0
$\begingroup$

You've pobably heard that low frequency sound waves almost always tend to pass through big walls while high frequencies never make it. The pattern is clearly consistent, but what is an actual formula that relates the cutoff frequency to the dimensions of the wall (with no sounds going around the edges)?

Also, does it work the same for rough surfaces or is there more high-pitched sound that gets absorbed due to the resonance of small porous structures?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I don't think you would find a formula that only relates the wall dimensions to the cutoff frequency. The wall materials and construction method will play a crucial role. You might be surprised by all the little things building designers can add in to minimize sound transfer. $\endgroup$
    – JMac
    Commented Oct 3, 2018 at 20:01
  • $\begingroup$ You can still measure averages or simply assume a common wall material like plaster or wood, there's no reason to be outlandish and assume something like a diamond wall or a liquid xenon wall. I already sound proof walls for audio recording so I'm not surprised at a material, but really those are still functions of elementary aspects. Diffusion panels work because of an elementary shape, porous structures can be estimated as roughness, sponge-like structures can be low-density with flexibility or tensile strength. It's all the same basic properties in engineering. $\endgroup$
    – Vane Voe
    Commented Oct 3, 2018 at 20:47

1 Answer 1

Reset to default
0
$\begingroup$

For sound waves, the standard "rough estimate" is that if an object's size is of order ~1 wavelength, then it will affect the passage or transmission of sound waves. So an object 1 foot in size will create a shadow behind it for frequencies of 1000Hz and greater; a 10 foot object will do the same for 100Hz and greater, and so on.

The shape and orientation of the object relative to the incoming sound waves will strongly affect this, so for good estimates you have to run a finite-element model or better still, perform experiments.

Improve this answer
$\endgroup$
11
  • $\begingroup$ Thanks that's a good estimate, really all I need is an estimate for a virtual environment too. Do you have any insight into approximating the sound waves based just on the planar angle of a sound source? $\endgroup$
    – Vane Voe
    Commented Oct 4, 2018 at 1:18
  • $\begingroup$ I've also found this oc.nps.edu/~bird/oc2930/acoustics/soundchannel.html It suggests that merely thickness has a correlation to cutoff frequency. Water is certainly no wall, but it is a commonly encountered body that sounds pass through, and it suggests the deeper the water, the lower the frequency that passes through. $\endgroup$
    – Vane Voe
    Commented Oct 4, 2018 at 1:51
  • $\begingroup$ Actually I'm realizing there is ambiguity in interpreting that formula. I've seen something like that before, but I can't be sure of the interpolation you suggest from only 2 examples. So for every exponential factor of 10 that the surface area increases, the frequency passing through decreases by an exponential factor of 10? But what is it decreasing from? Infinity? The starting point is unclear to me, I don't understand if this is applied to a specific starting frequency of 1000Hz or if any frequency decreases by the base 10 exponent of the surface area. $\endgroup$
    – Vane Voe
    Commented Oct 4, 2018 at 3:12
  • $\begingroup$ no, it comes from the fact that a 1000Hz tone has a wavelength of one foot, 100Hz is 10 feet, 10Hz is 100 feet. $\endgroup$
    – niels nielsen
    Commented Oct 4, 2018 at 3:38
  • $\begingroup$ Okay, I suppose that makes sense-ish...how many meters are in a Hertz? Can I just assume you're using the speed of a wave in air and use f = c/lambda? How can I estimate linear modifications of this formula based on the damping coefficient of a material? Say I'm using glass. Is the cutoff now going to be a shadow of around 10Hz or 10,000Hz? $\endgroup$
    – Vane Voe
    Commented Oct 4, 2018 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.