talk about the velocity associated with any wave, there are two velocities, namely particle velocity and wave velocity. SI unit of velocity is m etre second power minus one.

Particle velocity: The velocity with which the particles of the medium vibrate in order to transfer the energy in the form of a wave is called particle velocity.

Wave velocity: The velocity with which the wave travels through the medium is called wave velocity. In other words, the distance travelled by a sound wave in unit time is called the velocity of a sound wave.

Velocity = distance travelled by time

If the distance travelled by one wave is taken as one wavelength (λ) and, the time taken for this propagation is one time period (T), then, the expression for velocity can be written as

Therefore V is equal to lambta by T equation 5.1

Therefore, velocity can be defined as the distance travelled per second by a sound wave. Since, Frequency (n) is equal to 1, equation (5.1) can be written as

V = n λ equation (5.2)

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 greater than v L greater than V G

5.1.5 Factors affecting velocity of sound

In the case of solids, the elastic properties and the density of the solids affect the velocity of sound waves. Elastic property of solids is characterized by their elastic moduli. The speed of sound is directly proportional to the square root of the elastic modulus and inversely proportional to the square root of the density. Thus, the velocity of sound in solids decreases as the density increases whereas the velocity of sound increases when the elasticity of the material increases.

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 is proportional to square root of one by 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 increase with the increase in temperature. V ∝ √T. Velocity at temperature T is given by the following equation: V T = (v 0 + 0.61 T) meter second power minus one. VT is equal to V0 plus zero point six one T with in the bracket metre second power minus one

Here, V 0 is the velocity of sound in the gas at 0° C. For air, V 0 = 331 meter second power minus one. Hence, the velocity of sound changes by 0.61 metre second power minus one 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.

Speed of sound waves in different media are given in table 5.1

Serial Number	Nature of the medium	Name of the Medium	Speed of sound(in meter second power minus one)
1	Solid	Copper	5010
2	Iron	5950
3	Aluminum	6420
4	Liquid	Kerosene	1324
5	Water	1493
6	Seawater	1533
7	Gas	Air (at 0 degree Celsius)	331
8	Air (at 20 degree celsius)	343

Example Problem 5.1

1. At what temperature will the velocity of sound in air be double the velocity of sound in air at 0 degree Celsius

Solution : Let T degree celsius be the required temperature. Let V 1 and V 2 be the velocity of sound at temperatures T 1 Kelvin and T 2 kelvin respectively. T 1 = 273 Kelvin (0 degree celsius) and T 2 = (T degree Celsius +273) Kelvin

V 2 by V 1 = square root of T 2 by T 1 = square root of 273 + T divided by square root of 273 = 2

Here it is given that V 2 by V 1 = 2

So, 273 + T divided by 273 = 4

T = (273 times 4) minus 273 = 819 degree Celsius.

5.2 REFLECTION OF SOUND

When you speak in an empty room, you hear a soft repetition of your voice. This is nothing but the reflection of the sound waves that you produce. Let us discuss about the reflection of sound in detail through the following activity.

When sound waves travel in a given medium and strike the surface of another medium, they can be bounced back into the first medium. This phenomenon is known as reflection. In simple the reflection and refraction of sound is actually similar to the reflection of light. Thus, the bouncing of sound waves from the interface between two media is termed as the reflection of sound. The waves that strike the interface are termed as the incident wave and the waves that bounce back are termed as the reflected waves, as shown in Figure 5.3

5.2.1 Laws of reflection

Like light waves, sound waves also obey some fundamental laws of reflection. The following two laws of reflection are applicable to sound waves as well.

1. The incident wave, the normal to the reflecting surface and the reflected wave at the point of incidence lie in the same plane.
2. The angle of incidence ∠ i is equal to the angle of reflection ∠ r.

These laws can be observed from Figure 5.4.

