That was the qualitative explanation my link above gave. At the end of it there was another link to a more rigorous treatment that kind of explained the math behind it.
The pure truth here is that this is a complicated boundary physics problem. You need to see how the pressure wave that comes at the end is released outside and spreads (fluid mechanics) and what is the reaction (effectively a reflected wave) to this essential change in medium from that of the inside the air column geometry. I suppose only if one put some numbers to it to see exactly how high/low the pressure gets to compare to atmospheric you can get an idea why the end region is essentially feeling it is all a different medium (and the suction created) and all this is non trivial but an essential distinction between inside and outside the pipe (even if still open) .
Did you go to the next page after the quoted blue statement? You will see there that continuity of flow and pressure results in a reflection (much like there is nowhere to go for molecules but in such directions or redistribution that effectively generate a new wave back). It is air inside and outside and no real wall in between at the end but it does behave like some kind of wall due to the different geometry and the redistribution of air this causes at the boundary to satisfy essential things like conservation of mass (no more air is generated or lost in the whole system) and lack of discontinuity in pressure (smoothness in gradient no infinite forces).
See here about the reflection of a wave when it changes medium.
http://www.acs.psu.edu/drussell/Demo...t/reflect.html
Imagine that someone is blowing air inside the pipe that needs to get out eventually so this is a substantial source of alteration of local properties and deserves full fluid mechanics analysis of the boundary.
Read what the link says;
Earlier in
http://www.animations.physics.unsw.e...-impedance.htm
"The specific acoustic impedance z is the ratio of sound pressure to particle velocity, and z = ρv , where ρ is the density and v the speed of sound. (See Acoustic impedance, intensity and power to revise). So for our duct with cross sectional area A, provided that the wave is strictly one dimensional and traveling in one direction, the acoustic volume flow is just U = Au .
For this very special case, we define the characteristic acoustic impedance Z0, where
Z0 = p/U = p/Au = z/A so
Z0 = ρv/A
Of course, there usually are reflections from the other end of the pipe, whether open or closed. So there is a sum of waves traveling to the right and left and quite often these give strong resonances, which is after all how musical wind instruments work. For ducts however, the characteristic acoustic impedance Z0, is the dimensional quantity that scales the acoustic impedance. To see what the impedance spectra of simple ducts and musical instruments look like, see What is acoustic impedance.
It is possible to have no reflections, however. If the pipe length L is very long, then it takes a long time for the reflection to return. So, for the time 2L/v, the duct is effectively infinite. Further, when the echos do eventually return, they are much attenuated by losses at the wall. (I mention this because, in our research laboratory, we use acoustically infinite acoustical wave-guides as calibrations for acoustic measurements. See this web site http://newt.phys.unsw.edu.au/music/ for the research lab, this page http://newt.phys.unsw.edu.au/jw/acou...asurement.html for measurement techniques or this scientific paper http://newt.phys.unsw.edu.au/jw/repr...SmithWolfe.pdf for some technical information, if you’re interested.)
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"Let's suppose that we send a sine wave along the pipe towards the right. Let the pressure due to this wave be p = p>sin (kx − ωt) , where the subscript in the amplitude is to remind us that the wave is traveling to the right (See The wave equation for sound). The flow due to this wave is U = U>sin (kx − ωt) and, as we saw above,
p>/U> = Z0.
Outside the pipe (at x = 0 but looking out, which we'll write as x = 0+), there is sound radiation away from the pipe. Let's write that the pressure* looking into the radiation field (at x = 0+) is p(0+) = prad sin (− ωt).g(x) , and the flow due out of the pipe due to this wave is U(0+) = Urad sin (− ωt).h(x) . As we also wrote above
prad/Urad = Zrad ,
remembering that Zrad is almost entirely imaginary, so that prad is almost 90° ahead of Urad in phase—so at least one of them has to be a complex number.
* We can't write a simple one dimensional wave equation here, because the radiation diverges. The geometry very near the pipe end is actually somewhat complicated, but it's not important to this argument. Away from the end, i.e. for kr >> 1, where r is the distance for the end of the pipe, the amplitudes of both prad and Urad go like 1/r and they are in phase: the local geometry approaches that of a plane wave.
Continuity of pressure. Now exactly at x = 0, at the end of the pipe, the pressure just inside the pipe must equal that just outside. (If we didn't satisfy continuity of pressure, we'd have an infinite pressure gradient and so infinite force.) We also have
Continuity of flow: the flow coming out of the pipe (at x = 0−) must equal the flow radiating outwards (at x = 0+). (Continuity of pressure is required because we are not creating or destroying air.) Of course, we can't satisfy those conditions with these two waves: the magnitude of Z0 is much greater than that of Zrad so, if we set the flows equal, the pressures can't be equal.
What happens is no surprise: there is a reflection at the open end: let's write the reflected wave, which travels to the left, as p = p<sin (kx + ωt) and the flow due to this wave is U = U<sin (kx + ωt). So, at x = 0, we have to satisfy the continuity of pressure and flow conditions:
p> + p< = prad and U> − U< = Urad .
(Why the minus sign? Because U> and Urad are flowing out of the pipe while U< is flowing in.) We can use the equations above for Z0 and Zrad to rewrite the equation above right as p>/Z0 − p</Z0 = prad/Zrad. This and the equation above left are two equations that can be solved to give the reflection coefficent at the end:
Reflection coefficient = p</p> = (Zrad − Z0)/(Zrad + Z0) = − (1 − Zrad/Z0)/(1 + Zrad/Z0)
where we remember that the reflection coefficient is in general a complex number (Zrad is almost completely imaginary). Now, except at very high frequencies, |Zrad| << Z0, so the reflection coefficient is almost −1: we have reflection with a change in phase of π in pressure, but no change in phase in the flow. This gives us a pressure node (two waves out of phase add approximately to zero) and flow antinode at an open end (two waves in phase add to give a maximum).
So, we have a reflection at the open end of a pipe: very useful the standing waves in musical instruments, the voice and elsewhere. Examples, explanations and animations are given in Open and Closed Pipes and Pipes and Harmonics.
Reflection at a closed end is somewhat easier. If the pipe is ideally closed, then the conservation of flow must mean that U> and U< add to zero, which means that they are out of phase by π. Their pressures are in phase, so the closed end of a pipe has an antinode in pressure and a node in flow.
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I just put all this here to motivate you to go and really look at the link for a cleaner attempt to analyze this than the hand waving arguments about why reflection is unavoidable for essential physical reasons from fluid mechanics.
Last edited by masque de Z; 10-13-2016 at 03:29 PM.