Science : Chapter 5 : Acoustics
|Effect of density||V ∝√(1/D)|
|Effect of temperature||Vt = (vo + 0.61 T) ms-1|
|Speed of Sound||distance travelled / Time taken = 2d/t|
|Veloocity = 2d/t||Distance travelld by sound / time taken|
|Source and listener move toward each other||n’ = (v+vL/v-vS) n|
|Source and listener move away from each other||n’ = (v-vL/v+vS) n|
|Listerner move twoards the stationary source||n’ = (v+vL/v) n|
|Listener move away from the stationary source||n’ = (v-vL/v) n|
|Source move towards staionary listener||n’ = (v/v-vS) n|
|Source move away from staionary listener||n’ = (v/v+vS) n|
|Frequency||n = 1/Time period (T)|
|Wavelength||λ = 1/frequency (n)|
|Amplitued||A = D (Distance)/F(Frequency)|
|Time Period||T = 1/frequency (n)|
|Velocity (of a wave) or wave velocity||Velocity (v) = Distance (D)/Time taken(T)|
V = λ x 1/T
Also, V = λn [⇒v=1/T]
|Intensity of sound||I = Work (W) / Area (A) x time (t)Also, I = Power (P) / Area (A)|
I. Choose the correct answer
1. When a sound wave travels through air, the air particles
- vibrate along the direction of the wave motion
- vibrate but not in any fixed direction
- vibrate perpendicular to the direction of the wave motion
- do not vibrate
Ans ; vibrate along the direction of the wave motion
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
- 330 ms–1
- 660 ms1
- 156 ms–1
- 990 ms–1
Ans ; 330 ms–1
3. The frequency, which is audible to the human ear is
- 50 kHz
- 20 kHz
- 15000 kHz
- 10000 kHz
Ans ; 20 kHz
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?
- 330 ms–1
- 165 ms–1
- 330 × √2 ms–1
- 320 / √2 ms–1
Ans ; 330 × √2 ms–1
5. If a sound wave travels with a frequency of 1.25 × 104 Hz at 344 m s–1, the wavelength will be
- 27.52 m
- 275.2 m
- 0.02752 m
- 2.752 m
Ans ; 0.02752 m
6. The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?
- none of these
Ans ; none of these
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
- 17 m
- 20 m
- 25 m
- 50 m
Ans ; 25 m
II. Fill in 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 travels from south to north, the particles of the medium would be vibrating in __________.
Ans ; 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) __________.
Ans ; 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 __________.
Ans ; 2068 Hz
III. State whether the following statements are true or false: If false correct the statement.
1. Sound can travel through solids, gases, liquids and even vacuum. ( False )
- Sound can travel through solids, gases, liquids and cannot through vacuum.
2. Waves created by Earth Quake are Infrasonic. ( True )
3. The velocity of sound is independent of temperature. ( False )
- The velocity of sound is dependent of temperature.
4. The Velocity of sound is high in gases than liquids. ( False )
- The Velocity of sound is less in gases than liquids.
IV. Match the items in column-I to the items in column-II:
|Column – I||Column – II|
|1) Infrasonic||a) Compressions|
|2) Echo||b) 22 kHz|
|3) Ultrasonic||c) 10 Hz|
|4) High pressure region||d) Ultrasonograph|
Ans ; 1 – c, 2 – d, 3 – b, 4 – a
V. Assertion and reason type questions:
Mark the correct choice as
- if both the assertion and the reason are true and the reason is the correct explanation of the assertion.
- if both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
- if the assertion is true, but the reason is false.
- if 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.
- Ans : iv) 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.
- Ans : ii) If both the assertion and the reason are true and the reason is not the correct explanation of the assertion
VI. Very short answer questions.
1. What is a longitudinal wave?
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.
2. What is the audible range of frequency?
Audible waves : These are sound waves with a frequency ranging between 20 Hz and 20,000 Hz. These are generated by vibrating bodies such as vocal cords, stretched strings etc.
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 ms–1?
λ = 0.20 m; V= 331 ms–1
V = nλ
|nn||= V/λ= 331ms–1/0.20m= 331/0.20= 1655 Hz.|
5. Name three animals, which can hear ultrasonic vibrations.
VII. Answer briefly
1. Why does sound travel faster on a rainy day than on a dry day?
During rainy days, the moisture content is more in the atmosphere and speed or velocity of sound.
