# Doppler effect frequency

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1. Jan 18, 2015

### arpon

As we know, when the observer is moving away from the source, then the apparent frequency is,
$f_{observer} = f_{source} (\frac{v_{sound} - v_{observer}}{v_{sound}})$
But, if $v_{observer} > v_{sound}$ , $f_{observer}$ becomes negative.

2. Jan 18, 2015

### jfizzix

With sound waves being vibrations in air, if you move faster than sound away from the source, those pressure waves hit your eardrums in the reverse order. Windiness aside, the sound would be heard as being played backwards.

3. Jan 18, 2015

### arpon

And what it would be, if the observer is stationary and the source is moving faster than sound, i.e.
$f_{observer} = f_{source} (\frac {v_{sound}}{v_{sound} - v_{source}})$ , and $v_{source} > v_{sound}$ ;

4. Jan 19, 2015

### jfizzix

you would hear the sound in regular order:
The first pressure wave the source emitted would be the first thing that hits your eardrum, and the second would be the second, and so on

However, the sound would be slowed down:
The pressure waves would hit your eardrum less often because each time the wave has to travel a longer distance from the source at the same speed.

5. Jan 20, 2015

### arpon

When the source is moving faster than sound and moving towards the observer, the source has to go ahead of the observer to reach sound to the observer. So, in this case, I think, the formula for the source coming to the observer is not applicable. Only, when the source is moving away the observer, the sound can be heard.

6. Jan 20, 2015

### A.T.

Right, and after the super sonic source passed the stationary observer he would hear the previously played sounds in reverse order.

Why not?

7. Jan 20, 2015

### arpon

I think, the observer will hear two sounds, one coming from in front of him and the other coming from behind him.

Sound coming from in front of the observer (frequency $f_1$) will be fast but played backward. And sound coming from behind the observer (frequency $f_2$) will be slow but played in regular order.
$f_1 = f_{source} (\frac {v_{sound}}{v_{source} - v_{sound}})$
$f_2 = f_{source} (\frac {v_{sound}}{v_{source} + v_{sound}})$
And, look, for sufficient velocity of the source,$f_1$ may be less than $f_{source}$ !