Phase Velocity Derivation I spend few hours trying to derive phase velocity of sinusoidal wave $$\cos(kx - \omega t).$$ I know that it must be equal to $\omega \over k$ but after banging my head for few hours and trying to find solution on internet I gave up. 
 A: The wave equation is $z(x,t)= \cos(kx - \omega t)$ and a graph of the displacement of a particle from its equilibrium position $z$ against the position of the particle from an origin $x$ at a given time $t$ is shown below.
You can liken the graph to a photograph of the wave taken at one instant of time; it is called a wave profile.   
The graph actually shows shows two such photographs (wave profiles) taken at a time $t$ and at a later time $t+\Delta t$.  

By inspection of the movement of the peaks $A$ and $A'$, or the troughs $C$ and $C'$, or the positions of zero displacement $B$ and $B'$ etc,  you can surmise that the wave is travelling in the positive x-direction.  
The important thing is that the displacements of the particle at different times which you considering are the same; peak and peak, trough and trough etc.
So you have $z(x,t) = z(x+\Delta x , t + \Delta t)$ or $\cos(kx - \omega t) = \cos(k[x+\Delta x] - \omega [t+\Delta t])$ and a solution of this equation is $kx - \omega t = k[x+\Delta x] - \omega [t+\Delta t] \Rightarrow \dfrac {\Delta x}{\Delta t} = \dfrac {\omega}{k}$ and this is called the phase velocity.  

Another way of derivation is to say that you want to find a condition such that $kx-\omega t$ (the "phase" of the wave) is a constant ie you are tracking the passage of a crest at a given time to a crest at a later time etc.  
Now differentiate  $kx-\omega t = \rm constant$ with respect to time.  
$k \dfrac {dx}{dt} - \omega =0 \Rightarrow \dfrac{dx}{dt} = \dfrac {\omega}{k}$ and you have the phase velocity.
A: The phase velocity denotes the velocity at which a peak of the sinusoidal pattern is moving.  Consider the wave as $\cos[\psi(x,t)]$, where $\psi=kx-\omega t$. There will always be a peak at $\psi=0$, since that corresponds to $\cos0=1$.  We want to know how fast that peak at $\psi=0$ is moving.*  But the condition $\psi=0$ is just
$$kx-\omega t=0\\
x-\frac{\omega}{k}t=0.$$
And the last is the just the equation of something moving with velocity $\omega/k$.
*It can be any other peak as well; that just shifts the relation between $x$ and $t$ by an overall constant, leaving the velocity the same.
