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This question follows from a previous SE question asking how the thickness of a material affects the acoustic transmission coefficient.

This website seems to suggest that the equations defining the compliance and inertance components of specific acoustic impedance ($z$) are dependent on a certain assumption of scale:

Both use the idea of a compact region: a region whose dimensions are much smaller than the wavelengths we are considering.

Further searching suggests this simplifying approach is also sometimes called acoustical compactness:

The ‘size’ of the body at a given frequency is called its compactness and is characterized by the parameter $ka$ where $a$ is a characteristic dimension, or by the ratio of characteristic dimension to wavelength $a/λ$. A compact source, one with $ka ≪ 1$, radiates like a point source, while non-compact bodies must be treated in more detail, as we saw in the case of a sphere in §2.1.


Does this mean the usual equations describing acoustic impedance cannot be applied when a media layer has thickness $a \ll \lambda$ ? I can imagine that pressure starts to become more complicated to describe at this scale, but how exactly is the assumption involved?

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No, it does not. consider 1/4" thick lead sheeting: it does an excellent job both absorbing incident sound of certain frequencies and reflecting other frequencies. plane wave incidence of sound waves and the related equations for reflection and transmission as functions of impedance ratios work well in that regime.

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  • $\begingroup$ BTW the usual (engineering) assessment of applicability of the equations I referenced above is if the characteristic length is of order ~ 1 wavelength; meaning for example that a flat plate one foot across radiates well at 1000Hz. $\endgroup$ – niels nielsen Dec 21 '17 at 6:03
  • $\begingroup$ Thanks Niels, that's very close to what I'm looking for, and incidentally I agree with you so far. However, the previous question was asking about very thin plates (I think they said 100 Tungsten atoms, or ~ $1.35 \times 10^{-8} \ m$ thick.) Given this more specific example, does your answer mean this very thin sheet would only begin to impede waves with frequency $\gt$ 74 MHz? I guess I can also imagine the pressure wave deforming the sheet as well, instead of causing it to vibrate? $\endgroup$ – D. Betchkal Dec 21 '17 at 18:31
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    $\begingroup$ I never dealt with sheets that thin, so I don't know how to analyze the problem. but this might help: look up "ribbon microphone" on wikipedia. these are devices in which an extremely thin sheet of metal is suspended in a very strong magnetic field. sound waves impinging upon the ribbon of metal cause it to vibrate in the field and a very tiny current is thus developed within the ribbon. tiny wires carry off the current to an amplifier. these fragile mics are used in recording studios, particularly for recording saxophones, and there must be some literature on their physics. $\endgroup$ – niels nielsen Dec 21 '17 at 18:58
  • $\begingroup$ A very helpful suggestion. Apparently ribbon microphones operate like a pressure-gradient ear. With low mass, it appears the ribbon can (fairly closely) move with the particles in the medium. From wikipedia: "Ribbon microphones are also called 'velocity microphones' because the induced voltage is proportional to the velocity of the ribbon and thus of the air particles in the sound wave, unlike in some other microphones where the voltage is proportional to the displacement of the diaphragm and the air." Do you think the nearly bi-directional pickup pattern suggests there isn't much impedance? $\endgroup$ – D. Betchkal Dec 29 '17 at 19:51
  • $\begingroup$ the output impedance of a ribbon mic is extremely low and, as you note, the ribbon is well-coupled to the air surrounding it. i conclude from this that the acoustic impedance of the ribbon is also very low. best regards, niels $\endgroup$ – niels nielsen Dec 29 '17 at 19:54

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