Phase vs. Group Velocity.

1. Oct 10, 2004

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?

2. Oct 10, 2004

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.

3. Oct 10, 2004

CollectiveRocker

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

4. Oct 10, 2004

arildno

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

5. Oct 10, 2004

CollectiveRocker

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

6. Oct 10, 2004

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".

7. Oct 10, 2004

CollectiveRocker

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

8. Oct 10, 2004

arildno

Multiply your phase velocity with k.

9. Oct 10, 2004

CollectiveRocker

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

10. Oct 10, 2004

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

11. Oct 10, 2004

CollectiveRocker

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

12. Oct 10, 2004

CollectiveRocker

And that will give me the group velocity?

13. Oct 10, 2004

arildno

Yes, it will
Differentiate, if you dare..

14. Oct 10, 2004

CollectiveRocker

product rule?