In the above Figure 5.4, the sound waves that travel towards the reflecting surface are called the incident waves. The sound waves bouncing back from the reflecting surface are called reflected waves. For all practical purposes, the point of incidence and the point of reflection is the same point on the reflecting surface. A perpendicular line drawn at the point of incidence is called the normal. The angle which the incident sound wave makes with the normal is called the angle of incidence, 'i'. The angle which the reflected wave makes with the normal is called the angle of reflection, 'r'.

Acoustical wonder of Golconda fort (Hyderabad, Telangana)

The Clapping portico in Golconda Fort is a series of arches on one side, each smaller than the preceding one. So, a sound wave generated under the dome would get compressed and then bounce back amplified sufficiently

5.2.2 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 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 = minus 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 (Figure 5.5).

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 (Figure 5.6). As the particles of the rarer medium are free to move, a rarefaction is produced at the interface. Thus, a compression is reflected as a rarefaction and a rarefaction travel from right to left.

More to know: What is meant by rarer and denser medium?

The medium in which the velocity of sound increases compared to another medium is called rarer medium. (Water is rarer compared to air for sound). The medium in which the velocity of sound decreases compared to another medium is called denser medium. (Air is denser compared to water for sound)

5.2.3 Reflection of sound in plane and curved surfaces

When sound waves are reflected from a plane surface, the reflected waves travel in a direction, according to the law of reflection. The intensity of the reflected wave is neither decreased nor increased. But, 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 to the other focus, no matter where it strikes the wall. This principle is used in designing whispering halls. In a whispering hall, the speech of a person standing in one focus can be heard clearly by a listener standing at the other focus.

5.3 ECHOES

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. If you shout or clap near a mountain or near a reflecting surface, like a building you can hear the same sound again. The sound, which you hear is called an echo. It is due to the reflection of sound. One does not experience any echo sound in a small room. This does not mean that sound is not reflected in a small room. This is because smaller rooms do not satisfy the basic conditions for hearing an echo.

5.3.1 Conditions necessary for hearing echo

1. The persistence of hearing for human ears is 0.1 second. This means that you 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 second.

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

velocity is equal to distance travelled by sound divided by time taken

V is equal to 2d by t

d is equal to vt by two

Since, t is equal to 0.1 second, then d is equal to v into zero point one divided by two is equal to v divided by twenty.

Thus, the minimum distance required to hear an echo is 1 by 20th part of the magnitude of the velocity of sound in air. If you consider the velocity of sound as 344 the minimum distance required to hear an echo is 17.2 m.

5.3.2 Applications of echo

Some animals communicate with each other over long distances and also locate objects by sending the sound signals and receiving the echo as reflected from the targets. 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. This is a safe testing tool, as it does not use any harmful radiations. Echo is used to determine the velocity of sound waves in any medium.

5.3.3 Measuring velocity of sound by echo method

Apparatus required: A source of sound pulses, a measuring tape, a sound receiver, and a stopwatch.

Procedure:

1. Measure the distance 'd' between the source of sound pulse and the reflecting surface using the measuring tape.
2. The receiver is also placed adjacent to the source. A sound pulse is emitted by the source.
3. The stopwatch is used to note the time interval between the instant at which the sound pulse is sent and the instant at which the echo is received by the receiver. Note the time interval as 't'.
4. Repeat the experiment for three or four times. The average time taken for the given number of pulses is calculated.

Calculation of speed of sound: The sound pulse emitted by the source travels a total distance of 2d 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 sound wave is given by speed of sound equal to distance travelled divided by time taken = 2 times d divided by t.

5.4 APPLICATIONS REFLECTION OF SOUND

5.4.1.Sound board

These are basically curved surfaces (concave), which are used in auditoria and halls to improve the quality of sound. This board is placed such that the speaker is at the focus of the concave surface. The sound of the speaker is reflected towards the audience thus improving the quality of sound heard by the audience.

5.4.2.Ear trumpet

Ear trumpet is a hearing aid, which is useful by people who have difficulty in hearing. In this device, one end is wide, and the other end is narrow. The sound from the sources falls into the wide end and are reflected by its walls into the narrow part of the device. This helps in concentrating the sound and the sound enters the ear drum with more intensity. This enables a person to hear the sound better.