2. Why does an empty vessel produce more sound than a filled one?
The sound is produced by the vibration of the vessel. More the vibration amplitude and frequency more is the noise. The empty one will be the condition allowing more amplified vibration because of more free space inside it (less molecules) that’s why empty vessels make more noise. There is a lot of space for the surface to vibrate. But when it is filled then the surface is constrained by an outward force. More is the pressure less is sound. That is why solid body makes least 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 ms–1)
Speed of the sound wave as function of temperature is given by
V = (V0 ) (√(1) + Tc/273)
Where Tc is the temperature in ºcelcius.
So the speed of sound wave in air at 46ºc is given by
V = 331 (√(1) + 46/273)
= 357.8 m/s.
4. Explain why, the ceilings of concert halls are curved.
When a person is talking at one focus, his voice can be heard distinctly at the other focus. It is due to the multiple reflections of sound waves from the curved walls.
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.
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 : n = 200 Hz; V = 400 m/s
|= 400 m/s / 200 Hz|
|λ||l = 2 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 ms–1, what will be the height of the cloud?
Given, Speed = 330 m/s; Time = 9.8 sec
|Let Height||= Distance|
|Speed||= Distance / time|
|We know distance is||= Speed × Time|
|Distance of the cloud||= 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?
Given, Frequency (v) = 600 Hz Time period = ?
|We know that frequency||= 1/T|
|T||= 1/600 Hz = 0.0016 Sec.|
Thus, time interval between two consecutive compression of the given wave = 0.0016 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?
Given, V = 1400 m/s-1, T = 1.6 sec,
|Let distance||= speed × time|
|2d||= V × t|
|d||= V × t/2|
|= 1400 × 1.6 / 2 = 700 × 1.6|
|d||= 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 second respectively. What is the speed of sound in the air?.
t1 = 0.9 s; t2 = 1.1 s; d = 680 m then V= ?
|d1 = Vt1 / 2||d2 = Vt2 / 2|
By adding the equation 1 and 2, we get
|d1 + d2but d1+d2d||= ½ ( Vt1 + Vt2 )= d= ½ (Vt1 + Vt2)|
|Then factor by V in equation 3|
|d680 m680 mV||= V/2 (t1 + t2)= V/2 (0.9 + 1.1)s= V/2 (2) s= 680 m/s.|
Therefore the velocity of air is 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?
|This is a question of Speed , distance and time which can be solved by using the formula|
|Speed||= distance/ time|
|As sound signal sent by one boat reaches the other boat.|
|Here distance between two observers||= 4.5 Km= (4.5 × 1000) m = 4500 m|
|Total time taken by sound signal to reach otherAccording to the formula speed of sound in water||= 3 sec= 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 1s.
What is the depth of sea given that the speed of sound in water 1450 ms–1?
|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 / 2725||= D= D|
|The distance is 725 m.|
IX. Answer in Detail
1. What are the factors that affect the speed of sound in gases?
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: vT = (vo + 0.61 T) m s–1 Here, vo is the velocity of sound in the gas at 0° C. For air, vo = 331 m s–1. Hence, the velocity of sound changes by 0.61 m s–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 reflection of sound? Explain.
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
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, a compression is reflected as a 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 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.
c) Reflection at curved surfaces.
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.
3. a) What do you understand by the term ‘ultrasonic vibration’?
The vibrations whose frequencies are greater than 20000Hz are called Ultrasonic Vibrations.
b) State three uses of ultrasonic vibrations.
- Ultra sonic vibrations are used in SONAR to measure the depth of sea (or ocean) and to locate under water objects like Submarines, sea – rocks and shipwrecks
- It is used for scanning and imaging the position and growth of a foetus and presence of stones in the gall bladder and kidney.
- It is used for homogenizing milk in milk plants where fresh milk is agitated with desired quantity of fat and powdered milk to obtain toned milk.
c) Name three animals which can hear ultrasonic vibrations.
Bats can hear ultrasonic sounds having frequencies up to 1,20,000 Hz. Some animals like dogs and dolphins can hear sounds having frequencies up to 40.000 Hz.
4. What is an 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. 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.
a) State two conditions necessary for hearing an echo.
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 atleast 0.1 s. Thus, the minimum time gap between the original sound and an echo must be 0.1 s.
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||= 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|
Thus 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 m s–1, the minimum distance required to hear an echo is 17.2 m.
b) What are 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. This is a safe testing tool, as it does not use any harmful radiations.
c) How can you calculate the speed of sound using echo?
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 = 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?
- both a and b
- data not sufficient
Ans : Light has a longer wavelength and has the 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.
An echo is heard when the time interval between the original sound and the reflected sound is at least 0.1s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease.