Past exam of the mathematical methods II course of the University of Oxford / 2017 / 10 / Solution

For a short length $\delta x$, the kinetic energy is $\frac{1}{2}\mu y_t^2\delta x$. For small slope, its increase in length is $[\sqrt{1+y_x^2}-1]\delta x\simeq \frac{1}{2}{y}_x^2\delta x$, so its elastic potential energy is $\frac{1}{2}T{y}_x^2\delta x$. For a travelling wave $y=f(x-vt)$, $y_t=-v{f}^{\prime }$ and $y_x=f^{\prime }$. Using $v^2=T/\mu$ givesWriting a pulse as $f(x)=Ag(x/w)$ yields $E=(T{A}^2/w)\int [g^{\prime }(s){]}^{2}ds$. Thus energy is proportional to amplitude squared for a fixed shape and width. Width must also be retained when comparing pulses on different strings.