In a medium, the energy per unit area per unit time delivered perpendicuar to the direction of the wave propagation s caled the intensity of the wave. It is denoted by $I$.

### Intensity of a wave

Definition of Intensity of a wave:

If the energy $E$ is delivered in the time $t$ rom area $A$ perpendicular to the wave propagation, then

$I=\frac{E}{At} \qquad{1}$

Unit: $Joule/m^{2}-sec$ or $watt/m^{2}$

Dimensional formula: $[MT^{-3}]$

We know that the total mechanical energy of a vibrating particle is

$E=\frac{1}{2}m \omega^{2} a^{2}$

Where $\omega$ is the angular frequency and $a$ is the amplitude of the wave.

$E=\frac{1}{2}m (2\pi n)^{2} a^{2} \qquad \left( \omega=2\pi n \right)$

$E=2 \pi^{2} m n^{2} a^{2} \qquad(2)$

Where $m$ is the mass of the vibrating particle.

Now substitute the value of $E$ from equation $(2)$ to equation $(1)$. So the intensity of the wave

$I=\frac{2 \pi^{2} m n^{2} a^{2}}{At} \qquad(3)$

If the wave travels the distance $x$ in time $t$ with velocity $v$, Then

$t=\frac{x}{v}$

Now substitute the value of the above equation in equation $(3)$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{Ax}$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{V}$

Where $V$ is the volume of the corresponding medium during the wave propagation in time $t$.

$I=2 \pi^{2} \rho v n^{2} a^{2} \qquad \left(\because \rho=\frac{m}{V} \right)$

It is clear that for wave propagation in a medium with a constant velocity, i.e. wave's intensity is directly proportional to the square of amplitude and frequency both.

$I\propto a^{2}$ and $I \propto n^{2}$