5.4.3.Megaphone

A megaphone is a horn-shaped device used to address a small gathering of people. It's one end is wide, and the other end is narrow. When a person speaks at the narrow end, the sound of his speech is concentrated by the multiple reflections from the walls of the tube. Thus, his voice can be heard loudly over a long distance.

5.5 DOPPLER EFFECT

The whistle of a fast-moving train appears to increase in pitch as it approaches a stationary listener, and it appears to decrease as the train moves away from the listener. This apparent change in frequency was first observed and explained by Christian Doppler (1803 to 1853), an Austrian Mathematician and Physicist. He observed that the frequency of the sound as received by a listener is different from the original frequency produced by the source whenever there is a relative motion between the source and the listener. This is known as Doppler effect

This relative motion could be due to various possibilities as follows:

• The listener moves towards or away from a stationary source
• The source moves towards or away from a stationary listener
• Both source and listener move towards or away from one other
• The medium moves when both source and listener are at rest

DEFINITION
Whenever there is a relative motion between a source and a listener, the frequency of the sound heard by the listener is different from the original frequency of sound emitted by the source. This is known as "Doppler effect".

For simplicity of calculation, it is assumed that the medium is at rest. That is the velocity of the medium is zero. Let S and L be the source and the listener moving with velocities v S and v L respectively. Consider the case of source and listener moving towards each other (Figure 5.7). As the distance between them decreases, the apparent frequency will be more than the actual source frequency.

Let n and n dash be the frequency of the sound produced by the source and the sound observed by the listener respectively. Then, the expression for the apparent frequency n' is

n dash is equal to v plus v1 by v minus vs into n

Here, v is the velocity of sound waves in the given medium.

Let us consider different possibilities of motions of the source and the listener. In all such cases, the expression for the apparent frequency is given in table 5.2.

Case Number.	Position of source and listener	Note	Expression for apparent frequency
1	Both source and listener move They move towards each other	a) Distance between source and listener decreases. b) Apparent frequency is more than actual frequency.	N dash equal to v plus v1 divided by v minus vs in to n
2	Both source and listener move They move away from each other	a) Distance between source and listener increases. b) Apparent frequency is less than actual frequency. c) v S and L become opposite to that in case1.	N dahs equal to v minus v1 by v plus vs in to n
3	Both source and listener move They move one behind the other Source follows the listener	a) Apparent frequency depends on the velocities of the source and the listener. b) v S becomes opposite to that in case2.	N dah equal to v minus v1 by v-vs into n
4	Both source and listener move They move one behind the other Listener follows the source	a) Apparent frequency depends on the velocities of the source and the listener. b) v S and v L become opposite to that in case3.	N dash is equal to v plus v1 by v plus vs into n
5	Source at rest Listener moves towards the source	a) Distance between source and listener decreases. b) Apparent frequency is more than actual frequency. c) v S = 0 incase1.	N dash is equal to v plus v1 by v into n
6	Source at rest Listener moves away from the source	a) Distance between source and listener increases. b) Apparent frequency is less than actual frequency. c) v S = 0 incase2.	N dash is equal to v minus v1 by v into n
7	Listener at rest Source moves towards the listener	a) Distance between source and listener decreases. b) Apparent frequency is more than actual frequency. c) v L = 0 incase1.	N dash is equal to v by v minus vs into n
8	Listener at rest Source moves away from the listener	a) Distance between source and listener increases. b) Apparent frequency is less than actual frequency. c) v L = 0 incase2.	N dash is equal to v by v plus vs into n

Suppose the medium (say wind) is moving with a velocity W in the direction of the propagation of sound. For this case, the velocity of sound, 'v' should be replaced with (v + W). If the medium moves in a direction opposite to the propagation of sound, then 'v' should be replaced with (v minus W).

