# Wavelength and Frustum shaped resonant cavity

Given a right conical frustum resonator top radius r, bottom radius R and height h, what is the relationship of the wavelength w that we wish to create resonance with, to r, R, and h? Assume for now that the sides of the chamber are perfectly reflective and that the wave, acoustic or electromagnetic, is introduced on the sloping side of the frustum. Much more interested in the sources than the answer but both would be nice! Thank you!

• Is this what you mean by "frustrum," or do you mean something else? – Kyle Kanos Aug 6 '14 at 20:27
• Acoustic wave on surface or in volume? Reflection on top and bottom? Looking for first harmonic only? It seems that the wavelength will be $h/2$ so you have nodes at top and bottom - but I expect it will not be a sinusoid so the more interesting question is what the resonant frequency is... – Floris Aug 6 '14 at 20:29
• This is what I mean by frustrum, basically a cone with the pointy end lopped off. Top, bottom and sloping sides would all be reflective, looking for resonance inside. Again more interested in how to derive. jwilson.coe.uga.edu/emt725/Frustum/Formulas.html – Randy Howk Aug 6 '14 at 21:12
• is it open on the end or closed? – Gödel Aug 7 '14 at 2:30
• closed on both ends EM wave (microwave or laser) – Randy Howk Aug 7 '14 at 20:45

Everywhere I go says pretty much the same thing:

An open conical tube, that is, one in the shape of a frustum of a cone with both ends open, will have resonant frequencies approximately equal to those of an open cylindrical pipe of the same length.

The resonant frequencies of a stopped conical tube — a complete cone or frustum with one end closed — satisfy a more complicated condition:

$kL$ = $n\pi - arctan(kx)$

where the wavenumber $k$ is

$k$ = $2\pi\frac fv$ and $x$ is the distance from the small end of the frustum to the vertex. When $x$ is small, that is, when the cone is nearly complete, this becomes

$k(L+x)$ approximately equal to $n\pi$

leading to resonant frequencies approximately equal to those of an open cylinder whose length equals $L + x$. In words, a complete conical pipe behaves approximately like an open cylindrical pipe of the same length, and to first order the behavior does not change if the complete cone is replaced by a closed frustum of that cone.

This is a Stanford paper on the subject: ftp://ccrmaftp.stanford.edu/pub/Publications/Theses/GaryScavoneThesis/thesis.pdf

So, as far as I can tell, it depends on the length and not the radius. However, in the Stanford paper it does talk about the changing radius of the frustum, and its effects on the rest of the wavefront (but that was only in terms of air pressure, not general wave physics).

• Thanks! this is helpful I am looking for for EM, like microwaves or a laser behind a one way mirror with the emitter embedded in the sloped part of the frustrum and a closed on both ends, but this is a good start. – Randy Howk Aug 7 '14 at 14:54
• My pleasure, what you are looking for is esoteric, good luck with your research! – Gödel Aug 7 '14 at 21:06
• Aaargh! Apparently NASA had a similar idea AND the physics chops to follow up on it! iflscience.com/space/warp-drives-mars-and-back-time-lunch – Randy Howk Apr 27 '15 at 18:42