Right, so this is odd:(adsbygoogle = window.adsbygoogle || []).push({});

In my course there are derivations of transmission for 2 cases, normal incidence of an (acoustic) wave on an unbounded interface, and oblique incidence of an (acoustic) wave on a flexible unbounded interface.

So my idea was, the second will be an expanding case of the first.

However, they seem to contradict. (Skip below for the mathematical derivation. Stay here if you feel more capable of giving a purely physical, intuitive explanation.)

Whereas the first case seems to find the mass factor of the interface to be more important for higher frequencies, and the stiffness of the medium (=the interface suspension) for lower frequencies, it's the other way around for the second case!

This seems to be so because in the second case, instead of a constant stiffness s, we have a bending stiffness which is linked to 'free flexural waves' (I don't know what this is?). This bending stiffness seems to be frequency depending somehow, and this is probably what causes the different effect.

But how can this be? In my eyes, nothing seems to have changed from case 1 to 2, except that the direction of incidence has changed. Surely, this cannot give a wholly different behaviour?

At the beginning of case 2, my course says:

This seems to suggest it behaves by the same principle. Having established the principle of applying the blocked surface pressure as the forcing field on a fluid-loaded structure, we may now apply it to the case of an unbounded, thin, uniform, elastic plate upon which acoustic plane waves of frequency omega are incident at an arbitrary angle phi.

In the first case, the derivation is done by modelling the interface as a mass m, linked to a suspension medium, modelled as a damper with factor r and a spring with factor s.

This way, I get a transmission of the form :

So, when my frequency omega is smaller than the resonance frequency of the medium, then the stiffness term with s is the most powerful.Code (Text):tau = 1/[ [(omega*m-s/omega)/(2 rho c)]² + [1+r/(2rhoc)]² ]

For higher frequencies, the mass term becomes more important.

In the second case, something similar is done, only here the stiffness is being replaced by something called the bending stiffness D, and there's seems to be an equivalence in the formula's:

D kz^4 ~ s , where kz is the normal factor of the wave number.

However, since kz is frequency dependent, this gives us a totally different picture:

When omega is smaller than the 'critical frequency' (=the equivalent of the resonance frequency), the mass term is the most important. When omega gets higher, the stiffness term takes over.

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# Homework Help: Interfaces and waves

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