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CollectiveRocker
Oct10-04, 01:24 PM
If a question says: The phase velocity of ripples on the liquid surface is (2πS / λp)^(1/2), where S is the surface tension and p is the density of the liquid. Find the group velocity of the ripples. I know that the phase velocity = omega/k, and group velocity = delta omega/delta k. Do I just take the deriviative of the phase velocity with respect to S?

arildno
Oct10-04, 01:27 PM
1. Is S k?????????????
2.\frac{d}{dk}(\frac{\omega}{k})=\frac{d\omega}{dk }??????????????
is that what you're saying?
Then think again.

CollectiveRocker
Oct10-04, 01:29 PM
S is the surface tension of the liquid. Is there another way to find the group velocity?

arildno
Oct10-04, 01:31 PM
You have the definition; wherever have you gotten the idea that the surface tension S is the wavenumber "k"??????

CollectiveRocker
Oct10-04, 01:34 PM
I realize that S is not k. Yet how do I do the problem?

arildno
Oct10-04, 01:41 PM
Since you have the phase velocity, you may find the frequency \omega
The group velocity is then, by your definition, the derivative of \omega with respect to "k".

CollectiveRocker
Oct10-04, 01:49 PM
How can we find omega if we don't know what k is?

arildno
Oct10-04, 01:54 PM
Multiply your phase velocity with k.

CollectiveRocker
Oct10-04, 02:01 PM
then isn't (2πS / λp)^(1/2) a constant?

arildno
Oct10-04, 02:04 PM
No, because your wavelength satisfies identically the relation:
\lambda{k}=2\pi
since your expression for ph.vel. is proportional to the square root of the wavelength, your frequency will be proportional to the square root of the wavenumber

CollectiveRocker
Oct10-04, 02:09 PM
this probably sounds really idiotic on my part. I just need to take (dw/dk) of k(2πS / λp)^(1/2), right?

CollectiveRocker
Oct10-04, 02:09 PM
And that will give me the group velocity?

arildno
Oct10-04, 03:01 PM
Yes, it will
Differentiate, if you dare..:wink:

CollectiveRocker
Oct10-04, 04:20 PM
product rule?