Waves on deep water with surface tension T and density ρ are governed by the dispersion relation
$$ω^2 = gk + \frac{T}{\rho}k^3.$$
Calculate the phase and group velocities of the waves. Show that the phase velocity reaches a minimum at a wave number $k_c$ that you should calculate. What is the group velocity for this wave number?
the phase velocity is the speed of a point of constant phase which is $c_p^2=\frac{w^2}{k^2}$
the group velocity of a wave is the velocity with the overall shape of the waves' amplitudes $v_g =\frac{dω}{dk}$
But I am confused since the c can take either direction, does it mean that I need to consider both $c_p=\frac{ω}{k} or-\frac{ω}{k}$.
And when I got $c_p=\sqrt{\frac{g}{k}+\frac{T}{\rho}k}$ then how should I get the minimun phase velocity.
what is the motion of the individual wave crests relative to the centre of a wave packet for $k<k_c$ and for $k>k_c$?
Best Answer
The phase speed is given by $$ \frac{\omega^2}{k^2}=\frac{g}{k}+\frac{T}{\rho}k\implies c_p^2 = \frac{g}{k}+\frac{T}{\rho}k $$ so Phase speed is $$ c_p = \pm\sqrt{\frac{g}{k}+\frac{T}{\rho}k} $$ The group speed is given by $$ 2\omega\frac{\partial \omega}{\partial k} = g + \frac{3T}{\rho}k^2 $$ so we have $$ v_g = \frac{g + \frac{3T}{\rho}k^2}{2\omega} $$ To determine the minimum phase speed we are essentially compute $$ \frac{\partial c_p}{\partial k} = 0 $$ so we use the first relation $$ 2c_p\frac{\partial c_p}{\partial k} = -\frac{g}{k^2}+\frac{T}{\rho} $$ This is zero for a trial case, $c_p=0$, but in general we require $$ -\frac{g}{k^2}+\frac{T}{\rho} =0 \implies k_c^2 = \frac{g\rho}{T} $$ so the minimum phase speed occurs for $$ k_c = \sqrt{\frac{g\rho}{T}} $$