5
$\begingroup$

We know that a moving source of sound waves produces Doppler shift, so that sound waves are closer-together in the front and farther apart behind, creating a difference in pitch.

Let's consider a situation where I generate a sound pulse (lasting one second). If I'm stationary, then you'll be hearing my pulse for exactly 1 second. If I'm moving towards you, then it will be heard for some time less than a second. If I'm moving away from you, then it will be greater than a second. For now, let's assume that I'm moving towards you.

If the sound waves are traveling at a constant speed and are closer together, then you'll hear all of it in a shorter amount of time. So, as my speed approaches very close to Mach 1, the waves would be so close together that the sound you hear will be almost instantaneous (something like a "bang!").

At Mach 1, all the waves would be stacked right on top of each other, thus creating a sonic boom. Now, as my speed exceeds Mach 1, (proceeding with the same logic), the sound should probably be reversed. Because, the wave that's been produced recently would be in front of its ancestor, so that information from the latest wave is received first, followed by the previous wave, followed by its previous wave, and so on.

So when we reach Mach 2, we should hear the original sound pulse backwards. Because, as we approached Mach 1 from Mach 0, the pitch increased, the waves bunched and we heard a bang! So, as we approach Mach 2 from Mach 1, the pitch should decrease back to the original pitch (because the time delay between the latest wave and the previous wave should increase, since the source is moving relatively faster) and now you should hear the same pulse backwards (a good example might be that if I shout "God" at Mach 2, you'll probably hear "Dog").

Getting deeper with this logic, as my speed exceeds Mach 2, the pitch of the sound you hear should be further reduced.

Summing everything up,

Mach 0.0 - sound is normal
Mach 0.5 - sound is sped up and high pitched
Mach 1.0 - sound is a boom!
Mach 1.5 - sound is sped up, high pitched and reversed
Mach 2.0 - sound is reversed
Mach 2 + - sound is slowed down, lower pitched and reversed

Am I right with my logic?

Improve this question
$\endgroup$
14
  • $\begingroup$ @KyleKanos - Yes, I did some skimming on Mach wave and Mach numbers. As for reversed, what I mean is that it would be like playing the audioclip in reverse. Imagine saying "Dog" and people hearing "God". THIS is the closest I could find that shows the effect that I was thinking of, but it goes into no detail about the actual sounds produced. $\endgroup$
    – Shimizoki
    Commented Jul 2, 2015 at 19:02
  • $\begingroup$ @KyleKanos - It is quite possible I have not, but from what I am seeing from the link I provided is that it merely states that there will be a sonic boom and the object will pass before you hear it. Your link does not mention anything in the way of sound (Just once again mentioning Mach front) Unless of course you are saying that all sound generated by a object that is moving M>2 is the same... You are correct, I don't understand. $\endgroup$
    – Shimizoki
    Commented Jul 2, 2015 at 19:17
  • $\begingroup$ Sound is a pressure wave. Traveling faster than the speed of sound compresses those waves into one larger wave. The faster you go, the louder the "boom" is. $\endgroup$
    – Kyle Kanos
    Commented Jul 2, 2015 at 19:19
  • 4
    $\begingroup$ I'm really confused by the close votes here. The sound is reversed, and the OP seems to understand this better than most users on this site. $\endgroup$
    – user10851
    Commented Jul 3, 2015 at 20:48
  • 1
    $\begingroup$ Dear user1179554: your question is a good one and your reasoning is probably sound: I've never even thought about this before. The only reason I can think of for the close votes is that it is a kind of "check my work" question. The way to do that in keeping with the rules is to ask a short question: almost your title alone is enough although there are some comments showing where other users found this unclear. THEN answer your own question with your reasoning: this is allowed here. It's a great way to get a review of your understanding. This is great stuff BTW. $\endgroup$
    – Selene Routley
    Commented Jul 4, 2015 at 0:58

0

Reset to default

Browse other questions tagged or ask your own question.