# Thread: Displacement amplitude of sound wave at minimum audible intensity

1. Hi!

In my wave mechanics class, we derived that the time-average intensity of a sound wave, is given by
,
where is the wave speed, is the density, is the displacement amplitude and is the angular frequency.

The derivation seemed allright, but in the next sentence the lecturer stated the minimum audible intensity of sound is (a figure supported by other sources, and concluded from the above formula that (with , , and ). This value equates to about a tenth of the radius of a hydrogen atom, so surely there must be some flaw to the argument. I and my fellow students supposed that the problem lies in the formula being derived from a macroscopic point of view, where the air is treated as a continuous medium, rather than a discrete medium consisting of molecules (which effect must certainly be present at displacement amplitudes corresponding to the low intensity of .

Would anybody care to shed some light on this issue?

Thanks!

2.

3. Your question is confusing. You have a distance s0 in units of W/m^2. ????????????

4. How silly of me; the unit was of course supposed to be . I have corrected it now. Thanks for pointing it out!
However, my original question remains the same; what is the flaw of the argument?

5. Assuming your figures to be correct, it is quite possible that all the molecules in some small region might be displaced by a small amount relative to where they would have been in the absence of the sound wave. That the displacement is small in relation to the size of a molecule doesn't make such a scenario impossible. Are you proposing that such small displacements aren't possible, or that they are possible but would not give rise to audible sound?

6. I am proposing that such a small displacement would not be detectable by the cilia in the ear.

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