# Doppler Effect

1. Feb 15, 2006

### 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).

Last edited by a moderator: Feb 15, 2006
2. Feb 15, 2006

### Integral

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

3. Feb 15, 2006

### d_leet

Observers aren't always stationary though. The effect will be different if the wall approaches him than if he approaches the wall.

4. Feb 15, 2006

### dekoi

So which is the correct answer??

5. Feb 15, 2006

### Staff: Mentor

Why? (to b, not to a)

6. Feb 15, 2006

### d_leet

Because if the sound is moving then the waves will be closer together or farther apart than if the sound is staionary.

7. Feb 16, 2006

### Staff: Mentor

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

8. Feb 16, 2006

### d_leet

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.