# 2.2 Energy in a mechanical wave (ticket #6227)

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Calculate the energy in a wave on a string.

## 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 ${\phantom{\rule{veryverythinmathspace}{0ex}}}^{2}$ for the string fragment $m=\mu dx$

Why is this $dx$ and not $ds$ ? Lets consider that the string is not perturbed, then it is horizontal and hasmass as given. When the string is perturbed it stretches a little bit - but the mass does not increase.

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 $=T\left(ds-dx\right)$ is done. This is equal to the potential energy stored in the stretched string segment. So the potential energy in this case is: $U=T\left(ds-dx\right)$ Now $\begin{array}{c}ds={\left(d{x}^{2}+d{y}^{2}\right)}^{1/2}\\ =dx{\left[1+{\left(\frac{\partial y}{\partial x}\right)}^{2}\right]}^{1/2}\end{array}$ Recall the binomial expansion ${\left(1+A\right)}^{n}=1+nA+\frac{n\left(n-1\right){A}^{2}}{2!}+\frac{n\left(n-1\right)\left(n-2\right){A}^{3}}{3!}+\dots$ so $ds\approx dx+\frac{1}{2}{\left(\frac{\partial y}{\partial x}\right)}^{2}dx$ $U=T\left(ds-dx\right)\approx \frac{1}{2}T{\left(\frac{\partial y}{\partial x}\right)}^{2}dx$ 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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