# 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?