Why can't a standing wave form a displacement node at the open end of a pipe?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.