Solved problems

1. A source producing a sound of frequency 90 Hz is approaching a stationary listener with a speed equal to (1by10) of the speed of sound. What will be the frequency heard by the listener?

Solution: When the source is moving towards the stationary listener, the expression for apparent frequency is

2. A source producing a sound of frequency 500 Hz is moving towards a listener with a velocity of 30 metre second power minus 1. The speed of the sound is 330 metre second power minus1. What will be the frequency heard by listener?

Solution : When the source is moving towards the stationary listener, the expression for apparent frequency is

n dash is equal to v by v minus v s whole bracket into n

n dash is equal to 330 by 330 minus 30 whole bracket into 500 = 550 Hz

3. A source of sound is moving with a velocity of 50 meter second power minus 1 towards a stationary listener. The listener measures the frequency of the source as 1000 Hz. what will be the apparent frequency of the source when it is moving away from the listener after crossing him? (Velocity of sound in the medium is 330 meter second power minus 1)

Solution: When the source is moving towards the stationary listener, the expression for apparent frequency is

n dash = v divided by v – v s whole bracket multiplied by n

1000 = 330 divided by 330 minus 50 whole bracket multiplied by n

n = 1000 times 280 divided by 330

n = 848.48 Hz.

The actual frequency of the sound is 848.48 Hz. When the source is moving away from the stationary listener, the expression for apparent frequency is

n dash = open bracket v divided by v plus v s close bracket multiplied by n = open bracket 330 divided by 330 + 50 closed bracket multiplied by 848.48 =736.84 Hz

4. A source and listener are both moving towards each other with a speed v by 10 where v is the speed of sound. If the frequency of the note emitted by the source is f, what will be the frequency heard by the listener?

Solution: When source and listener are both moving towards each other, the apparent frequency is

n dash = open bracket v + v l divided by v minus v s close bracket multiplied by n

n dash = open bracket v + v by 10 divided by v minus v by 10 close bracket multiplied by n

n dash = 11 divided by 9 multiplied by f = 1.22 f

5. At what speed should a source of sound move away from a stationary observer so that observer finds the apparent frequency equal to half of the original frequency?

Solution: When the source is moving away from the stationary listener, the expression for the apparent frequency is

n dash = open bracket v by v + v s close bracket multiplied by n

n by 2 = open bracket v by v + v s close bracket multiplied by n

v s = v

5.5.1 Conditions for no Doppler effect

Under the following circumstances, there will be no Doppler effect and the apparent frequency as heard by the listener will be the same as the source frequency.

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

5.5.2 Applications of Doppler effect

(a) To measure the speed of an automobile
An electromagnetic wave is emitted by a source attached to a police car. The wave is reflected by a moving vehicle, which acts as a moving source. There is a shift in the frequency of the reflected wave. From the frequency shift, the speed of the car can be determined. This helps to track the over speeding vehicles.

(b) Tracking a satellite
The frequency of radio waves emitted by a satellite decreases as the satellite passes away from the Earth. By measuring the change in the frequency of the radio waves, the location of the satellites is studied.

(c) RADAR (Radio Detection and Ranging)
In RADAR, radio waves are sent, and the reflected waves are detected by the receiver of the RADAR station. From the frequency change, the speed and location of the aeroplanes and aircrafts are tracked.

(d) SONAR
In SONAR, by measuring the change in the frequency between the sent signal and received signal, the speed of marine animals and submarines can be determined.

Points to Remember

• Wave velocity is the velocity with which the wave travels through the medium.
• 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.
• Infrasonic waves are sound wave with a frequency below 20 Hz. A human ear cannot hear these waves.
• Ultrasonic waves are sound waves with frequency greater than 20 kHz, A human ear cannot detect these waves.
• Reflection of sound waves obey the laws of reflection.
• when a compression hits the boundary of a rarer medium, it is reflected as a rarefaction.
• An echo is the sound reproduced due to the reflection of the original sound wave.
• The minimum distance between the source and the reflecting surface should be 17.2 m to hear an echo clearly.
• "The apparent frequency" is the frequency of the sound as heard by the listener.

