1
$\begingroup$

Would the math formula for resonant air columns open at both ends apply to a hollow tube shaped like a donut?

No ends are closed because there are no ends. Would it be considered an infinitely long open air column?

What would the math formula be to determine the fundamental?

$\endgroup$
2
$\begingroup$

A donut shaped pipe can have a standing wave whose wavelength is equal to an integer fraction of the circumference of the pipe - you would have one wave traveling clockwise, and another traveling anticlockwise, and where they result in node and antinodes depends on their relative phase.

So for a pipe of length $L$ bent into a torus, the radius $R = \frac{L}{2\pi}$ and the wavelengths would be

$$\lambda_n=\frac{L}{n}$$

The frequency is of course the speed of sound $c$ divided by the wavelength $\lambda$, so

$$f_n = \frac{n c}{L}$$

Or in terms of the major radius of the torus,

$$f_n = \frac{n c}{2\pi R}$$

Where the fundamental frequency is the value obtained when n=1

$\endgroup$
0
$\begingroup$

If the pipe is sufficiently thin relative to the wavelength of the vibrations, it makes no difference whether it is straight or bent. For example long organ pipes are often made with sharp 90 or 180 degree bends to fit them into the available space, and this has no practical effect on the sound the produce.

For a pipe open at both ends, the relative phase of the velocity at each end is arbitrary. At resonance, the two velocities can be either in phase or out of phase.

If you bend the pipe and connect the ends, the velocities at the two "ends" much be in phase with each other. If they are not, the pressure at that point will not stay constant at zero.

So if the open pipe modes have frequencies 0, f, 2f, 3f, etc., the donut modes will be 0, 2f, 4f, etc.

However, apart from the zero frequency, the donut modes can have their nodes and antinodes at any position around the ring (the donut doesn't have "ends" like an open pipe). So a resonance with a frequency 2f, 4f, can be considered as an arbitrary linear combination of two modes with the same frequency, but with nodes and antinodes at different positions around the donut.

$\endgroup$

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.