|Answer very briefly|
|Question 1.||What is a longitudinal wave?||Answer : If the particles of the medium vibrate along the direction of propagation of the wave.|
|Question 2.||What is the audible range of frequency?||Answer :Audible range of frequency is from 20 Hz to 20000 Hz.|
|Question 3.||What is the minimum distance needed for an echo?||Answer : 1. The minimum distance required to hear an echo is 1/20th part of the magnitude of the velocity of sound in air. |
2. If you consider the velocity of sound as 344 ms-1, the minimum distance required to hear an echo is 17.2 m.
|Question 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?||Answer : Speed v = 331 m/s
Let the wave length be λ = 0.2 m
Let the frequency be n
∴ Frequency n = v/λ = 331/0.2 = 3310/2
∴ Frequency = 1655 Hz
|Question 5.||Name three animals, which can hear ultrasonic vibrations.||Answer : Mosquitos, bats and dogs are the three animals that can hear ultrasonic vibrations.|
|Question 1.||Why does sound travel faster on a rainy day than on a dry day?||Answer : When humidity increases, the speed of sound increases. That is why you can hear sound from long distances clearly during rainy seasons.|
|Question 2.||Why does an empty vessel produce more sound than a filled one?||Answer : In an empty vessel, multiple reflections of sound takes place. Hence more sound is produced in an empty vessel than filled one.|
|Question 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)||Answer :
Velocity of sound, v0 = 331 ms-1 |
Air temperature, T = 46° C
Velocity of sound in air temperature vT = (v0 + 0.61T) ms-1
= 331 + (0.61 × 46)
= 331 + 28.06
vT = 359.06 ms-1.
|Question 4.||Explain why the ceilings of concert halls are curved.||Answer :
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.
|Question 5.||Mention two cases in which there is no Doppler effect in sound?||Answer :
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.
|Answer in Detail|
|Question 1.||What are the factors that affect the speed of sound in gases?||Answer :
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: Vr = (vo + 0.61 T)ms-1. Here, vo is the velocity of sound in the gas at 0° C. For air, vo = 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.
|Question 2.||What is mean by the reflection of sound? Explain.||Answer :
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
(b) Reflection at the boundary of a denser medium
(c) Reflection at sound in curved surfaces
(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.
(c) Certain creatures like dog, bats, dolphins and mosquito can detect the waves.
|Question 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.
(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 / Timetaken
Since, t = 0.1 second, then d=v×0.12=v20
(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 / Timetaken =2d/t.