Finding group velocity from phase velocity of surface ripples

[CollectiveRocker]

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]

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]

S is the surface tension of the liquid. Is there another way to find the group velocity?

 

[arildno]

You have the definition; wherever have you gotten the idea that the surface tension S is the wavenumber "k"?

 

[CollectiveRocker]

I realize that S is not k. Yet how do I do the problem?

 

[arildno]

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]

How can we find omega if we don't know what k is?

 

[arildno]

Multiply your phase velocity with k.

 

[CollectiveRocker]

then isn't (2πS / λp)^(1/2) a constant?

 

[arildno]

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]

this probably sounds really idiotic on my part. I just need to take (dw/dk) of k(2πS / λp)^(1/2), right?

 

[CollectiveRocker]

And that will give me the group velocity?

 

[arildno]

Yes, it will
Differentiate, if you dare..

 

[CollectiveRocker]

product rule?