5
$\begingroup$

I understand why amplitude doesn't affect the speed of the sound AFTER the 'leading compression'. The extra force provided on one stage of the cycle is countered on the other stage. But shouldn't the 'leading compression' travel faster if amplitude is greater? Because wave motion hasn't been set up yet shouldn't all compressions face the same amount of resistance while moving forward?

$\endgroup$
  • 5
    $\begingroup$ The acoustic wave equation is linear, so amplitude does not matter in determining the speed of wave propagation. However, implicit in its derivation is that the material response is linear, which is only valid for small compressions. So it's probably true that amplitude does affect the speed of sound at sufficiently high intensities, although I don't know how high it has to be to break the linearity assumption. $\endgroup$ – DumpsterDoofus May 13 '14 at 13:43
  • $\begingroup$ Possibly-related question: to what extent (if any) does the speed of a nonlinear shock wave—a thunderclap, a sonic boom, a high-energy explosive blast—depend on its amplitude? $\endgroup$ – rob May 14 '14 at 0:04
  • $\begingroup$ Dumpster Doofus so for small compressions the speed at which the pressure difference evens out is constant? Is there an intuitive way to understand this rather than using the wave equation? Thanks. $\endgroup$ – user42991 May 14 '14 at 6:13
2
$\begingroup$

The phase speed of a sound wave is dependent on the wave amplitude. This is how and why a sound wave can steepen into a shock wave. In fact, without dissipation, even the sound waves produced from talking would steepen uncontrollably to a point similar to a shock, but would break before forming a shock (since dissipation is required to initiate shock formation).

In the stationary frame of the wave, a compressional wave increases the fluid speed, which allows a trailing wave to catch a leading wave. If you think of a compressional wave pulse as a series of finite steps of increasing pressure, then each trailing step is at a higher pressure than the corresponding adjacent leading step. If there is no energy dissipation, then the trailing waves will not only catch up to the leading waves, they can overtake them leading to a gradient catastrophe or wave breaking.

Mathematical Justification

The phase speed of a sound wave is given by: $$ C_{s}^{2} = \frac{ \partial P }{ \partial \rho } = \partial_{\rho} \ P $$ where P = pressure and $\rho$ = mass density. If we assume an adiabatic equation of state (i.e., P $\propto$ $\rho^{\gamma}$), then we see that: $$ C_{s} = \sqrt{ \frac{ \partial P }{ \partial \rho } } = \left\{ P_{o} \gamma \rho^{\gamma - 1} \frac{ \partial \rho }{ \partial \rho } \right\}^{1/2} \\ = \left\{ \frac{ \gamma P_{o} \rho^{\gamma} }{ \rho } \right\}^{1/2} = \left\{ \frac{ \gamma P(\rho) }{ \rho } \right\}^{1/2} $$ One can see that the $C_{s}$ is dependent upon the density/pressure and we know that sound waves are longitudinal (and compressive) oscillations in density/pressure. Since $\gamma$ $\geq$ 1, one can show that $C_{s}^{2}$ scales with density and pressure. Thus, higher pressure/density corresponds to higher sound speeds.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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