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Incident sound intensity on a infinite rigid wall with an opening

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maryhomes

Mechanical
Jul 30, 2013
1
Dear all,

I have a question which is a bit general acoustics, but definitely something for mechanical engineers with a heart for acoustics.

My question is whether it is logic that a plane wave, incident from an infinite air medium onto a rigid wall with an opening of 1.5m by 1m (nothing is in the opening, after the opening there is a again an infinite air medium - rigid wall separates the full space), results in a non-constant incident intensity level curve measured and averaged over the opening. I am wondering some time now why this is not constant? I thougth that the intensity and power in a plane wave are constant i.e. that is what I understood from my acoustics class.

The non-constant intensity profile is as follows : a constant intensity level of around 55dB for plane waves from 10 - 70Hz. Around 80Hz a sudden increase happens in the intensity values till around 100 Hz till a level of 85dB is reached. Hereafter the 85 dB remains more or less constant till 1000 Hz. For these simulations a 1Pa amplitude pressure sine waves is taken for the whole frequency range.

I would be very grateful I someone could give me more insight why this is so,....

Thank you very much,

Mary
 
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Is this for school? Have you looked at waveguide equations?

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
At 10 Hz the wavelength is 30 m and a 1m hole looks like a point source. At 1000 Hz the wavelength is 30cm and a 1 m hole looks like a very complex set of sources.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
I think the point is that once your plane wave has hit the opening, it is no longer a plane wave.

- Steve
 
The "cut-off" wavelength of an opening is not a binary thing. Most gap attenuators in the RF will still propagate below the cut-off, just not as much. I would assume that even though the gap is 30x smaller than the opening, this just means that that opening is propagating ~30x less than if the opening were larger than the wavelength.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Just a nit pick. How can 30x anything be less? Do you mean 1/30?

Ted
 
Well, duh... 1/30th as large just does not have the same punch

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
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