Past exam of the mathematical methods II course of the University of Oxford / 2021 resit / 10 / b / Solution

A reference string segment $dx$ has mass $\mu dx$ and transverse velocity $y_t$. Its kinetic energy is $dK=\mu y_t^{2}dx/2$. In the small-slope approximation its extra arc length isThe work against constant tension $T$ supplies the corresponding elastic potential energy $dU=T(d\ell -dx)$. To quadratic order, the energy density per unit reference length is thereforeFor the travelling wave $y=A\cos (kx-\omega t+\varphi )$, the wave equation implies ${\omega }^2=(T/\mu )k^2$. Its kinetic energy density and elastic energy density are equal, soThe energy per wavelength of a transverse string wave comes from integration over any full wavelength $\lambda =2\pi /|k|$:The printed expression for energy per wavelength is missing a factor $\lambda$. Its right-hand side is the mean energy density, with units of energy per length. For a concrete counterexample within the small-slope regime, take $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $T=1\mathrm{N}$, $A=0.01\mathrm{m}$ and $\omega =\pi {\mathrm{s}}^{-1}$. Then $\lambda =2\mathrm{m}$ and direct integration gives $E_{\lambda }=10^{-4}{\pi }^2\mathrm{J}$, whereas the printed right-hand side is $5\times 10^{-5}{\pi }^2\mathrm{J}{\mathrm{m}}^{-1}$. It cannot equal an energy.