Waves

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Question
CBSEENPH11020660

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60o with the source observer is at rest. the apparent frequency observed by the observer (velocity of sound in air 330 ms-1) is 
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  • 100 Hz

  • 103 Hz

  • 106 Hz

  • 97 Hz

Solution

B.

103 Hz

Given, as a source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance. Such that, source is moving with a speed of 19.4 m/s  at angle 60o with source- observer line as shown in figure
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The apparent frequency heard by observer
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Question
CBSEENPH11020588

A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hr. He finds that traffic has eased and a car moving ahead of him at 18 km/hr is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be,

  • 1332 Hz

  • 1372 Hz

  • 1412 Hz

  • 1454 Hz

Solution

C.

1412 Hz

Both the observer and the source are moving.
Therefore, using the formula of apparent frequency,
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Question
CBSEENPH11020561

A train moving t a speed of 220 ms-1 towards a stationary object. emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is
(speed of sound in air is 330 ms-1)

  • 3500 Hz

  • 4000 Hz

  • 5000 Hz

  • 3000 Hz

Solution

C.

5000 Hz

From Doppler's shift, we know for this case
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Question
CBSEENPH11020687

A transverse wave os represented by y = A sin (ωt - kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

  • π A /2

  •  π A

  • 2πA

  • A

Solution

C.

2πA

Wave velocity 
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Question
CBSEENPH11020801

A transverse wave propagating along x-axis is represented by:

straight y left parenthesis straight x comma space straight t right parenthesis space equals space 8.0 space sin space open parentheses 0.5 πx minus 4 πt minus straight pi over 4 close parentheses
where x is in metres and t is in seconds. The speed of the wave is

  • 4 straight pi m/s
  • 0.5 straight pi space straight m divided by straight s
  • straight pi over 4 straight m divided by straight s
  • 8 m/s

Solution

D.

8 m/s

The standard transverse wave propagating along x-axis can be written as
                     straight y equals straight a space sin space left parenthesis kx space minus space ωt plus space straight ϕ right parenthesis
The given equation is
 straight y left parenthesis straight x comma space straight t right parenthesis space equals space 8.0 space sin space open parentheses 0.5 space straight pi space straight x space minus space 4 πt minus straight pi over 4 close parentheses space space space space space space space space... left parenthesis 1 right parenthesis
The standard wave equation can be written as,
          straight y space equals space straight a space sin space left parenthesis kx minus ωt plus straight ϕ right parenthesis                         ...(2)
where straight alpha is amplitude, k the propagation constant and straight omega the angular frequency,
comparing the Eqs. (i) and (ii), we have 
                       straight k space equals space 0.5 space straight pi comma space space space straight omega space equals space 4 straight pi
therefore    Speed of transverse wave 
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