I Are 1D Longitudinal Waves Reflected at an Open End?

[greypilgrim]

Hi.

I found this intuitive explanation on StackExchange why a sound wave is reflected at the end of an open pipe. The basic idea is (better have a look at the link, there are diagrams) that when a pressure maximum travels along the pipe and leaves at the end, suddenly particles can flow away into directions that were previously limited by the walls, creating a minimum in pressure. Then air flows back, this rebound creates a pressure maximum just outside the tube which then reflects the next pressure maximum coming from inside.

This argument seems to only work in at least two dimensions, so is there no reflection at an open end in 1D? Does it even make sense to theoreticize about gases in 1D? I'm not even sure if pressure can be defined then (but surely particle density can).

What also confuses me about this argument is that it suggests that pressure is fluctuating at the open end, while for standing waves it is normally considered a node.

 

[A.T.]

... so is there no reflection at an open end in 1D? Does it even make sense to theoreticize about gases in 1D?

Consider a 1D mass-spring system, with a free end. Where would the wave energy go, if the wave wasn't reflected?

See for comparison the reflection of transversal waves at a free end:
https://www.physicsforums.com/threads/waves-travelling-between-strings.1079845/post-7255862

 

[jack action]

Imagine a pipe with one closed end and one open end. The pipe is filled with columns of air molecules, side by side. All molecules are at atmospheric pressure $P_{atm}$. The local pressure $P_{loc}$ on each column can be set separately. The local pressure ratio $P_r$ can be evaluated as $P_r = \frac{P_{loc}}{P_{atm}}$.

Let's increase the pressure on the first column gradually, say at $P_r = 1.25, 1.50, 1.75, 2.00$, and then go back down again at $P_r = 1.75, 1.50, 1.25, 1.00$.

1. The first column increases to $1.25$ while all other columns remain at $1.00$;
2. The first column increases to $1.25$ while the first column now increases to $1.50$;
3. After a while, we get the following pressure wave:
   1. column 1 at $1.00$
   2. column 2 at $1.25$
   3. column 3 at $1.50$
   4. column 4 at $1.75$
   5. column 5 at $2.00$
   6. column 6 at $1.75$
   7. column 7 at $1.50$
   8. column 8 at $1.25$
   9. column 9+ at $1.00$
4. If we keep the pressure on column 1 at $1.00$, this pressure wave will travel at the speed of sound through all the columns until it reaches the closed end. Note that we not talking about the molecules moving, only the pressure wave is;
5. At the closed end, this pressure wave will bounce back toward the open end. As it is going back, if a column has two pressure waves crossing each other, they will add up.

This example was with an increase in pressure (compression), but it also works with a decrease in pressure (expansion, i.e. $P_r \lt 1.00$)

Reflection of a pressure wave on a closed end (bottom graph "pressure"):

Opposing pressure waves (blue & red, with both compression and expansion pressures) crossing each other, resulting in the total pressure (black):

Now, what is different if we replace the closed end with an open end? An open end will reflect the opposite wave, i.e., a compression wave will reflect an equivalent expansion wave, and an expansion wave will reflect an equivalent compression wave.

Reflection of a pressure wave on an open end (bottom graph "pressure"):

The reason a compression wave bounces back on a closed end is to send a message back to the inlet: «There is no way out here, we're pushing everything back out.» With an expansion wave, the message is «We can't pull anything in from here, we're pulling whatever you took back in.»

With an open end, the molecules can exit freely, creating a rarefaction from a compression wave. So they send back this expansion wave saying, «We have an exit here, let us help pull this way.» With an expansion wave, it pulls the molecules in, which creates that compression wave, sending the message «We have an inlet here, let us help push toward your way.»

Creating the nodes and anti-nodes of a standing wave coincides with the pressure and reflected waves crossing paths and either maximizing (two compression or two expansion waves) or cancelling (one compression and one expansion waves) themselves. Knowing they travel at the speed of sound, with a fixed frequency of the input vibration, it is just a matter of adjusting the pipe length to locate nodes and anti-nodes such that it enters into resonance. The number of nodes within the pipe will determine the mode.

