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Why does not the high frequency sound propagate as far?

The dispersion curve $\omega(k)$ is almost linear, right?

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The basic phenomenon is that high frequency sound is more strongly attenuated than low frequency sound. The mechanism for sound attenuation is viscous damping. The absorption coefficient is $$ \gamma= \frac{\omega^2}{2\rho c^3}\left[ \frac{4}{3}\eta + \zeta + \kappa\left(\frac{1}{c_v}-\frac{1}{c_p}\right) \right], $$ where $\omega$ is the frequency, $\rho$ is the density, $c$ is the speed of sound, $\eta$ is shear viscosity, $\zeta$ is bulk viscosity, and $\kappa$ is thermal conductivity.

Physically, what is happening is that sound is a longitudinal oscillation of the fluid velocity. Viscosity tries to equalize velocity, and damps any oscillation. The damping force is proportional to gradients of velocity $\eta\nabla u$, and the energy dissipated scales as $\eta(\nabla u)^2$. In Fourier space this is $\dot E \sim \eta k^2 u^2\sim \eta\omega^2u^2/c^2$. Since $E\sim \rho u^2$ the damping $\dot{E}/E\sim \eta\omega^2/{\rho c^2}$. If we write this in terms of the absorption length there is one extra power of $1/c$, which gives the formula above.

Bulk viscosity and thermal conductivity work analogously. Again, the damping is proportional to $\omega^2$.

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  • $\begingroup$ Nice concise answer $\endgroup$ – Floris Jan 17 '16 at 19:42
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In most media, high frequencies will be attenuated more strongly than low frequencies. Even if the loss mechanisms are the same, for a certain amplitude a higher frequency wave will have a greater velocity (displacement of particles, not wave velocity), and thus greater "drag" per cycle. If the loss mechanism is the same, then the wave will lose the same fraction of energy per wavelength - but if the wavelength is shorter, that means you travel less far for a certain attenuation. And so long waves win. In fact, according to Stokes' law of sound attenuation, attenuation increases with frequency squared - this makes sense as you get the viscous dissipation due to the higher velocity (proportional with frequency), and then another term because of the shorter wavelength.

The Stokes equation for incompressible fluids is

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

where $\eta$ is the dynamic viscosity coefficient of the fluid, $\omega$ is the sound's frequency, $\rho$ is the fluid density, and $c$ is the speed of sound in the medium. The coefficient $\alpha$ appears in the expression for amplitude with distance:

$$A(x) = A(0) e^{-\alpha x}$$

If the fluid's compressibility cannot be neglected (as is the case for water), an additional term is added that captures the effect of the volume viscosity $\zeta$, giving a more complex expression

$$\alpha = \frac{2\left(\eta+\frac34\zeta\right) \omega^2}{3 \rho c^3}$$ An example of the attenuation of (sea) water as a function of frequency is given in this graph (from http://resource.npl.co.uk/acoustics/techguides/seaabsorption/ainslie.gif)

[![enter image description here][1]][1]

The light blue line is for "pure" water, and it follows Stokes' Law over many decades of frequency. There are some additional terms due to the fact that sea water is not, in fact, "just" water. I confess I find the axis labels odd (wrong, actually) - they are marked in dB/km which is a logarithmic scale, but it seems that the attenuation given makes more sense as a linear scale (two decades more attenuation for one decade increase of frequency agrees with Stokes). But it's from a 1968 paper and I doubt the author is interested in publishing a correction at this time.

Another interesting historical note. In his original derivation, Stokes made the observation (bottom of page 302)

[...] but as we do not possess any means of measuring the intensity of sound the theory cannot be tested, nor the numerical value of $\mu$ determined, in this way.

Food for thought - deep insights like this were developed before people could do the kinds of experiments that give rise to a curve like the one reproduced above. A sign of stunning intellect.

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