5
$\begingroup$

When a droplet of water hits the surface of a lake it produces ripples that travel outward. My conception is that it travels at the speed of sound in water (since sound are nothing less that perturbations on the medium).

But ripples travel much slower than that. What's happening? Is my conception of propagation of ripples wrong?

Perhaps that discrepancy is related to the difference between group velocity and phase velocity?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
14
$\begingroup$

The restoring force for waves on the surface of water is gravity, is why such waves are called "gravity waves" (not gravitational waves!). Gravity waves on deep water have $$ \omega_{\rm gravity}(k)= \sqrt{{gk}}, $$ which is independent of the density of the water.

The restoring force for sound waves in water is the water's elastic properties and $$ \omega_{\rm sound}= \sqrt{\frac{\kappa}{\rho}}k $$ where $\kappa$ is the bulk modulus and $\rho$ is the density. The waves therefore have completely different origins and properties.

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ Great answer. I have a question though: wouldn't the restoring force on the surface wave be the water cohesion? $\endgroup$
    – Pedro Pinho
    Feb 27, 2021 at 21:16
  • 4
    $\begingroup$ @ Pedro Pinho. There is small effect due to surface tension, but it is only more imortant than gravity (i.e the water's weight wanting to make the water flat) when the wavelength is a few millimeters or less. The Wikipedia article on capillary waves has details. $\endgroup$
    – mike stone
    Feb 27, 2021 at 21:19
  • 1
    $\begingroup$ Interesting. What is $k$ in your first equation? $\endgroup$
    – Gert
    Feb 27, 2021 at 22:04
  • 2
    $\begingroup$ It is the wavenumber: $k=2\pi/\lambda$. And $\omega$ is the angular frequency $\omega=2\pi f$. $\endgroup$
    – mike stone
    Feb 27, 2021 at 22:35
  • $\begingroup$ Aren't there some non-linear effects (on deep water)? What is the "range" of the model? $\endgroup$
    – Peter Mortensen
    Feb 28, 2021 at 7:35

Your Answer

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

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