Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section D / D13 / Solution

For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.