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!
 A: 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.
(from) http://medlibrary.org/medwiki/Acoustic_resonance
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).
