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What is the relationship between the velocity of a wave source and the amplitude of a shock wave?

Physics Asked on June 13, 2021

I’m having trouble understanding the kind of interference leading to shock wave formation and how it relates to the velocity of the wave source.I came across this image:

It illustrates the formation of a shock cone from spherical wavefronts at supersonic wave source velocity. From what i see, even if the wavefronts were much more closer than the image depicts, overlap would still not occur at every point on the shock wavefront. Does this imply that amplitude of a shockwave is not uniform? How would wave source velocity affect this(Is there a mathematical relationship)?

I’m not concerned with the velocity of propagation of the shock wavefront or with the changes in temperature,pressure and density upstream and downstream of the shock wavefront. I’m only concerned with the amplitude of the shock wave and the mechanism of interference as it relates to the velocity of the wave source.

One Answer

In the picture, the slow-moving wavefronts as well as the fast-moving wavefronts have their source in the engine in the plane (lets assume the rate at wich wavefronts are emitted represent the frequency of the plane's motor). In the picture, the plane is moving at a higher speed than that of sound in the right direction. The wavefronts of the motor move both to the left in this case. If you watch a circle (wavefronts, which are continuously emitted from the motor of the plane) emerging from the plane, the left side is moving at a speed higher than the speed of sound from the plane (relative to the plane). The fast velocity with which this sound removes itself from the plane is $v_{fsp}=v_s + v_p$.
The slower moving sound (the right side of the circles) of this wavefront (also moving to the left) is $v_{ssp}=v_p-v_s$.

It's obvious that the fast (to the left moving) wavefronts can't interfere with the slow-moving wavefronts because these fast wavefronts are ahead of them.

Except where the fast and slow wavefronts (from different moments of their emission) meat at the edges of the cone. There the slow part and the fast part of the wavefronts enhance each other.
This enhancement (a supersonic bang) moves over the ground so you can hear the bang. The higher the frequency of the motor, the more energy is contained in it, and the bang increases in intensity.
After the bang, you can't hear the plane anymore because the sound waves emerging from the plane can't reach you. From the frame of the plane, the waves are traveling faster than the speed of sound but according to an observer on the ground, these waves can't reach him/her.

So only the bang is heard, whose intensity is higher if the frequency of the motor is higher or the plane is going faster.

If the plane is moving at the speed of sound the cone becomes a straight line. In this case, a bang is also heard, and after passage, there is silence as the speed of sound is relative to the plane, not to the ground.

If the plane is flying below the speed of sound you can imagine no bang is heard, but the plane is heard flying by with the usual Doppler effects.

Answered by Deschele Schilder on June 13, 2021

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