As a mechanical wave propagates in a medium, it transfers energy form one particle to another and the successive particles get the disturbance.
In Figure we consider a transverse wave in a string travelling in positive x-direction of our coordinate system. We consider a particular point \(p\) in the string which is disturbed by the wave.
The string on the left of the point \(p\) exerts tension on the point which has two components namely \(F_\text{x}\) and \(F_\text{y}\). The transverse force \(F_y\) exerts transverse force on the particle and hence does work on the particle.
Note that the component \(F_x\) is the tension the string would have in undisturbed condition of the string. The wave has constant amplitude and constant frequency.
The slope at the point \(p\) is the negative of the ratio \(F_y /F_x\) because \(F_y\) is negative but the slope is positive. The slope is also equal to the derivative of wave function with respect to position \(x\) keeping time \(t\) constant, therefore the slope at the point \(p\) is
\[\begin{align*} \frac{{\partial y}}{{\partial x}} &= \frac{{ - {F_y}}}{{{F_x}}}\\ {\rm{or,}}\quad {F_y} &= - {F_x}\frac{{\partial y}}{{\partial x}} \tag{1} \label{1} \end{align*}\]
The power is the rate of doing work and the instantaneous power at the point \(p\) is the product of downward force \(F_y\) and the downward velocity \(v_y\) at that point. So,
\[P = {F_y}{v_y} = - {F_x}\frac{{\partial y}}{{\partial x}}{v_y} \tag{2} \label{2}\]
The wave function for a simple harmonic wave travelling in positive x-direction is \(y = A\cos (kx - \omega t)\) and you can find;
\[\begin{align*} \frac{{\partial y}}{{\partial x}} &= - Ak\sin (kx - \omega t)\\ {\rm{and,}}\quad {v_y} &= \frac{{\partial y}}{{\partial t}} = A\omega \sin (kx - \omega t) \end{align*}\]
Therefore, substituting the values of \(\partial y /\partial x\) and \(v_y\) in Eq. \eqref{2}, you'll get the power which is
\[P = {F_x}{A^2}\omega k{\kern 1pt}{\sin ^2}(kx - \omega t) \tag{3} \label{3}\]
We know from the expression of wave speed that \(F_x = {v^2}\mu \) and \(k = \omega/v\), the above equation can be rewritten as
\[P = {A^2}\omega^2\mu v {\kern 1pt}{\sin ^2}(kx - \omega t) \tag{4} \label{4}\]
The value of \({\sin ^2}\) function oscillates between \(0\) and \(1\) and hence its average value is \(1/2\). So, the average value of \({\sin ^2}(kx - \omega t)\) in the above equation is \(1/2\) and the average power is
\[{P_{av}} = \frac{1}{2}{A^2} \omega^2 \mu v \tag{5} \label{5}\]
The above equation shows the the average rate of energy transfer, that is average power of a simple harmonic or sinusoidal wave along a string is proportional to the square of amplitude, square of angular frequency, linear density of the string and the wave speed.
