Derivation Phase Velocity of a Wave

Level 2 requires school mathematics. Suitable for pupils.

Updated by Alexander Fufaev on 12/19/2022

The phase velocity $ v_{\text p} $ of a wave is the velocity with which a point of the wave moves. Here we want to express the phase velocity using the angular frequency $\omega$ and the angular wavenumber $k$.

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Phase velocity of a point A of the wave.

So consider any point on the wave, for example the top of a wave crest (point A in illustration 1). We want to find out how fast this point moves from A to B.

The velocity is distance between A and B per time. This distance corresponds by definition to the wavelength $\lambda$. And the time after which the point $A$ arrives at $B$ is by definition the period $T$. Thus the phase velocity is:

Phase velocity as wavelength per period

$$ \begin{align} v_{\text p} ~=~ \frac{\lambda}{T} \end{align} $$

The angular wavenumber $k$ is the angle travelled per length. Within one wavelength $\lambda$ the angle $2\pi$ is travelled: $ k = \frac{2\pi}{\lambda} $. Rearranged for the wavelength yields:

$$ \begin{align} \lambda ~=~ \frac{2\pi}{k} \end{align} $$

The angle frequency $\omega $ is the angle traveled per time. Within a period $T$ the angle $2\pi$ is covered: $\omega = \frac{2\pi}{T} $. Rearranged for the period we get:

$$ \begin{align} T ~=~ \frac{2\pi}{\omega} \end{align} $$

Substitute the wavelength 2 and period 3 into formula 1 to express the phase velocity with angular wavenumber $k$ and angular frequency $\omega$:

$$ \begin{align} v_{\text p} ~=~ \frac{2\pi}{k} \, \frac{\omega}{2\pi} \end{align} $$

Here the factor $2\pi$ is cancelled and you get the formula:

Phase velocity expressed with frequency and angular wavenumber

$$ \begin{align} v_{\text p} ~=~ \frac{\omega}{k} \end{align} $$

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Derivation Phase Velocity of a Wave