How do we derive a relation between the displacement amplitude and the intensity of a sound wave? I saw a formula on the internet regarding the same i.e
But I can't find a way that would help me in deriving this expression.
All related suggestions are appreciated.
Sound wave velocity: $$ v_s = \sqrt{\frac{B}{\rho}};\tag{1} $$ where $\rho$ the mass density of medium, $B$ is the boulk modulous defined as ratio of pressure change $\Delta P$ with percentage change of volume ($\frac{\Delta V}{V}$) $$ B = -\frac{\Delta P }{\Delta V / V} = -V\frac{d P}{d V}. $$ Assume the area of the cross-section of propagation is $A$, and $V= A dx$, the volume unit without deformation. The deformation $\Delta V = A \Delta s$, $s$ is the disantace displacement. $$ \Delta P (x) = B \frac{\Delta V}{V}\big\vert_x = B\frac{\partial \Delta s}{\partial x}.\tag{2} $$
Sinusoidal equation for displacement: $$ \Delta s(x, t) = \Delta s_0 \sin (kx-\omega t). \tag{3} $$ where angular frequecy $\omega = 2\pi f = k v_s$.
Using Eq.(2) converting displacement $\Delta s(x,t)$ to pressure deviation $\Delta P(x,t)$
\begin{align} \Delta P(x,t) &=B\frac{\partial \Delta s}{\partial x} \\ &= B k\Delta s_0 \cos k(x-vt). \end{align}
The work done per wave length $F = A\Delta P$:
\begin{align} W &= \int_{0}^{\frac{2\pi}{k}} F d\Delta s \\ &= \int_{0}^{\frac{2\pi}{k}} A B k\Delta s_0 \cos (kx-\omega t) k\Delta s_0 \cos (kx-\omega t) dx\\ &=A B k^2 \Delta s_0^2 \int_{0}^{\frac{2\pi}{k}} \cos^2 (kx-\omega t) dx\\ &= \frac{\pi}{k} A B k^2 \Delta s_0^2 \end{align}
Intensity (period $T =\frac{2\pi}{\omega}$, $B = \rho\,v_s^2$, and $v_s = \frac{\omega}{k}$ ) \begin{align} I &= \frac{W}{A T} =\frac{\pi}{k}\frac{ B k^2 \Delta s_0^2}{2\pi/\omega}\\ &= \frac{1}{2} \rho v_s^2 k \omega \Delta s_0^2 \\ &=\frac{1}{2} \rho v_s \omega^2 \Delta s_0^2\\ &=2\pi^2 \,\rho\, v_s f^2 \Delta s_0^2 \end{align} In the last equation, $\omega$ is replaced by frequency $\omega = 2\pi f$.
