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.

  • $\begingroup$ Note that in 1-dimension \begin{equation} \text{phase}\boldsymbol{=} \phi \boldsymbol{\stackrel{def}{\equiv}} k\cdot x\boldsymbol{-}\omega\cdot t\qquad k,\omega \in \mathbb{R} \tag{01}\label{01} \end{equation} and in 3-dimensions \begin{equation} \text{phase}\boldsymbol{=} \phi \boldsymbol{\stackrel{def}{\equiv}} \mathbf{k}\boldsymbol{\cdot} \mathbf{x}\boldsymbol{-}\omega\cdot t\qquad \mathbf{k}\in \mathbb{R}^{3},\omega \in \mathbb{R} \tag{02}\label{02} \end{equation} $\endgroup$
    – Frobenius
    Commented Dec 22, 2018 at 8:05
  • $\begingroup$ $$y = \cos(kx - ωt)\\ x = \frac1k (\arccos y +ωt) \\ \frac{\partial x}{\partial t} = \frac ωk$$ $\endgroup$
    – Shub
    Commented Dec 26, 2022 at 16:27

2 Answers 2


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 $t+\Delta t$.

enter image description here

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.

  • $\begingroup$ I liked that that your soulutions are quite formal because it leaves a little possibility for an error but I missing some steps in these solutions. In first solution you calculated $\Delta x \over \Delta t$. This is only approximation of speed. It is accurate only if speed is constant (so the ${dx(t) \over dt} = const $). In second solution you first taken $x$ to be variable and then you differentiated it by $t$. This should be equal to zero. If it is not then $x$ should really be function of $t$ i.e. $x(t)$ but in your solution it is not. $\endgroup$ Commented Dec 22, 2018 at 12:32
  • 1
    $\begingroup$ @Trismegistos : For constant $k,\omega$ \begin{equation} \dfrac{\Delta x}{\Delta t}\boldsymbol{=}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{=}\dfrac{\omega}{k} \end{equation} $\endgroup$
    – Frobenius
    Commented Dec 22, 2018 at 13:26
  • $\begingroup$ @Trismegistos When you are looking at the peak of a travelling wave you will notice that the position of the peak $x$ changes as the time $t$ charges, so $x$ is a function of $t$. It is the displacement of the particle from its equilibrium position $z$ ie the peak being observed which is kept constant. $\endgroup$
    – Farcher
    Commented Dec 22, 2018 at 14:02

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.


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