We often seem to only hear the lower beats of music far away, whilst the higher frequency sounds seem to diminish more quickly - remaining unheard. I know that sounds with higher frequency have shorter wavelengths, whilst lower frequencies have longer wavelengths. However, this does not necessarily mean that the sounds travel at different speeds. Under normal pressure and density, sounds are known to travel 331 m/s. How does this information allow me to answer whether the Inverse Square Law is applicable in this real world situation or not? Why can we hear low frequencies from far away, but not high frequencies?

I read very briefly on Stoke's Law just then as I was typing this post. Does it have any relevance to this problem?

Much thanks.


1 Answer 1


Yes this is related to Stoke's law for sound attenuation, which states that a plane wave decreases amplitude exponentially with a factor $\alpha$ given by:

$$\alpha = \frac{2\eta\omega^2}{3\rho V^3}$$

where you can see that the dependence on the frequency squared $\omega$ of the sound will yield a higher coefficient of attenuation for higher frequency sounds comparatively to lower ones.

So between two plane waves with different frequencies $\omega_H=2\omega_L$, the higher one will attenuate four times faster. That is, if they have at a starting point the same amplitude, after a certain characteristic distance $d=\frac{3\rho V^3}{2\eta\omega_L}$ the wave with higher frequency $\omega_L$ will be $e^3 \approx 20$ times weaker.

For air, using $\rho$=1.225e-3$kg/m^3$, $V$=331$m/s$ and $\eta$=1.8e-5$kg/ms$, and taking the A pitch standard as the high frequency $\omega_L$=440Hz and its lower octave as the lower frequency $\omega_L$=220Hz, the characteristic distance will be 76 732 $km$. And the distance at which their relative amplitudes will be in a one half ratio would be 17 729 $km$.

  • 1
    $\begingroup$ Probably equally significant for "hearing sound from far away" is diffraction: in the near field approximation (valid for sound diffracting around real world obstacles) low frequencies will diffract around them more easily. Thus a sound "from far away" which is likely to encounter trees, buildings etc. will be attenuated - much more so if the frequency is higher. The observed sound intensity will be a function of the three effects. Note also that intensity scales with amplitude squared... $\endgroup$
    – Floris
    Jul 17, 2015 at 15:22
  • $\begingroup$ @rmhleo : So for my experiment; I have these values: ρ=1.225e-3kg/m3, V=331m/s and η=1.8e-5kg/ms and the lower frequency is 500 Hz, whilst the higher frequency is 10000 Hz. I calculated α for both: 1.35e-7 and 2.7e-6 respectively. So this means for the lower frequency, its amplitude decreases exponentially with a factor of 1.35e-7 ?? How do I calculate the characteristic distance? Thanks very much for your help. $\endgroup$
    – user80922
    Jul 19, 2015 at 6:19
  • $\begingroup$ Sorry, I did not cover all in the answer. The law of attenuation, expressed mathematically is $A(d)=A_0e^{-\alpha d}$, where $A$ is the amplitude at distance $d$ from the point at which the wave's amplitude was $A_0$. The characteristic distance I mention is an arbitrary definition I made (see the formula above) taken as the step for which the amplitude of the lower wave decreases $e$ times. That is $\frac{A_L(d)}{A_0}=1/e=e^{-\alpha_L d_c}$ where you can check that $d_c$ yields the expression above. The relative attenuation at distance d is: $A_H/A_L = A_0_H/A_0_L e^{-(\alpha_H-\alpha_L) d} $ $\endgroup$
    – rmhleo
    Jul 19, 2015 at 8:39

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