Doppler effect with moving medium but no relative motion

[Soren4]

I would like to be sure about one case of the use of Doppler effect with sound waves.

If the medium (in the case of sound air) is moving, but there is no relative motion between the observer and the source there is no Doppler effect at all. (And the absence of relative motion is frame-indipendent)

So imagine we have source and observer steady relative to ground and air moving. The speed of sound in frame of reference of the ground changes, but this does not mean that frequency of the wave changes, just that it wavelenght is different. So in such case frequency does not change at all, but, $v_{sound}$ changes. And this imply that $\lambda$ changes too.

Is all of this correct, or is there something conceptually wrong?

I'm a bit confused because I found a similar situation in a exercise, which I write here.

A guitar player is playing in front of crowd. There is wind blowing at speed $v$ from the stage to the crowd. If the frequency of sound waves is $f$ what is the frequency perceived by the crowd?

If all I said is correct then there should not be any change in frequency, nevertheless in the solution of this exercise I found this $$f'=f(\frac{c+v}{v})$$ So it is interpreted as the crowd comes closer to the source.

I think this resolution is wrong, because it contradicts what said above.

I would really appreciate any suggestion

 

[beamie564]

Why is there $c$ ? What does light have to do in this case?

 

[fresh_42]

Why don't take it simple? You cannot experience the Doppler-effect inside an ambulance.

 

[Doc Al]

I'm a bit confused because I found a similar situation in a exercise, which I write here.

A guitar player is playing in front of crowd. There is wind blowing at speed $v$ from the stage to the crowd. If the frequency of sound waves is $f$ what is the frequency perceived by the crowd?

If all I said is correct then there should not be any change in frequency, nevertheless in the solution of this exercise I found this $$f'=f(\frac{c+v}{v})$$ So it is interpreted as the crowd comes closer to the source.

I think this resolution is wrong, because it contradicts what said above.

Where did you get this exercise and solution? I agree with you: The solution is incorrect.

 

[DaTario]

I think the wind breaks the isotropy in the wave propagation.

 

[Doc Al]

I think the wind breaks the isotropy in the wave propagation.

Not sure what you mean. Have you done the problem? (I think it's easiest to apply the equations in the frame of the moving air.)

 

[DaTario]

Not sure what you mean. Have you done the problem? (I think it's easiest to apply the equations in the frame of the moving air.)

I mean that, in the reference frame of the source, it seems correct to say that the wave propagates with different velocities in opposite directions which are not perpendicular to the wind´s direction.

I am trying to think as in the problem of the boat going up and down the river, given that the boat can only travel with fixed velocity in relation to the water.

Best Regards,
DaTario

P.S.: BTW, I believe the solution is correct.

 

[Doc Al]

I mean that, in the reference frame of the source, it seems correct to say that the wave propagates with different velocities in opposite directions which are not perpendicular to the wind´s direction.

That's certainly true.

I am trying to think as in the problem of the boat going up and down the river, given that the boat can only travel with fixed velocity in relation to the water.

OK, but I don't see the relevance to the Doppler problem.

P.S.: BTW, I believe the solution is correct.

Did you apply the standard Doppler formulas?

 

[DaTario]

OK, but I don't see the relevance to the Doppler problem.

Consider the two sections of a given spherical crest emitted by the source at rest, the first one going toward the observer and the other in the opposite direction. As the observer moves toward the source he sees each one with different velocities.

Did you apply the standard Doppler formulas?

 

[Doc Al]

Consider the two sections of a given spherical crest emitted by the source at rest, the first one going toward the observer and the other in the opposite direction. As the observer moves toward the source he sees each one with different velocities.

OK, but I still don't see the relevance. (Only the sound heading toward the observer counts here.)

Try this: The air is still, but the observer and source both move in the same direction with speed v (with respect to the air). How will the observed frequency compare to the source frequency?

 

[DaTario]

OK, but I still don't see the relevance. (Only the sound heading toward the observer counts here.)

Try this: The air is still, but the observer and source both move in the same direction with speed v (with respect to the air). How will the observed frequency compare to the source frequency?

It must be:
$f_o = f_s \frac{v-u_o}{v - u_s} = f_s \frac{v-u}{v - u} = f_s$

It was a nice shot in the discussion, but speaking non locally, we have two different situations. If there were more observers distributed in different positions, all in rest relatively to our first observer, he, our first observer, would have to think carefully in order to infer thar the others are hearing or not the same frequency he is hearing.

Now, let me try to explain my point. I was trying to infer that in the reference frame of the observer, when he is moving toward the source, the isotropy is destroyed for the velocities of the sections of the crest depends on its directions.