Group velocity of single wave?

[kasse]

Homework Statement

Find the group velocity of the wave $$f(x) = cos(b\sqrt{\omega}x - \omega t)$$

2. The attempt at a solution

What? I thought group velocity was an interference phenomenon...The velocity of the beats produced when two waves with slightly different frequencies interfere. What is meant by the group velocity of one wave?

Anyway, I know that $$v_{g} = \frac{d\omega}{dk} = \frac{d(k^{2}/b^{2}}{dk} = 2v$$ which is the correct answer. I still don't understand what I've found however...

 

[Hootenanny]

The group velocity of a wave function is the velocity of which the envelope (or changes in amplitude) propagate through space. And yes, the definition which you quoted above is only valid for wave packets (superposition of two or more waves), but what's to say that your wave function above, does not represent a wave packet?

 

[kasse]

I thought waves written in the form $$f(x) = cos(kx - \omega t)$$ represents one single wave and that the wave velocity equals the group velocity for such waves. Maybe this does not count when k depends on $$\omega$$?

 

[Hootenanny]

I thought waves written in the form $$f(x) = cos(kx - \omega t)$$ represents one single wave and that the wave velocity equals the group velocity for such waves. Maybe this does not count when k depends on $$\omega$$?

Seems you have figure out yourself .

If k=k(ω), that is if the [angular] wavenumber is a function of angular frequency or ω = ω(k) then the wavefunction f(x,t) represents a wavepacket rather than an individual wave.