I want to know how the geometry,more specificaly a thickness of simple wall affects how much will sound be reflected and how much will pass through.

I tried to search this on google for one and half hours but I didnt found single article or even single graph dedicated to this topic.I got bunch of thickness vs absorbtion stuff but that is not of interest for me,absorbtion and reflection are different things,perfect absorber have zero reflection.

Every article I found only talker about acoustic impedance of solid material compared to air,but it never mentions the thickness of the solid material.My point is,tungsten have 230000 times bigger impedance than air,but if you make wall of tungsten that is very thin,like lets say 100 atoms thick,then it will reflect sound less than if the wall was 1 meter thick.

If I know the impedance of air,if I know the intrinsic impedance of the material,then how do I calculate how much is reflected and how much transmitted at specific freqency and specific wall thickness?

Can you please give me links to articles or studies/white papers that are about wall thickness impact on reflectance? What words should I type in google so it give me results that are about reflectance/transmittance vs wall thickness instead of articles that talk purely about either absorbtion vs wall thickness or just mention that materials with increasing impedance difference cause increasing sound reflection but never mentions the critical aspect,the key factor that this all depends on the thickness of that wall not just intrinsic acoustic impedance of the material its made from.


You say

but if you make wall of tungsten that is very thin,like lets say 100 atoms thick,then it will reflect sound less than if the wall was 1 meter thick.

I definitely know I'm incapable of explaining the nuances of various transmission processes over such a huge scale of thicknesses ($1e^{-10} \ m$ to $1 \space m$), but it sounds like you're interested in attenuation, the dissipation of acoustical energy within a media as a function of wave propagation distance.

There's a pretty helpful equation on wikipedia from Chen and Holm 2003.

It expresses attenuation for an arbitrary viscoelastic media as a function of the wave propagation distance, $\Delta x$, angular frequency $\omega$, and an experimentally-derived, frequency-dependent parameter, $\eta$.

$$P(x + \Delta x) = P(x)e^{-\alpha(\omega)\Delta x}, \alpha(\omega) = \alpha_0 \omega^{\eta}$$

Reliance on the experimentally-derived parameter might make it less useful, but then again it's rare you see exact frequency-specific values reported for acoustic reactance either. The above equation does give you a way of calculating

how much [energy is] transmitted at specific frequency and specific wall thickness?


The following answer now addresses the acoustic impedance of thin materials directly.

Looking into microphones (such as ribbon microphones, which can be quite thin, i.e., $6 \times 10^{-6} \ m$), it is apparent that impedance is still treated the usual way.

The total mechanical impedance of the sheet is still:

$$Z = j(\omega m - \frac{1}{\omega L})$$

Where $m$ is the mass of the sheet (easily estimated) and $L$ is Young's modulus. The latter parameter is less easily estimated and appears to be the crux of the problem in your case. There appears to be an approach developed by Wagner et. al 2013 that is appropriate for nanosheets.

Then you can estimate transmission the usual way by impedance mismatch, remembering that if you need frequency-dependent values:

$$v = \frac{f}{j(\omega m - \frac{1}{\omega L})}$$

  • $\begingroup$ No,I am not interested in attenuation,I am interested in reflection,specificaly how does thickness of wall affect the amount of sound energy that gets reflected.If you have sound in air,and it hits wall made of tungsten that is infinately thick,99.999% of the sound gets reflected,if that wall is 10 atoms thick,it passes through it like hot knife through butter.The intrinsic impedance of tungsten didnt change,it allways was orders of magnitude higher than that of air,but there is this second factor,and that is wall thickness. $\endgroup$ – wav scientist Nov 11 '17 at 6:47
  • $\begingroup$ $Z$ is dependent on density, too, right? perhaps there's some assumption about the beginning thickness that's relevant to the use of these equations? that's all I can think of. $\endgroup$ – D. Betchkal Nov 11 '17 at 19:26
  • $\begingroup$ wav scientist, I've updated my answer to answer your question. $\endgroup$ – D. Betchkal Jan 4 '18 at 16:59
  • $\begingroup$ I dont understand equations,could you explane that equation is simple way thats easy to understand? As far as I know,the impedance comes down to three factors 1. Density 2. Speed of sound 3. Thickness $\endgroup$ – wav scientist Jan 6 '18 at 17:37
  • $\begingroup$ @wavscientist the most basic way I can think of explaining acoustic impedance is that mass ($m$) and stiffness (as I've written it here, $\frac{1}{L}$) both change the reflection of acoustic waves. For very thin materials, like you've described the mass is quite small so waves are unlikely to be impeded except at high frequencies (large $\omega$). Quantifying stiffness of very thin materials is something I've never tried. More on this topic. $\endgroup$ – D. Betchkal Jan 6 '18 at 21:58

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