Explaining the Ripple Effect in a Circular Wave Pulse

[Mindscrape]

Here is the problem:
If a pebbel is tossed into a pond, a circular wave pulse propagates outward from the disturbance. If you look closely you will see a fine structure in the pulse consisting of surface ripples moving inward though the circular disturbance. Explain this effect in terms of group and phase velocity if the phase velocity of ripples is given by v_p = \sqrt{2 \pi S/ \lambda \rho}, Where S is the surface tension and p is the density of the liquid.

I am not really sure where to start. Should I find the envelope velocity and compare it to the high frequency velocity. I know it will have something to do with superposition, but Fourier analysis seems like it is the wrong approach.

 

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Can someone help me out here?Answer:The explanation for this effect can be found by considering the group velocity and phase velocity of the wave pulse. The group velocity of a wave is the velocity of the envelope of the wave, and it is given by v_g = \frac{d\omega}{dk}, where \omega is the angular frequency and k is the wavenumber. In contrast, the phase velocity of a wave is the velocity of the individual ripples and it is given by v_p = \sqrt{\frac{2\pi S}{\lambda \rho}}, where S is the surface tension and \rho is the density of the liquid. In this case, since the phase velocity of the ripples is given, we can calculate the group velocity of the wave pulse. We can see that since the group velocity is the velocity of the envelope of the wave, it is slower than the phase velocity. This means that the individual ripples will move inward through the circular disturbance at a faster rate than the wave itself is propagating outward. This is why you can see the fine structure in the pulse consisting of the surface ripples moving inward.