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 

  • 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

The apparent frequency heard by observer
straight f subscript straight o space equals space straight f subscript straight s open square brackets fraction numerator straight V over denominator straight v minus straight v subscript straight s space cos space 60 to the power of straight o end fraction close square brackets space equals space 100 space open square brackets fraction numerator 330 over denominator 330 minus 19.4 space straight x begin display style 1 half end style end fraction close square brackets

equals space 100 space open square brackets fraction numerator 330 over denominator 300 minus 9.7 end fraction close square brackets space equals space 100 open square brackets fraction numerator 330 over denominator 320.3 end fraction close square brackets

equals space 103.02 space Hz

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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,
 straight f space equals space straight f subscript straight o space open parentheses fraction numerator straight v plus straight v subscript straight o over denominator straight v space plus straight v subscript straight s end fraction close parentheses
space space equals space 1392 space open square brackets fraction numerator 343 plus 10 over denominator 343 plus 5 end fraction close square brackets
space space equals space 1392 space open square brackets 353 over 348 close square brackets space Hz
space space equals 1412 space Hz

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
straight n apostrophe space equals space straight n open parentheses fraction numerator straight v plus straight v subscript straight s over denominator straight v minus straight v subscript straight s end fraction close parentheses
space equals space 1000 open parentheses fraction numerator 330 plus 220 over denominator 330 minus 220 end fraction close parentheses
equals space 1000 open parentheses 550 over 110 close parentheses space equals space 5000 space Hz

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 
straight v space equals straight lambda over straight T space equals space fraction numerator ωλ over denominator 2 straight pi end fraction
Maximum space particle space velocity space left parenthesis straight v subscript max right parenthesis subscript straight p space equals space Aω
Given comma space straight v equals space left parenthesis straight v subscript max right parenthesis subscript straight p
fraction numerator ωλ over denominator 2 straight pi end fraction space equals Aω
rightwards double arrow space straight lambda space equals space 2 straight pi space straight A

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 
                                         straight v equals straight omega over straight k equals fraction numerator 4 straight pi over denominator 0.5 straight pi end fraction
space equals space 8 space straight m divided by straight s

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