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Let us consider a spherical cavity of radius $R$ and a point-like source of field at its center.

In the following I will use the jargon of acoustics, but I think that one can think of the propagation of electromagnetic waves, or any other field.

Let us assume that the membrane of the loudspeaker oscillates at a frequency $f$ and that the speed of sound of the medium within the sphere -say air- is $c$. See attached picture.

Let us further assume that the spherical shell is perfectly reflecting.

I am wondering what happens upon changing the ratio $\lambda/R$. In particular: what happens if $\lambda > R$? Can the loudspeaker radiate sound in the medium? Can the energy be transferred from the loudspeaker to the medium? How is -qualitatively- the pressure field in the medium: flat, a complex time-dependent interference pattern, or localized around the loudspeaker (as evanescent fields in electromagnetisms)?

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AndreaPaco
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  • I am more comfortable with EM verbiage. Size controls reflectivity and bandwidth. Efficient, low reflectivity, transmission is an issue of impedance matching see https://physics.stackexchange.com/questions/449947/why-are-longwave-radio-towers-so-tall/450072#450072 The really difficult problem is that the bandwidth disappears exponentially fast with the size of the radiator, hence the need for a separate bass/sub-woofer, for example. – hyportnex Jul 25 '23 at 14:52
  • I don't have the answer up my sleeve but to my understanding, this boils down to calculating the radiation impedance the loudspeaker "sees" (or radiates into). If I am not utterly in error I believe the quantity of interest here is not $\lambda$ but $\lambda / 2$ since a standing wave could be sustained in the radial direction for this length. So, below the region $2 R \ll \lambda / 2$, you go into the "pressure zone" (a term borrowed from room acoustics) where no standing waves are supported but the source still radiates. I haven't gone through the maths but I believe this would result (cont) – ZaellixA Jul 25 '23 at 17:19
  • (cont.ed) somewhat spatially uniform radiation impedance for the source, with uniform being the limit. Still, there would be some impedance where the acoustic energy would be transferred. You would most probably end up with a large impedance mismatch which could drastically reduce the radiated acoustic energy. I am sorry I don't have any numbers or maths to throw. If I manage to go through some derivation I'll post it as an answer. – ZaellixA Jul 25 '23 at 17:22

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