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Doppler Effect

  1. Feb 15, 2006 #1
    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).
    Last edited by a moderator: Feb 15, 2006
  2. jcsd
  3. Feb 15, 2006 #2


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    Treat the wall as a source which is approaching the observer. Remember that observers are aways stationary.
  4. Feb 15, 2006 #3
    Observers aren't always stationary though. The effect will be different if the wall approaches him than if he approaches the wall.
  5. Feb 15, 2006 #4
    So which is the correct answer??
  6. Feb 15, 2006 #5


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    Why? (to b, not to a)
  7. Feb 15, 2006 #6
    Because if the sound is moving then the waves will be closer together or farther apart than if the sound is staionary.
  8. Feb 16, 2006 #7


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    No. In an approaching situation, the waves are compressed no matter what.
  9. Feb 16, 2006 #8
    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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