Samacheer Kalvi Class 10 Science Chapter 5 Acoustics book back questions and answers — complete solutions for all MCQ, short answer and long answer questions. This chapter covers sound waves, characteristics of sound, reflection of sound, reverberation, echo, sonar, ultrasound and the human ear as per the Tamil Nadu Grade 10 Science syllabus. All book back questions answered with clear explanations.
Acoustics — key concepts & quick answers
What is sound and how does it travel? Sound is a form of energy produced by vibrating bodies. It travels as longitudinal mechanical waves through a material medium (solid, liquid or gas) and cannot travel through a vacuum.
What is the audible range of frequency for humans? The human ear can hear sound roughly in the frequency range 20 Hz to 20,000 Hz (20 kHz).
What is the difference between infrasonic and ultrasonic sound? Infrasonic sounds have frequencies below 20 Hz; ultrasonic sounds have frequencies above 20,000 Hz. Both lie outside the normal human hearing range.
What is an echo? An echo is the repetition of a sound heard after reflection of the sound waves from a distant surface. The reflected sound must reach the ear at least 0.1 second after the original sound.
What is the relation between speed, frequency and wavelength? v = f λ, where v is the wave's speed, f its frequency and λ its wavelength.
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MCQI. Multiple Choice Questions1 mark each
Q.1 When a sound wave travels through air, the air particles
✓ Answer: (A) vibrate along the direction of the wave motion
Q.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
✓ Answer: (A) 330 m s-1
Q.3 The frequency which is audible to the human ear is (upper limit)
20 kHz. The typical audible range for humans is about 20 Hz to 20,000 Hz (20 kHz).
Q.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?
✓ Answer: (C) 330 x √2 m s-1 (approx. 466.2 m s-1)
Q.6 The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?
✓ Answer: (D) none of these
Q.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
✓ Answer: C, 25 m
FillII. Fill in the Blanks1 mark each
Q.1Rapid back and forth motion of a particle about its mean position is called Vibration▾
✓ Answer
Vibration. Vibration is defined as the rapid, repetitive back and forth motion of a particle or object around its equilibrium or mean position. This oscillatory movement is a fundamental concept in the study of waves and sound. The speed and amplitude of these vibrations determine characteristics of the resulting wave, such as frequency and loudness.
Q.2If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating in both north and south▾
✓ Answer
Both north and south. In a longitudinal wave, the particles of the medium oscillate back and forth in a direction parallel to the direction in which the wave is traveling. If the energy of a longitudinal wave is propagating from south to north, the individual particles of the medium will move both towards the north and towards the south from their mean positions, creating compressions and rarefactions along the path of the wave.
Q.3A 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.▾
✓ Answer
Answer: 500 Hz. For a source moving towards a stationary observer, apparent frequency n' = [v/(v - vs)] n = [330/(330 - 33)] x 450 = 500 Hz.
Q.4A 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.▾
✓ Answer
Answer: 2068 Hz. For a source moving towards a stationary observer, n' = [v/(v - vs)] n = [1220/(1220 - 40)] x 2000 = 2068 Hz approximately.
T/FIII. True or False1 mark each
#
Statement
Answer
Correction (if False)
1
Sound can travel through solids, gases, liquids and even vacuum.
False
Sound waves cannot travel through vacuum.
2
Waves created by earthquake are infrasonic.
True
-
3
The velocity of sound is independent of temperature.
False
The velocity of sound depends on temperature.
4
The velocity of sound is high in gases than liquids.
False
The velocity of sound is higher in liquids than in gases.
MatchIV. Match the Following1 mark each
Column A
Column B
Infrasonic
10 Hz
Echo
Reflection of sound
Ultrasonic
Frequencies greater than 20 kHz
High pressure region
Compressions
A&RV. Assertion & Reasoning2 marks each
Q.1Assertion: The change in air pressure affects the speed of sound.▾
✓ Answer
The assertion is false. At constant temperature, pressure alone does not affect the speed of sound in an ideal gas because v = sqrt(gamma RT), so it depends mainly on temperature and the nature of the gas, not on pressure alone.
Q.2Assertion: Sound travels faster in solids than in gases.▾
✓ Answer
The assertion is true. Sound travels faster in solids than in gases mainly because solids have much greater elasticity. If the reason states that this is only because solids have greater density, the reason is false.
