How do waves following a shockwave catch up to the shock wave?

[Jake DAprile]

TL;DR Summary

Im trying to understand how a wave traveling behind a shock wave can catch up to that wave, such as in a blast.

I was in an argument about a jet engine and I was arguing that since there is a cutoff in terminology what would kill someone approaching a engine is not technically sound, but a shock wave, (I'm probably wrong about this, but that's not the question). That got me wondering how waves can catch up to each other and amplify such as in a blast. I'm studying chemistry, not physics, so forgive me if half of this is wrong.

 

[davenn]

Summary: I am trying to understand how a wave traveling behind a shock wave can catch up to that wave, such as in a blast.

I'm not sure that they do catch up
and speaking of which, what waves are you referring to ?

Do you have an example of a case where they do ?

 

[Dr. Courtney]

In a free field, the trailing waves don't catch up to the leading shock front, so the duration of the disturbance gets longer as the shock wave travels further from the source.

 

[sophiecentaur]

I'm not sure that they do catch up
and speaking of which, what waves are you referring to ?

They don't "catch up" but they travel outwards with a different Vee shaped wavefront. The angle is determined by the actual Mach number of the object causing the waves. The speeds are different (shockwave and resulting sound wave) but the frequencies (initially) will be the same. You have different speeds and the same frequencies so the wavelengths will be different. This means that the direction of the sound wave has to change from the angle of the shock wave Vee. Higher the Mach number, the sharper the Vee shape.
It's similar to the change of direction due to wave refraction at an interface between substances with different wave speeds (Snell's law etc); you have phase continuity across the transition.
As mentioned above, the pulse will get wider and wider which is due to a dispersion effect, I think.

 

[davenn]

In a free field, the trailing waves don't catch up to the leading shock front, so the duration of the disturbance gets longer as the shock wave travels further from the source.

They don't "catch up" but they travel outwards with a different Vee shaped wavefront.

on a plane sound shockwave yes
for an explosion shockwave (that was also mentioned in the OP), no it's going to be circular, hemispherical,
spherical depending on the situationThanks for the backup, guys, I was 99% sure of that
But I have, on occassion, known to be wrong

Dave

 

[sophiecentaur]

on a plane sound shockwave yes
for an explosion shockwave (that was also mentioned in the OP), no it's going to be circular, hemispherical,
spherical depending on the situation

In a free field, the trailing waves don't catch up to the leading shock front, so the duration of the disturbance gets longer as the shock wave travels further from the source.

Both comments are very relevant. Clearly something going slower than another 'thing' can't catch it up so the explanation must be that the sound wave is created as energy from a spherical shock wave is transferred into regular sonic waves over a period of time and that, I guess, must mean that the two waves (pulses) exist in the same space for a while (travelling at different speeds). The time taken for the transfer will spread out the length of the sound pulse until there is no more supersonic energy.
I read in a recent PF thread that the sound pulse from a supersonic craft is received at each reception point from contributions along a length of the shock wave. This lengthens the sound pulse.