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Standing waves are causing great confusion for me. I have read many answers on stackexchange. However, I still don't understand standing waves within closed and open tubes.

Problem 1. I struggle to understand why there must be an antinode on the open end. However, I understand why there must be a node at the closed end. My intuition here, is that the sound wave reflects at this point, and overlaps with the incoming sound wave causing destructive interference. Hence a node. Now the antinode at the open end is not so trivial. I understand that there is nothing blocking it so it can have maximum displacement. However, why does it have to have this maximum displacement. Could a node just as well form at the open end? When discussing with my teacher, he said that it is possible to have a node at the open end, but you wouldn't hear anything. Is this correct? Could it be that you can have a node, but then it no longer exists as a harmonic?

Problem 2. All diagrams I have seen, draw the antinodes tocuhing the walls of the pipe. However, I struggle to understand why this must be the case. Could I not have a small standing wave that doesn't touch the walls of the pipe. That is the yellow lines amplitude is shrunken such that they don't touch the walls of the pipe. What is limiting this? enter image description here

For example this. Here the lines do not touch the tube.

Edit: Consider sending a wave with a wrong frequency such that there is no formation of a standing wave. Can we have a node at the open end then?

A comment on the linked post states there can either be an antinode or a node at the open end but you wouldn't hear a sound if they were nodes (or at least it wouldn't be very harmonic). Is this correct? This aligns with what my teacher has said.

I am trying to understand which statement is correct.

  1. That is there can be either an antinode or a node at the open end, but a node results in it not being a harmonic.

  2. That it must always be an antinode. If this is true, please explain why this is the case.

Farcher has explained my latter problem superbly. However, I still have confusion about problem 1. I understand that an antinode can form on the open end as it has no restrictions. However I do not understand why there must be an antinode. Shouldn't no restrictions correlate to either being able to exist at the open end?

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You need to read my answer to the post Why are the closed and open ends of an organ pipe nodes and anti-nodes? first to understand that those yellow lines and red lines are graphs without the axes or the corresponding labels shown.

So in your last diagram ignore the blue pipe other than it giving you an indication of where in the pipe the displacement is measured.
The vertical dimension has noting to do with the magnitude of the horizontal displacements of the particle in the tube.

enter image description here

The two graphs (green and red) indicate the displacements of the particles at times separated by half a period.

All you diagrams show displacements of particles and it is perhaps not surprising that a particle at a fixed end does not move as it has nowhere to go?

Indeed the antinode at a open end is difficult to comprehend but think of it as sound reflect from the interface between a restricted (by the walls of the pipe) volume in the pipe and an unrestricted volume outside the pipe.

There is never a displacement node at an open end however there is a pressure node there.

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  • $\begingroup$ Consider sending a wave with a wrong frequency such that there is no formation of a standing wave. Can we have a node at the open end then? $\endgroup$ Aug 3 at 19:08
  • $\begingroup$ @QuinGardinerBax Sound waves are longitudinal and so the oscillations are left to right and right to left in the diagram but that is very difficult to draw. The graph might give the impression that it is a transverse wave but it is not. $\endgroup$
    – Farcher
    Aug 3 at 22:43
  • $\begingroup$ If there is no standing wave then where would a node come from? $\endgroup$
    – Farcher
    Aug 3 at 22:43
  • $\begingroup$ On the linked post. Jay walker states that "They can be either but you wouldn't hear a sound if they were nodes (or at least it wouldn't be very harmonic)". This suggests there can be a node at the open end. Is this correct? $\endgroup$ Aug 4 at 20:35
  • $\begingroup$ You have explained my other question superbly. However, I still don't understand why there must be an antinode. $\endgroup$ Aug 4 at 20:37
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As Farcher already answered your second question I will just try to answer your first question: As Farcher already explained in this post Why are the closed and open ends of an organ pipe nodes and anti-nodes? there are two ways to describe a sound wave. You can describe it using pressure or the displacement of the air particles. The way you talked about it is via the displacement of air particles. The pressure desciption is 90 degrees out of phase with the displacement description which Farcher shows very nicely in the first diagram of the post mentioned above. As d_b explained in Why antinodes must be present at the free ends of an open organ? (Neglecting end correction) for the pressure to be continous in space (meaning its value doesn't suddenly jump up or down at a point) the pressure at the end of the pipe needs to be equal to the surrounding air pressure outside of the pipe. This means that the pressure wave has a node at this point (because the pressure value doesn't differ from the usual air pressure without sound, which is (approximately) equal to the pressure of the air outside the pipe). And because the pressure wave is 90 degrees out of phase with the air particle displacement wave, this means that the displacement wave has an antinode at this point (because nodes and antinodes always differ by a phase of 90 degrees). Therefore, to answer your question, I would argue that at the open end, there always has to be an antinode of the displacement wave (which is the wave you were talking about). I think the comment you mentioned from the other post, stating that it could also be a node which wouldn't sound harmonic, is wrong.

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