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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 ${\phantom{\rule{veryverythinmathspace}{0ex}}}^{2}$ 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 $$\begin{array}{c}ds={(d{x}^{2}+d{y}^{2})}^{1/2}\\ =dx{\left[1+{\left(\frac{\partial y}{\partial x}\right)}^{2}\right]}^{1/2}\end{array}$$ 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 $$y=A{\mathrm{sin}}\left(\frac{2\pi x}{\lambda}-\omega t\right)$$ $$\frac{\partial y}{\partial t}=-\omega A{\mathrm{cos}}\left(\frac{2\pi x}{\lambda}-\omega t\right)$$ Lets evaluate it at time 0. $${\frac{\partial y}{\partial t}|}_{t=0}=-\omega A{\mathrm{cos}}\left(\frac{2\pi x}{\lambda}\right)$$ so $$\frac{dK}{dx}=\frac{1}{2}\mu {\omega}^{2}{A}^{2}{{\mathrm{cos}}}^{2}\left(\frac{2\pi x}{\lambda}\right)$$ now integrate $$\begin{array}{c}K={\int}_{0}^{\lambda}\frac{dK}{dx}dx\\ =\frac{1}{2}\mu {\omega}^{2}{A}^{2}{\int}_{0}^{\lambda}{{\mathrm{cos}}}^{2}\left(\frac{2\pi x}{\lambda}\right)dx\end{array}$$ In order to do this integral we use the following trig identity: $${{\mathrm{cos}}}^{2}A=\frac{{\mathrm{cos}}2A+1}{2}$$ so we get $$\begin{array}{c}K=\frac{1}{2}\mu {\omega}^{2}{A}^{2}{\left[\frac{x}{2}+\frac{\lambda}{8\pi}{\mathrm{sin}}\frac{4\pi x}{\lambda}\right]|}_{x=0}^{\lambda}\\ =\frac{1}{4}\mu \lambda {\omega}^{2}{A}^{2}\end{array}$$
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 $$\begin{array}{c}P=\frac{\Delta E}{\Delta t}=\frac{\frac{1}{2}\mu \lambda {\omega}^{2}{A}^{2}}{\tau}\\ =\frac{1}{2}\mu {\omega}^{2}{A}^{2}{\text{v}}\end{array}$$ Where I have used $\tau =1/\nu $ and $\lambda \nu =$ v thus $\tau =\lambda /{\text{v}}$
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