2.3 Energy in a mechanical wave

Energy transport

Lets calculate the energy in a wave of a string:

Consider a fragment of string so small it can be considered straight, as is shown in the figure

The the kinetic energy is $K=\frac{1}{2}m$ v ${}^2$ for the string fragment $m=\mu dx$

So we have $$K=\frac{1}{2}\mu dx{\left(\frac{\partial y}{\partial t}\right)}^2$$ and using this we can define the energy per unit length, ie. the kinetic energy density: $$\frac{dK}{dx}=\frac{1}{2}\mu {\left(\frac{\partial y}{\partial t}\right)}^2$$ When the string segment is stretched from the length $dx$ to the length $ds$ an amount of work $=T(ds-dx)$ is done. This is equal to the potential energy stored in the stretched string segment. So the potential energy in this case is: $$U=T(ds-dx)$$ Now $${\displaystyle \begin{array}{c}ds={(d{x}^2+d{y}^2)}^{1/2}\\ =dx{\left[1+{\left(\frac{\partial y}{\partial x}\right)}^2\right]}^{1/2}\end{array}}$$ Recall the binomial expansion $${(1+A)}^n=1+nA+\frac{n(n-1)A^2}{2!}+\frac{n(n-1)(n-2)A^3}{3!}+\dots$$ so $$ds\approx dx+\frac{1}{2}{\left(\frac{\partial y}{\partial x}\right)}^{2}dx$$ $$U=T(ds-dx)\approx \frac{1}{2}T{\left(\frac{\partial y}{\partial x}\right)}^{2}dx$$ or the potential energy density $$\frac{dU}{dx}=\frac{1}{2}T{\left(\frac{\partial y}{\partial x}\right)}^2$$ To get the kinetic energy in a wavelength, lets start with $$y=A\sin \left(\frac{2\pi x}{\lambda }-\omega t\right)$$ $$\frac{\partial y}{\partial t}=-\omega A\cos \left(\frac{2\pi x}{\lambda }-\omega t\right)$$ Lets evaluate it at time 0. $${\frac{\partial y}{\partial t}|}_{t=0}=-\omega A\cos \left(\frac{2\pi x}{\lambda }\right)$$ so $$\frac{dK}{dx}=\frac{1}{2}\mu {\omega }^2{A}^2{\cos }^2\left(\frac{2\pi x}{\lambda }\right)$$ now integrate $${\displaystyle \begin{array}{c}K={\int }_0^{\lambda }\frac{dK}{dx}dx\\ =\frac{1}{2}\mu {\omega }^2{A}^2{\int }_0^{\lambda }{\cos }^2\left(\frac{2\pi x}{\lambda }\right)dx\end{array}}$$ In order to do this integral we use the following trig identity: $${\cos }^{2}A=\frac{\cos 2A+1}{2}$$ so we get $${\displaystyle \begin{array}{c}K=\frac{1}{2}\mu {\omega }^2{A}^2{\left[\frac{x}{2}+\frac{\lambda }{8\pi }\sin \frac{4\pi x}{\lambda }\right]|}_{x=0}^{\lambda }\\ =\frac{1}{4}\mu \lambda {\omega }^2{A}^2\end{array}}$$

In similar fashion the potential energy can be found to be $$U=\frac{1}{4}\mu \lambda {\omega }^2{A}^2\text{.}$$ Deriving this will be assigned as a homework problem

So $$E=K+U=\frac{1}{2}\mu \lambda {\omega }^2{A}^2$$ Power $${\displaystyle \begin{array}{c}P=\frac{\Delta E}{\Delta t}=\frac{\frac{1}{2}\mu \lambda {\omega }^2{A}^2}{\tau }\\ =\frac{1}{2}\mu {\omega }^2{A}^2\text{v}\end{array}}$$ Where I have used $\tau =1/\nu$ and $\lambda \nu =$ v thus $\tau =\lambda /\text{v}$