Clear explanation of Doppler effect assimmetry

[DaTario]

Hi All,

I would like to know how can one explain the assimmetry in Doppler effect. I will illustrate what I mean.

If a source emits 440 Hz but is coming in my direction with a velocity V (measured in m/s) I will hear, let's say, 441 Hz. Now, if the source is at rest with respect to the medium and I am going in its direction with the same velocity (it is also a case of approximation, one could say), I will hear 442 Hz.

Why are these two frequencies different ?
Why one is not alowed to apply symmetry considerations here ?

Best wishes

DaTario

 

[mathman]

I am not familiar with the details of the calculation, but sound travels through a medium (say air), so it is not surprising if there is a difference in the result between the source moving and the receiver moving.

For e-m (light, etc.) it would be symmetric.

 

[DaTario]

I am not familiar with the details of the calculation, but sound travels through a medium (say air), so it is not surprising if there is a difference in the result between the source moving and the receiver moving.

For e-m (light, etc.) it would be symmetric.

Why is it not surprising ?

Best wishes

DaTario

 

[Archosaur]

Look at each case from your point of view.

In the first case, the source of the sound is moving toward you.
In the second case, the source of the sound and the air are moving toward you.

 

[DaTario]

Ok, I should say, then, that we should not expect symmetry because in one case we have wind and in the other we don't.

Correct?

Thank you

DaTario

 

[Archosaur]

Right. In one case, you have the "wind," but I'm not sure how to calculate what it would do.

 

[Drakkith]

Hi All,

I would like to know how can one explain the assimmetry in Doppler effect. I will illustrate what I mean.

If a source emits 440 Hz but is coming in my direction with a velocity V (measured in m/s) I will hear, let's say, 441 Hz. Now, if the source is at rest with respect to the medium and I am going in its direction with the same velocity (it is also a case of approximation, one could say), I will hear 442 Hz.

Why are these two frequencies different ?
Why one is not alowed to apply symmetry considerations here ?

Best wishes

DaTario

I don't know much about this, but is this really what happens? Or is this only in special circumstances?

 

[AndyNewton]

The asymmetry is easier to understand with a larger movement speed. This predicts that the doppler pitch change is greater when the sound source is moving towards the observer, and not so large if the observer is moving towards the source.

Let the sound source, at rest, generate a single frequency sound note of frequency F that would propagate in air at speed c and wavelength L.

Now make the sound source move at half the speed of sound towards a fixed observer:
In the time taken between emitting one wave pressure peak and the next, the first pressure peak propogates distance L towards the observer and also the sound source itself moves distance L/2 towards the observer (where it emits the next pressure peak).
So the moving source is at distance L/2 behind the previous peak when it emits the next peak, and the next, and the next...
The wavelength of the resulting soundwave is therefore L/2 (in the part that is directed towards the observer).
So the frequency of the soundwave in the part that is directed towards the observer is F times 2. The frequency doubles.

Or, make the observer move at half the speed of sound towards the sound source, with the source now fixed.
The wavelength of the sound generated is now simply L, of course, with frequency F and speed c.
A pattern of high pressure and low pressure regions is moving through the air at speed c, but the observer is also moving through the air at speed c/2 - heading into the oncoming pressure pattern. The relative speed between the observer and the pressure pattern is c + c/2 = c times 3/2.
The frequency at which the observer meets the high pressure regions is therefore increased by a factor of 3/2. The frequency goes up by 50%.

For an even more extreme example, let the source or observer movement speed be equal to the speed of sound. Using the same analysis method as above you will find that one movement gives a doubling of the observed frequency and the other movement gives something more dramatic!