What happens when a sound wave gets to the open end of a pipe? In a crude approximation, we could say that, outside the pipe, the pressure is atmospheric, so that it doesn't vary in time, so the acoustic pressure at the open is zero and the wave is reflected with 180° phase change. But of course some sound is radiated from the open end of the pipe so, even if the acoustic pressure is small compared to that inside the pipe, it isn't zero.
A related problem was first solved by William Strutt (aka Rayleigh). He considered a circular piston (radius a so area A = πa2) oscillating sinusoidally with amplitude xm and frequency ω in a large plane baffle, so its position with respect to the plane of the baffle is x = xmsin ωt. The acoustic flow at the surface of the piston is
The pressure p at the surface of the piston is the pressure due to the impedance Zrad of the radiation field. We should expect that the main contribution to this pressure is the inertance of the air very close to the piston: whereas more distant air is both very slightly moved and slightly compressed/rarefied, the air closest to the piston and just in front of it must move with it. Consider then a volume of air with density ρ, area A and length δ, and let its inertia determine (approximately) Zrad. The force required to move it is
F = mass*acceleration = (ρAδ)(dv/dt) = (ρAδ)(− ω2xmsin ωt)
But from the definition of pressure, F = pA, so the radiation impedance has a magnitude
and its phase is +90°: the radiation pressure is 90° ahead of the flow. For this piston in the infinite flange, δ turns out to be 0.85a.
Now we can consider this piston as approximating the flow at an open pipe. So, for a pipe ending in a large flange, there is an inertance due to the volume of air, with length δ ~ 0.85a, that moves with the wave at the end of the pipe. For a simple, narrow, open pipe, δ is 0.6 a. These are called end effects. For frequencies low enough tht δ << λ, these pipes behave approximately like ideally open pipes, but the radiation inertance at the open end gives them an extra length δ.
So, what happens to a wave when it gets to the end of a pipe? Suppose that our pipe is very long, but ends at x = 0, as sketched below. The pipe (x < 0) has a characteristic impedance Z0, and the radiation impedance at the end is Zrad. So the impedance at x = 0 depends on which way one 'looks'. 'Looking out' (at x = 0, but looking at x > 0), the impedance at the end of the pipe is Zrad. 'Looking in' (at x = 0, but looking at x < 0), the impedance at the end of the pipe is Z0. (As we've seen above, Z0 is real: the impedance of a very long pipe is resistive, while Zrad is largely intertive: its impedance is nearly completely imaginary.)
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 travelling 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,
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+) = pradsin (− ωt).g(x) , and the flow due out of the pipe due to this wave is U(0+) = Uradsin (− ωt).h(x) . As we also wrote above
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:
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.
In the sketch, a wavetrain p> in red is shown reflecting at the end of a pipe: p< is in blue and their sum, the standing wave, in black. The radiated wave is purple.
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.