# Doppler Effect

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:
$$f' \ = \frac{v + v_o}{v - v_s}$$

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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Integral
Staff Emeritus
Gold Member
Treat the wall as a source which is approaching the observer. Remember that observers are aways stationary.

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.

dekoi
So which is the correct answer??

berkeman
Mentor
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)

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.

berkeman
Mentor
No. In an approaching situation, the waves are compressed no matter what.

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.