12
$\begingroup$

I do not understand why standing sound waves can be formed in a one-side or two-side open tube. Consider a one-side open tube.

In particular how does the reflection of the wave at the open end occur? I found the following explanation.

The sound wave is a pressure wave and at the open end of the tube the pressure must be fixed to the atmosferic pressure and cannot vary, thus there is a node of pressure there.

I do not get why the pressure at the open end cannot vary from the atmospheric pressure.

Furthermore, if the outside pressure does not change, the sound wave (a pressure wave) is somehow trapped in the tube and do not exit from it. Nevertheless for example in a clarinet the sound wave do exit and propagate from the tube in air, I guess. Otherwise how could the sound be heard?

What is the physical reason why there must be pressure waves nodes at the open end of a tube (as in picture (b))?

enter image description here

$\endgroup$
  • $\begingroup$ It is all because of boundary conditions, which is pressure should be equal to atmospheric pressure where there is an open end, which means no pressure disturbance or displacement of particles at the open end, which means there should be a node at the open end, and from this you can derive everything else. $\endgroup$ – SamD97 May 23 '16 at 7:23
10
+50
$\begingroup$

From the Wikipedia article on sound:

In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement.

To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave,

$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}$$

(here $c$ is the speed of sound) but also the variation in the flow of air, that is, a particle displacement wave:

enter image description here

We consider the air in the tube to have a rest position, and the wave motion is expressed in terms of displacement from that position. If we denote by $ξ(x,t)$ the displacement of the air at position x at time t, then the wave equation for displacement is (I added a derivation at the end of this answer):

$$\frac{\partial^2 ξ}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 ξ}{\partial t^2}$$

As you cited, the pressure at the open end must be equal to the pressure of ambient air outside the pipe: this is just a boundary condition (I'll come back to this later, but a simple way to understand it is: if we didn't satisfy continuity of pressure, we'd have an infinite pressure gradient and so infinite force). However, for an open end, air can flow freely in and out: that is, the displacement wave does exit and propagate. This diagram shows what happens in an open (left) vs closed pipe (right) of the same length:

enter image description here

The red line is the amplitude of the variation in pressure, whereas the blue line is the amplitude of the variation in the flow of air. As you can see, the amplitude of the displacement wave is a maximum at the open end.

Now, to answer your questions, "why the pressure at the open end cannot vary from the atmospheric pressure?" and "how does the reflection of the wave at the open end occur?", I'll cite a passage from this wonderful explanation I've found (you should take a look at the animations if this is not clear enough, or alternatively at this video): Let's send a pulse of air down a cylindrical pipe open at the end. Then...

It reaches the end of the tube and its momentum carries it out into the open air, where it spreads out in all directions. Now, because it spreads out in all directions its pressure falls very quickly to nearly atmospheric pressure (the air outside is at atmospheric pressure). However, it still has the momentum to travel away from the end of the pipe. Consequently, it creates a little suction: the air following behind it in the tube is sucked out (a little like the air that is sucked behind a speeding truck).

Now a suction at the end of the tube draws air from further up the tube, and that in turn draws air from further up the tube and so on. So the result is that a pulse of high pressure air travelling down the tube is reflected as a pulse of low pressure air travelling up the tube. We say that the pressure wave has been reflected at the open end, with a change in phase of 180°.

For a more formal treatment of reflections at an open end, see this. Let me know if I can make myself more clear on any point.

Addendum: A derivation of the two wave equations.

The acoustic pressure is the local pressure deviation from the ambient, that is:

$$p(x,y)=P(x,t)- p_a$$

where $p_a$ is the ambient air pressure, and $P(x,t)$ is the absolute pressure. In this situation, Hooke's law states that

$$p=-B \frac {\partial ξ}{\partial x}$$

where the constant $B$ is the bulk modulus of air (i. e., air's resistance to uniform compression). Newton’s second law of motion implies that

$$\frac {\partial p}{\partial x} = - p_a \frac{\partial^2 ξ}{\partial t^2}$$

Combining these equations, we get

$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}$$

and

$$\frac{\partial^2 ξ}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 ξ}{\partial t^2}$$

where $c = \sqrt{B/{\rho}}$ (from Newton-Laplace equation), $\rho$ being air density. Source of this derivation: Dave Benson's Music: A Mathematical Offering.

Later addendum:

As @knzhou pointed out in his answer, it should be mentioned that for an open end, $p = 0$ is really only an approximation, because the volume of air just outside of the tube is not infinite. However, you can make a more accurate representation of the actual tube by working in terms of an effective length, which is a little longer than the geometrical end. This diagram shows the effective length for the fundamental vibrational mode of a flute:

enter image description here

The end correction is the amount by which the effective length exceeds the actual length, and under normal conditions it is usually somewhere around three fifths of the width of the tube.

Also, one could wonder what is actually happening at the boundary between the tube and the open air. All I know is that if inside the tube there is a plane wave, and when the displacement wave exits and propagates externally, it becomes a spherical one, then there must be some complicated effect going on between the two (as was suggested by @docscience in the comments). Perhaps a more knowledgeable person can explain it to us.

$\endgroup$
  • 1
    $\begingroup$ Excellent answer! You'll largely deserve the bounty! Minor point: in the addendum, do not call $ρ$ the ambient pressure, all the more as you later use $ρ$ to denote density... $\endgroup$ – L. Levrel May 23 '16 at 19:32
  • $\begingroup$ I've recently wondered about this myself and will have to research some of the references you've provided. But I expect the textbook explanations are all first order, simple approximations to what's really happening. It always seems if you look a little closer, experiment a little deeper that you'l find additional complexity involved. The somewhat soft boundary at the open end of a tube is probably such that the node (or antinode) does not physically exist at the geometric boundary but rather just inside (or outside) of the tube. Furthermore rather than being a point, I would expect it to .. $\endgroup$ – docscience May 23 '16 at 20:05
  • $\begingroup$ @L.Levrel Thanks for your comment, fixed it! And glad you liked the answer! $\endgroup$ – David Herrero Martí May 23 '16 at 20:28
  • $\begingroup$ @docscience I subscribe to your opinion, that's surely not the end of the story. Dave Benson's book is specially great for its treatment of the physics of musical instruments, you might like it. $\endgroup$ – David Herrero Martí May 23 '16 at 20:38
5
$\begingroup$

You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation.

Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The distance of this leakage is on the order of the radius of the tube; past then, the pressure is approximately the atmospheric pressure.

Actually solving the problem requires solving the full 3D wave equation with complicated boundary conditions. The few times this can be done explicitly, the result is a mess of special functions. However, to a good approximation, one can replace the full 3D solution with a 1D solution with its node displaced slightly out of the tube. This is called an 'end correction' and is important for the construction of musical instruments.

$\endgroup$
0
$\begingroup$

As your animation indicates, maximum displacement occur at the open end. That's at odds with your assertion that the open end is a node. I think the open end is an antinode, similarly, the ends of an open-open tube are antinodes.

$\endgroup$
  • $\begingroup$ You're right that the maximum displacement corresponds to an anti-node of displacement, but this does not correspond to anti-node of pressure (actually, pressure nodes lie on displacement anti-nodes, and vice-versa). $\endgroup$ – Kyle Oman Sep 17 '18 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.