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How does a moving sound source physically compress the waves emitted in front of it and lengthen the waves emitted behind it? Virtually all descriptions of the doppler effect provide nice visuals of waves bunched up in front of a moving source and spread out behind such a source. However, I would like a detailed description of how a moving source engages with the medium through which it travels to accomplish this. For example, does the bulk modulus of the medium change by virtue of a source moving through it? If so, does that factor into the frequency perceived by the listener?

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Watch this gif very carefully.

https://upload.wikimedia.org/wikipedia/commons/c/c9/Dopplereffectsourcemovingrightatmach0.7.gif

Supposing that the sound in question is a 100 Hz tone, every 0.01 sec the sound source will create a pressure maximum at its current location. The last pressure maximum will have propagated out into a sphere a few meters across in that time. If the source is moving, then it will now be closer to one side of that sphere than the other, having moved since it initiated the last pressure maximum 0.01 sec ago. The distance between successive wavefronts is preserved as they continue to expand, since the speed of sound is constant.

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  • $\begingroup$ Indeed, the speed of sound is constant in all directions, no change in bulk modulus. I also like Hewitt's video: youtube.com/watch?v=m3MkZjlacaI $\endgroup$ – Pieter Feb 18 '18 at 8:34
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I extensively studied a lot of acoustics as I learned about the functional parameters of speaker drivers. Combining this with a bit of physics, the Doppler effect can be represented using a speaker driver mounted on a car with its diaphragm facing the direction of travel.

When an audio signal is applied to this driver it's diaphragm begins to accelerate forwards, adding to it's initial velocity which it exhibits from the car. The diaphragm then reaches it's peak excursion and accelerates backwards, it's velocity is now subtracting from the initial velocity of the car.

With the diaphragm's velocity being offset by the velocity of the car, the resulting acoustic waves tend to have tighter regions of compression and rarefaction in the direction of travel, increasing the frequency of sound ahead of the car. And larger regions of compression and rarefaction opposite to the direction of travel, decreasing the frequency of sound behind the car.

This principle applies to all sound waves propagating from a moving body as the sound waves carry the momentum of the moving body relative to an outside observer who experiences the Doppler effect.

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Let the frequency be $f$ and then the wavelength in air frame is $\lambda=c/f$, where $c$ is the sound speed. The period is $T=1/f$.

Image the source emits a peak $1$ at $t$. Then the next peak, peak $2$ it emits will be at $t+T$.

When the source is at rest, then at $t+T$, peak 2 is at the position of the source and peak $1$ will be at (both forward and backward) $cT=c/f=\lambda$. So the wavelength is $\lambda$.

Now, when the source moves with velocity v. Then the position of the peak $2$ at $t+T$ will be $vT$. Hence the wavelength in the forward direction is (the distance between peak $1$ and peak $2$):

In the forward direction: $cT-vT=(c-v)/f$.

In the backward direction: $cT+vT=(c+v)/f$

Therefore, the frequency observed in the forward direction is $$\frac{c}{(c-v)/f}=\frac{c}{c-v}f$$

And in the backward direction, $$\frac{c}{(c+v)/f}=\frac{c}{c+v}f$$

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