Frequency of Periodic Wave

Theorem

Let $\phi$ be a periodic wave expressed as:

$\forall x, t \in \R: \map \phi {x, t} = \map f {x - c t}$

The frequency $\nu$ of $\phi$ can be expressed as:

$\nu = \dfrac 1 \tau$

where $\tau$ is the period of $\phi$.

By definition, $\nu$ is the number of complete wavelengths of $\phi$ to pass an arbitrary point on the $x$-axis in unit time.

Let $x_0$ be that arbitrary point.

By definition, $\tau$ is the time taken for one complete wavelength of $\phi$ to pass $x_0$.

So after unit time, $\dfrac 1 \tau$ wavelengths of $\phi$ pass $x_0$.

Hence the result.

$\blacksquare$

Let $\phi$ be a harmonic wave expressed as:

$\forall x, t \in \R: \map \phi {x, t} = a \map \cos {\omega \paren {x - c t} }$

The frequency $\nu$ of $\phi$ can be expressed as:

$\nu = \dfrac 1 \tau$

where $\tau$ is the period of $\phi$.