ShortVI. Short Answer Questions2 marks each
Q.1What is a longitudinal wave?▾
✓ Answer
A longitudinal wave is a type of mechanical wave where the particles of the medium through which the wave is passing vibrate or oscillate parallel to the direction of energy transfer or wave propagation. This parallel motion results in the formation of regions of compression (where particles are close together) and rarefaction (where particles are spread apart). Sound waves traveling through air are a classic example of longitudinal waves, as the air molecules vibrate back and forth along the same line that the sound is traveling.
Q.2What is the audible range of frequency?▾
✓ Answer
The audible range of frequency for a normal human ear is 20 Hz to 20,000 Hz (20 kHz).
Q.3What is the minimum distance needed for an echo?▾
✓ Answer
For a distinct echo, the reflected sound should reach the ear after at least 0.1 s. Since sound travels to the obstacle and back, 2d/v >= 0.1. Taking v about 344 m s-1, d >= 17.2 m. So the minimum distance is about 17 m.
Q.4What will be the frequency sound having 0.20 m as its wavelength, when it travels with a speed of 331 m s-1?▾
✓ Answer
Given speed v = 331 m s-1 and wavelength lambda = 0.20 m. Frequency f = v/lambda = 331/0.20 = 1655 Hz.
Q.5Name three animals, which can hear ultrasonic vibrations.▾
✓ Answer
Bats, dogs, and dolphins are among the animals that can perceive ultrasonic vibrations, which are sound waves with frequencies higher than the upper limit of human hearing (typically above 20 kHz). Bats use echolocation, emitting ultrasonic calls and interpreting the returning echoes to navigate and hunt insects in complete darkness. Dogs can hear frequencies well above those audible to humans, which is why dog whistles are effective. Dolphins also utilize ultrasonic sounds for communication and navigation in their aquatic environment.
ShortVI. Short Answer Questions2 marks each
Q.1Why does sound travel faster on a rainy day than on a dry day?▾
✓ Answer
Sound travels faster on a rainy day compared to a dry day primarily because of the increased humidity in the air. Rainy air contains a higher concentration of water vapor. Water vapor molecules are lighter than the nitrogen and oxygen molecules that make up the bulk of dry air. This increase in lighter molecules effectively reduces the overall density of the air mixture. Since the speed of sound in a medium is inversely proportional to the square root of its density, sound travels slightly faster in less dense, more humid air than in denser, dry air.
Q.2Why does an empty vessel produce more sound than a filled one?▾
✓ Answer
When an empty vessel is struck, the air inside it is set into vibration along with the vessel itself. This air column has a larger volume and can vibrate more freely with less damping. Damping refers to the gradual loss of energy from a vibrating system. In an empty vessel, the air can vibrate with greater amplitude and for a longer duration, leading to a louder sound. Conversely, when a vessel is filled with a liquid, the liquid absorbs a significant amount of the vibrational energy. This absorption causes the vibrations to be damped more quickly, resulting in a shorter duration and lower amplitude of vibration, and thus a fainter sound.
Q.3Air temperature in the Rajasthan desert can reach 46^\circ C. What is the velocity of sound in air at that temperature? (V0 = 331 m s-1)▾
✓ Answer
For air, vT = v0 + 0.61T. Here v0 = 331 m s-1 and T = 46°C. vT = 331 + (0.61 x 46) = 331 + 28.06 = 359.06 m s-1. The velocity of sound is approximately 359 m s-1.
Q.4Explain why, the ceilings of concert halls are curved.▾
✓ Answer
The ceilings of concert halls are often designed with curved surfaces, typically in a dome or parabolic shape, to enhance the acoustics. These curved surfaces act as reflectors that help in the uniform distribution of sound throughout the audience area. When sound waves strike a curved ceiling, they are reflected in a way that spreads them out evenly, preventing sound from concentrating in certain areas and becoming too loud, while also ensuring that no areas are too quiet or have poor sound quality. This controlled reflection helps to eliminate 'dead spots' where sound is weak and ensures that the audience can hear the music or speech clearly and with good fidelity, regardless of their seating position.
Q.5Mention two cases in which there is no Doppler effect in sound?▾
✓ Answer
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source. There are specific conditions under which this effect is not observed. Firstly, the Doppler effect is absent if both the source of the sound and the observer are stationary with respect to each other. In this scenario, the distance between them remains constant, and thus the frequency of the sound waves reaching the observer is unchanged. Secondly, the effect is also absent if the source and the observer are moving with the same velocity in the same direction. When they move together at the same speed and in the same direction, their relative velocity along the line connecting them is zero, meaning the distance between them does not change, and hence no frequency shift occurs.
