Finding the Group Velocity for Shallow Water Wave

[James Brady]

Homework Statement

Find the group velocity for a shallow water wave: $\nu = \sqrt{\frac{2\pi\gamma}{\rho\lambda^3}}$

Homework Equations

Phase velocity: $v_p = \nu\lambda$
group velocity: $v_g = \frac{d\omega}{dk}$

$k=\frac{2\pi}{\lambda}$
$\omega = 2\pi \nu$

The Attempt at a Solution

Get frequency in terms of wave number:

$\nu(k)=\sqrt{\frac{\gamma k}{\rho\lambda^2}}$
$\omega(k) = 2\pi\sqrt{\frac{\gamma k}{\rho\lambda^2}}$

$\frac{d\omega}{dk}=2\pi(\frac{1}{2}(\frac{\gamma k}{\rho \lambda^2})^{-1/2} \frac{\gamma}{\rho \lambda^2})$

We can rewrite the function, getting rid of k. Also did some canceling and moved the pi:$v_g=(\frac{\gamma 2\pi}{\rho \lambda^3})^{-1/2} \frac{\gamma\pi}{\rho \lambda^2}$ $v_g = \frac{\frac{\gamma\pi}{\rho \lambda^2}}{(\frac{\gamma 2\pi}{\rho \lambda^3})^{1/2}}$

Note that given the definition for phase velocity, we can write it as:

$v_p = \sqrt{\frac{2\pi\gamma}{\rho\lambda}}$

$v_g = \frac{\frac{\gamma\pi}{\rho \lambda^2}}{v_p \lambda} = \frac{\frac{\gamma\pi}{\rho \lambda^3}}{v_p}$

So this is where I'm stuck... The correct answer from the book is $\frac{3}{2}v_p$

 

[TSny]

Get frequency in terms of wave number:

$\nu(k)=\sqrt{\frac{\gamma k}{\rho\lambda^2}}$

Try expressing $\nu(k)$ completely in terms of $k$, rather than a mixture of $\lambda$ and $k$.

 

[James Brady]

$\frac{d\nu}{dk}=\frac{1}{2}(\frac{\gamma k^3}{4 \pi^2 \rho})^{-1/2}\frac{3k^2\gamma}{4\pi^2\rho}$

So I see we have that nice 3/2 term in there, but if my phase velocity term is correct, $v_p = \nu\lambda = \sqrt{\frac{2\pi \gamma}{\rho \lambda}} = \sqrt{\frac{\gamma k}{\rho}}$, then every thing other than the 3/2 coefficient must reduce to this...

So maybe this is just some algebra problem right now...?

$\frac{\frac{\gamma k^2}{4\pi^2 \rho}}{\frac{\gamma k^3}{4 \pi^2 \rho}^{1/2}} = \sqrt{\frac{\gamma k}{\rho}}$

Was writing out this question when I figured out that this does reduce to 3/2v_p. Just as long as we do $\frac{d \omega}{dk}$ (include the 2pi), instead of $\frac{d \nu}{dk}$

I figured I'd post my thoughts anyway. Thanks for tip @TSny.