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All resources on standing waves give the same explanation: waves are free to move in and out of the open end of a pipe and, therefore, this open end must be an antinode. This, of course, explains why standing waves can only form in pipes with length $(2n-1)\lambda$ where $n$ is an integer. This didn't satisfy me though, so I investigated what would happen if I tried to fit a standing wave in a pipe of length $n\lambda/2$ where $n$ is an integer and, naturally, what you get is a displacement node at the open end. Why is this not possible and why does air have to move at the open end? Why could you not just get cancellation and what is wrong with a longitudinal standing wave in a pipe with length of an integer number of half wavelengths?

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  • $\begingroup$ "so I investigated what would happen if I tried to fit a standing wave in a pipe of length nλ/2 where n is an integer and, naturally, what you get is a displacement node at the open end." What do you mean by this? Are you saying you essentially just made standing waves as if the end were closed? How did your analysis actually use the fact that one end is open? $\endgroup$ Oct 27 at 1:24
  • $\begingroup$ Yes, that’s precisely what I’m saying; I simulated standing waves as if the end was closed and I don’t understand why the boundary conditions of an open end don’t allow that. $\endgroup$
    – P0W8J6
    Oct 27 at 1:27
  • $\begingroup$ What about saying the air is free to move in and out of the end of the pipe is not sufficient for you? What about that does not make sense, specifically? $\endgroup$ Oct 27 at 1:29
  • $\begingroup$ Sure, the air could be free to move in and out but why does that stop the formation of a node? The air would be free to move in and out but merely wouldn’t do that in this specific case because of the constant phase cancellation of the reflected wave. $\endgroup$
    – P0W8J6
    Oct 27 at 1:32
  • $\begingroup$ It goes a little distance to answer my confusion but I am still stuck on a number of issues: first, a response on your linked question mentions that, if the open end was a displacement node and, by extension, a pressure antinode, there would be an infinite pressure gradient. I don’t understand how this would be the case and I don’t understand how this stops a standing wave from forming. What physically happens when I send a wave down an open ended pipe of an integer number of half wavelengths (which would form a node at the open end) and how does this stop a standing wave from forming? $\endgroup$
    – P0W8J6
    Oct 27 at 2:08
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Here is a physically intuitive way of thinking about this:

Because the end of the pipe is open (i.e., looking at the impedance of free air), a pressure maximum cannot form at the open end because there is a displacement degree of freedom there which a closed end doesn't have: the freedom to propagate all the way out of the pipe and into free space. This requires movement of the air in and out of the mouth of the pipe, which means there is a displacement maximum at the open end and a pressure minimum at the open end.

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It's good question: it clear why you must have a node at a closed end, and why you shouldn't have a node at the open end. But why do you have to have an anti-node? Couldn't you have neither, where the pressure amplitude is non-zero, but not maximal? I'll call it a "mixed-node" for sake of argument.

Let's call a node $|0\rangle$ and an anti-node $|1\rangle$. A mixed node would then be $|\alpha\rangle$ where $0\lt|\alpha|\lt1$, and may be complex. Since the sound wave equation is linear, the mixed node cane written as a mixture:

$$ |\alpha\rangle=\sqrt{1-|\alpha|^2}|0\rangle+\alpha|\alpha\rangle$$

The problem is, the open end projects out $|0\rangle$, so

$$\sqrt{1-|\alpha|^2} =0$$

thus:

$$\alpha =|\alpha|e^{i\phi}$$

With the right choice of $t=0$ we can make $\phi=0$, so:

$$\alpha=1$$

and

$$|\alpha\rangle = |1\rangle$$

Only the anti-node survives.

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    $\begingroup$ The OP is asking why the open end projects out $|0\rangle$ though. $\endgroup$ Oct 27 at 2:20
  • $\begingroup$ Thanks for your response. I have not yet studied wave equations but I follow along somewhat. What would happen physically if you had a maximal pressure antinode at the open end? $\endgroup$
    – P0W8J6
    Oct 27 at 2:20
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The resource answers are concerned with standing waves that (with some simplifying assumptions) are self-sustaining with the given boundary conditions.

Setting something up with no displacement at the free end is not impossible, but requires you to externally apply whatever pressure is needed to keep this other response going.

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