Animation of a standing wave (red) created by the superposition of a left traveling (blue) and right traveling (green) wave

​

 

[Baluncore]

This argument seems to only work in at least two dimensions, so is there no reflection at an open end in 1D? Does it even make sense to theoreticize about gases in 1D? I'm not even sure if pressure can be defined then (but surely particle density can).

The wave is a plane pressure wave, travelling along a tube. That is why it is called a 1D wave. The space in the tube is 3D, as is the space beyond the tube.

On reaching the open end of the tube, part of the wave energy becomes a radiating wave in 3D, the remaining energy is reflected by the impedance mismatch at the opening, as a 1D wave travelling back along the tube.

 

[A.T.]

The wave is a plane pressure wave, travelling along a tube. That is why it is called a 1D wave. The space in the tube is 3D, as is the space beyond the tube.

On reaching the open end of the tube, part of the wave energy becomes a radiating wave in 3D, the remaining energy is reflected by the impedance mismatch at the opening, as a 1D wave travelling back along the tube.

Note that the impedance mismatch only happens because the space is 3D, so it makes a difference for the pressure wave if there is a pipe or not. The wave propagation in the tube is along 1D only, but the motion of the particles is 3D.

If I understand the question correctly, @greypilgrim is thinking about a pure 1D medium, where the particles also move only along 1D. Here a pipe doesn't make a difference and there is no impedance mismatch at its end. Translating an open end to such a 1D gas is tricky, but you certainly can have a 1D mass-spring system with a free end (see post #2).

 

[tech99]

I suppose a wave on a single wire transmission line is 1D. As it travels along, it has a certain wave impedance, Zo. This is the ratio of electric to magnetic field and is dictated by the dimensions of the wire. All the time as it travels it sees ahead of itself an impedance equal to Zo. When it reaches the open end, it now sees an infinite impedance and that causes the reflection, because the current must now fall to zero.

 

[greypilgrim]

Consider a 1D mass-spring system, with a free end. Where would the wave energy go, if the wave wasn't reflected?

Well one could imagine the last oscillator absorbing all energy and its amplitude diverging until destruction, but I guess in an isotropic system it will just transfer the energy back into the chain, hence reflect it.

There's still this open question:

What also confuses me about this argument is that it suggests that pressure is fluctuating at the open end, while for standing waves it is normally considered a node.

I thought that an open end fixes the pressure to atmospheric there and hence create a node, but this explanation seems to describe an anti-node there (or slightly outside)?

 

[jack action]

I thought that an open end fixes the pressure to atmospheric there and hence create a node, but this explanation seems to describe an anti-node there (or slightly outside)?

A pipe with one closed end and one open end is the equivalent of a pipe twice as long with both ends open. Instead of going through the long pipe from one open end to the other, and having a node in the middle, the wave travels from the open to the closed end (where there is a node), bounces back, and returns to the original open end. So it is still effectively a pipe with "both ends open".

Watch the second video (Standing Wave Demo: Organ Pipes 2:33) in my previous comment to see the effect of blocking one end (the sound is one octave lower).

 

[greypilgrim]

I think you are talking about nodes of displacement, while I meant a node of pressure at an open end.

 

[jack action]

I think you are talking about nodes of displacement, while I meant a node of pressure at an open end.

I'm also talking about node of pressure.

For the first harmonic, the open pipe is equivalent to two closed pipes abutted together. Note that if the length of the open pipe $l_o$ is twice as long as the length of the closed pipe $l_c$ , then they have the same wavelength:
$$\lambda_o = 2l_o = 2(2l_c) = 4l_c = \lambda_c$$

 

[greypilgrim]

But your diagrams are clearly about displacement, not pressure. Here's the two side by side for a pipe with one closed end:

So just as I was saying, there should be a pressure node at the closed end. But the StackExchange explanation states that just outside the open end pressure fluctuates. That can't both be right, can it?

 

[jack action]

Sorry, I had it backward; your anti-node comment threw me off. But my logic still applies to what I wanted to say about the equivalent wavelengths:

Clearly, there is always a [pressure] node at an open end, where you have the atmospheric pressure.

Now I realize that when you were saying "this explanation" in your comment was actually meant for @A.T. and his explanation. I'm going to let him answer for it.