TEXTBOOK EVALUATION

Exercise I. Choose the correct 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 meter second power minus 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 meter second power minus 1 b) 660 meter second power 1 c) 156 meter second power minus 1 d) 990 meter second power minus 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 meter second power minus 1. What will be its value when temperature is doubled, and the pressure is halved?

a) 330 meter second power minus 1 b) 165 meter second power minus 1 c) 330 × √2 meter second power minus 1 d) 320 by √ 2 meter second power minus 1

5. If a sound wave travels with a frequency of 1.25 × 10 power 4 Hz at 344 meter second power minus 1, the wavelength will be

a) 27.52 meter b) 275.2 meter c) 0.02752 meter d) 2.752 meter

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 meter second power minus 1. The minimum distance between the sources of sound and the obstacle to hear the echo, should be

a) 17 meter b) 20 meter c) 25 meter d) 50 meter Exercise II. Fill up the blanks

1. Rapid back and forth motion of a particle about its mean position is called .

2. If the energy in a longitudinal wave travel from south to north, the particles of the medium would be vibrating in .

3. A whistle giving out a sound of frequency 450 Hz, approaches a stationary observer at a speed of 33 meter second power minus 1. The frequency heard by the observer is (speed of sound = 330 meter second power minus 1) .

4. A source of sound is travelling with a velocity 40 km per h towards an observer and emits a sound of frequency 2000 Hz. If the velocity of sound is 1220 km per h, then the apparent frequency heard by the observer is .

Exercise III. True or False (If false give the reason) 1. Sound can travel through solids, gases, liquids and even vacuum. 2. Waves created by earthquake are Infrasonic. 3. The velocity of sound is independent of temperature. 4. The Velocity of sound is high in gases than liquids. Exercise IV. Match the following

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

Exercise 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

a b c d

2) Assertion: Sound travels faster in solids than in gases.
Reason: Solid possess a greater density than that of gases.

a b c d Exercise VI. Answer very briefly

1. What is a longitudinal wave?

2. What is the audible range of frequency?

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

4. What will be the frequency sound having 0.20 m as its wavelength, when it travels with a speed of 331 m s power –1?

5. Name three animals, which can hear ultrasonic vibrations.

Exercise VII. Answer briefly

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

2. Why does an empty vessel produce more sound than a filled 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 meter second power minus 1)

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

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

Exercise VIII. Problem Corner

1. A sound wave has a frequency of 200 Hz and a speed of 400 meter second power minus 1 in a medium. Find the wavelength of the sound wave.

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 meter second power minus 1, what will be the height of the cloud?

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?

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 meter second power minus 1?

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?

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?

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 meter second power minus 1?

Exercise IX. Answer in Detail

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

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

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.

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?

Exercise 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

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.

REFERENCE BOOKS

1. Fundamental Physics by K.L. Gomber and K.L. Gogia
2. Fundamentals of sound and vibration by Franky Fahy and David Thomson
3. The theory of sound by Rayleigh and John William Strutt

INTERNET RESOURCE

1. http://people.bath.ac.uk/ensmjc/Notes/ acoustics.pdf

📝 Summary

1. Sound is produced by vibrations of different bodies and propagates as longitudinal waves through gaseous, liquid, or solid media, requiring a material medium for transmission.
2. As sound travels through a medium, particles vibrate along the direction of wave propagation, creating compressions and rarefactions with high and low pressure regions.
3. Sound waves are categorized by frequency: audible waves (20 Hz to 20,000 Hz) that humans can hear, infrasonic waves (below 20 Hz) like earthquake waves, and ultrasonic waves (above 20 kHz) detected by animals like bats and dolphins.
4. Acoustics is a branch of physics dealing with production, transmission, reception, control, and effects of sound, not limited to musical instruments and concert halls.
5. The speed of sound depends on the properties of the medium through which it travels, and sound cannot propagate in vacuum as demonstrated by the example of the Moon having no air.