Doppler Effect

  • Thread starter dekoi
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  • #1
dekoi
Suppose a person is travelling towards a wall with a tuning fork at frequency 'f' at a speed of 'v*'. Using the doppler effect equation:
[tex] f' \ = \frac{v + v_o}{v - v_s} [/tex]

What would the sign of v_o and v_s be? I don't understand, since the man is both the observer and the source. (Let - be receeding and + be approaching).
 
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  • #2
Integral
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Treat the wall as a source which is approaching the observer. Remember that observers are aways stationary.
 
  • #3
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Integral said:
Remember that observers are aways stationary.
Observers aren't always stationary though. The effect will be different if the wall approaches him than if he approaches the wall.
 
  • #4
dekoi
So which is the correct answer??
 
  • #5
berkeman
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d_leet said:
Observers aren't always stationary though. The effect will be different if the wall approaches him than if he approaches the wall.
Why? (to b, not to a)
 
  • #6
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berkeman said:
Why? (to b, not to a)
Because if the sound is moving then the waves will be closer together or farther apart than if the sound is staionary.
 
  • #7
berkeman
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No. In an approaching situation, the waves are compressed no matter what.
 
  • #8
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berkeman said:
No. In an approaching situation, the waves are compressed no matter what.
but if f is the frequency of the source.

then f(1+ v/v_sound) is noth the same thing as f/(1 - v/v_sound) the equations are different for each situation so the frequencies heard by the listener will be different in each case.
 

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