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I believe this question was asked in some form before, but I'm not clear on the answer. If a sound wave must equal air pressure when it exits a tube, why is it possible that at many points after the sound wave exits the pipe (is in open air) it does have greater pressure than the air (in the compression portion of the wave?

It does feel natural that as the air exits the pipe, it will revert to regular air pressure: I just don't understand why it is required, given that at many points after the pipe ends the wave does have high pressure. How does the air know that it's at the pipe's end? Why can't you have a wave leaving the pipe at a compression, and having pressure greater than air pressure?

Another way to state my question is like this: if a mechanism exists requiring a sound wave to be equal to air pressure on exiting a pipe, what removes that mechanism once the wave is further away from the pipe (in open air), allowing it to resume having pressures greater than air pressure?

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  • $\begingroup$ Do you mean a gauge pressure of zero, not an absolute pressure? Have you thought about how there can be pressure antinodes between the pressure nodes inside the pipe? $\endgroup$
    – Bill N
    Commented Feb 11, 2019 at 23:10
  • $\begingroup$ Related: physics.stackexchange.com/q/256008/2451 $\endgroup$
    – Qmechanic ♦
    Commented Feb 13, 2019 at 17:17

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The resonances of an air column can be characterized by the pipe's impedance as a function of frequency. Below is an example of a graph, from a group an UNSW (web page).

impedance as a function of frequency for a flute

To explain this theoretically, without approximations, you have to solve the wave equation. The solutions have air that vibrates beyond the end of the tube. Ultimately this is the only "why" answer there is.

However, the fact that it can be characterized as an impedance suggests a sort of heuristic explanation by appealing to analogy with the case of zero frequency, i.e., steady flow of air through the tube, which is possible in the case of a tube that's open at both ends. In this situation, the continuity equation (i.e., conservation of air) says that as air enters (exits) the end of the tube, it has to speed up (slow down). This happens gradually because the air is spreading out from an opening of finite size.

The electrical analogy would be two very large blocks of ohmic material with connected by a solid cylinder of ohmic material. The current gets concentrated and then dispersed as it comes into and out of the cylinder.

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to enlarge only slightly upon Ben Crowell's thorough and useful treatment of this topic: the simple reason that there are nonzero pressure zones immediately outside the open end of the pipe is that the pipe is radiating sound energy out the open end- in fact, the pressure right at the opening is typically nonzero itself, inasmuch as the resonant column of air protrudes slightly out the open end.

The impedance of the pipe with an open end is discontinuous at the open end and thereby presents a serious mismatch condition there, with part of the wave being reflected off the discontinuity to return back down the pipe (which contributes to the production of the standing wave inside) and part of it radiating out into the free space outside. As pointed out by Ben, the concept of impedances is very useful in this context.

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  • $\begingroup$ I'm realizing I may be too un-taught to grasp these answers. To take a stab at saying it in really simple words, is it correct to say that the pressure is equal to air pressure simply because when a compression sound wave reaches the end of a tube, it can spread out in the open air. This creates a new wave which can propagate new areas of pressure greater than air pressure (at points beyond the tube end)? $\endgroup$
    – WilliamII
    Commented Feb 12, 2019 at 15:17
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I did find these answers on Quora which I thought were good (especially the first two or three). https://www.quora.com/It-has-been-explained-to-me-that-a-node-exists-at-the-open-end-of-a-pipe-when-a-standing-wave-forms-because-the-pressure-is-fixed-at-the-ambient-pressure-Why-is-this-true-If-this-is-true-how-can-sound-waves

I'm pasting the one answer here so you can see it. I think the answers given earlier on this site were no doubt right on the money, but my own lack of understanding made it hard to grasp them.

That’s a very relevant question!

Maybe it was just badly worded in your course/textbook. Or maybe whoever told you that did not actually think of the problem too much. The pressure at the opening of a pipe can not be constant and exactly equal to ambient pressure, because then, as you say, there would be no sound propagating from the opening.

However, since the opening is in contact with ambient air, any change in pressure from inside the tube will very quickly dissipate into the void. Thus, even though the pressure at the opening will oscillate slightly (as it is fed pressure oscillations from inside), the amplitude of the pressure wave will be much smaller there than anywhere inside the pipe (because the pressure differential is dissipating quickly into all directions), effectively creating a node somewhere close to the opening of the pipe.

Remember that physics is an attempt to approximate messy reality with simple models.

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