Sound Wave Energy Loss due to thickness of aluminum

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SUMMARY

The discussion focuses on sound wave energy loss in aluminum discs, specifically comparing 1/8" and 1/4" thicknesses. It is established that transmission loss increases by 6 decibels for each doubling of frequency or wall mass per unit area, as noted in the referenced document. The attenuation coefficient for aluminum is crucial for determining specific energy loss values at desired frequencies. The example provided illustrates that using two 1/8" plates results in a reduction of intensity from 1 W/m² to 0.25 W/m², highlighting the low attenuation in metals for audible frequencies.

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  • Understanding of sound wave propagation and attenuation
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nst.john
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Hello everyone!

I previously opened a thread asking about sound wave energy loss in aluminum discs based on thickness. I am looking to find out more about exactly how much using a disc that is twice the thickness of the other, (one is 1/8" and the other is 1/4") sound energy would be lost using the thicker one compared to the thinner one. We had some problems using the 1/8" one and we may need to use the 1/4" but I don't want too much sound to be lost in the process. Thank you!
 
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A quick overview can be found at:

http://www.silex.com/pdfs/sound attenuation.pdf

The bottom of pg 12 states:
"...transmission loss increases 6 decibels for each doubling of frequency, or each doubling of the wall mass per unit area, up to a plateau frequency."

Remember that Google is your friend; that was the sixth hit searaching for 'sound attenuation versus frequency'
 
To find a specific number you will need to look up the attenuation coefficient for aluminium, in the frequency range of interest.
But as an example of how it works, assuming that you start with a beam of intensity 1 W/m2, if the 1/8" plates reduces the intensity to 0.5 W/m2, adding a second 1/8" plate will halve it again, to 0.25 W/m2.
However this is a quite extreme example. The attenuation in metals is quite low for audible frequencies.
 

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