# Homework Help: How does stationary sound waves in open tubes happen?

1. Jan 1, 2012

### louis058

This is not a question as such, but it's for answering a homework question.
I understand how stationary sound waves in a closed tube are formed, as they reflect off the closed end and the two waves travelling in opposite directions interfere to form a stationary resultant wave (that's how I was taught anyway), but how can it happen in an open tube?
There isn't anything to reflect off, but a stationary wave still forms?

2. Jan 1, 2012

### espen180

It is an effect of the boundary conditions imposed at the end of the tube. Usually we require that the gauge pressure vanishes at the end of an open pipe. This is an okay approximation for a thin tube.

The gauge pressure is the pressure relative to the ambient pressure (p=ptot-pamb).

3. Jan 1, 2012

### Chopin

To build on what espen180 said, the boundary conditions of the pipe (the fact that the pressure at the open end of the pipe must always be equal to the normal atmospheric pressure), end up causing a reflection just like there would be if the pipe was closed. The difference is, the reflection from an open tube is the inverse of the incident wave--that way, the sum of the incident and reflected wave always cancel out the pressure at the end of the tube, enforcing the boundary condition. Since the reflected wave is the inverse of the incident wave, they form standing waves at different frequencies than a closed tube, where the reflected wave is identical to the incident wave. That's why open tubes and closed tubes have different resonant frequencies.

4. Jan 2, 2012

### louis058

I understand what you're saying, but I'm still not sure why the reflection from an open tube would be the inverse and such. Is there an online resource that could explain this to me?

5. Jan 2, 2012

6. Jan 2, 2012

### Chopin

Look at it this way. The physical behavior of air (whether in a tube or not) can be described by a second-order differential equation, right? This is why pressure waves move through it at all. By solving the equations, we find that the solutions take the form of sinusoidal waves, moving with a velocity $v$, the speed of sound. We're allowed to make the frequency of those sinusoids anything that we want, and since the differential equation is linear, we're also allowed to add up two solutions, and the result will also be a solution. So out in the open air, it's possible to describe a 1000 Hz wave and a 300 Hz wave, moving in different directions through the air simultaneously.

In the tube, though, we have some additional physical constraints. For one, the waves can only move along the direction of the tube, so instead of waves being able to move in all directions, they can just go in two: forward through the tube, or backward through the tube. We can still have any frequency we want, but they must be going either forward or backward along the axis of the tube.

The second constraint is the boundary condition, which like in all differential equation problems, helps us to winnow down the set of allowable solutions from the wide-open field that we started with. In this case, the fact that the tube is open at the ends means that at these points, the pressure must at all times be equal to atmospheric pressure (which I'll call 0 for simplicity.) Intuitively this is because outside of the tube is very big, and we don't have any hope of imparting enough energy to it to appreciably change its pressure (this is the same reason you can model a heat bath as having a constant temperature, or an ideal voltage source as having a constant voltage, etc.)

So where we used to be free to pick any set of sinusoidal solutions we wanted and add them up, the introduction of this boundary condition means that we are no longer allowed to do that. Now we can only pick combinations that, when added up, leave us with 0 pressure at the two ends of the tube. If you do the math on that, you discover that the only way to enforce this condition at all times is for the solutions to be the sum of two sinusoids--one travelling in one direction, and the other travelling in the other direction, with their phases and frequencies set up just right. In those specific conditions, the two waves add up so that the ends always have 0 pressure, and the rest of the tube forms a standing wave. Therefore, these combinations of waves are the only allowable solutions to the equation under these boundary conditions.