NumericalVII. Numerical Problems3 marks each
Q.1A 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.▾
✓ Answer
Given frequency f = 200 Hz and speed v = 400 m s-1. Wavelength lambda = v/f = 400/200 = 2 m.
Q.2The 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?▾
✓ Answer
Light travels much faster than sound, so the delay is due to sound. Height of cloud = speed x time = 330 x 9.8 = 3234 m.
Q.3A 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?▾
✓ Answer
Frequency f = 600 Hz. The time period between successive compressions is T = 1/f = 1/600 = 0.00167 s. The 400 m distance does not change the time period of the wave.
Q.4An 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?▾
✓ Answer
The time given is for the wave to travel to the sea bottom and return. Depth d = vt/2 = (1400 x 1.6)/2 = 1120 m.
Q.5A 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?▾
✓ Answer
Let the distances from the man to the two walls be x and 680 - x. For the two echoes: 2x/v = 0.9 and 2(680 - x)/v = 1.1. Adding, 2 x 680 / v = 0.9 + 1.1 = 2.0. v = 1360/2.0 = 680 m s-1.
Q.6Two 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 = 4.5 km = 4500 m and time = 3 s. Speed of sound in water v = d/t = 4500/3 = 1500 m s-1 = 1.5 km s-1.
Q.7A 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?▾
✓ Answer
The 1 s is the total time for the signal to go down and return. Time to reach the sea bottom = 1/2 = 0.5 s. Depth = speed x time = 1450 x 0.5 = 725 m.
LongVIII. Long Answer Questions5 marks each
Q.1What are the factors that affect the speed of sound in gases?▾
✓ Answer
(i) Effect of density: Velocity of sound in a gas is inversely proportional to the square root of density, so it decreases as density increases. (ii) Effect of temperature: Velocity of sound in a gas increases with temperature. For air, vT = v0 + 0.61T, where v0 = 331 m s-1 at 0°C. (iii) Effect of relative humidity: When humidity increases, the speed of sound increases because humid air has lower effective density than dry air.
Q.2What 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▾
✓ Answer
Reflection of sound is the phenomenon where sound waves bounce back into the original medium upon striking the surface of another medium. This process is analogous to the reflection of light. a) When sound encounters the boundary of a rarer medium, which offers less resistance, a compression wave can be reflected back as a rarefaction. This occurs because the rarer medium cannot support the compression as effectively and essentially pushes back against it, inverting the wave. b) At the boundary of a denser or rigid medium, a compression wave is reflected back as a compression. This is due to the principle of equal and opposite reactions; the rigid surface exerts an equal and opposite force on the incoming sound wave, causing it to reflect without inversion. c) Reflection at curved surfaces has distinct effects. Convex surfaces tend to diverge the reflected sound waves, spreading them out and reducing their intensity over a larger area. This can be useful for making sound seem less harsh or for covering a wider space with softer sound. Conversely, concave surfaces act to converge the reflected sound waves towards a focal point. This can be used to concentrate sound energy, making it louder at the focus, which is employed in devices like parabolic microphones or whispering galleries.
Q.3a) 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.▾
✓ Answer
a) Ultrasonic vibrations are sound vibrations with frequency greater than 20 kHz, beyond the normal human hearing range. b) Uses include cleaning delicate objects such as jewellery and dental plates, finding the range and direction of submarines using SONAR, ultrasonic welding, and medical imaging. c) Dogs, bats and dolphins can hear ultrasonic vibrations.
Q.4What 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
An echo is the reflected sound heard distinctly after the original sound. a) Conditions: the time gap between original sound and echo must be at least 0.1 s, and the reflecting surface should be about 17 m or more away in air. b) Medical applications include obstetric ultrasonography, echocardiography and imaging of internal organs using reflected ultrasonic waves. c) If the distance to the reflecting surface is d and the echo time is t, the sound travels 2d. Therefore, speed of sound v = 2d/t.
Q.1Suppose that a sound wave and a light wave have the same frequency, then which one has a longer wavelength?▾
✓ Answer
Answer: b) Light. For the same frequency, wavelength lambda = v/f. Since the speed of light is much greater than the speed of sound, the light wave has the longer wavelength.
Q.2When 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.▾
✓ Answer
On a hotter day, the speed of sound increases, so the echo returns in a shorter time: t = 2d/v. If the fixed distance is just enough for an echo on a cooler day, the return time may fall below 0.1 s and the echo will not be heard distinctly. If the reflecting surface is far enough that t remains at least 0.1 s, the echo can still be heard.
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