Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 10

For a right-moving sinusoidal string displacement $y=A\cos (kx-\omega t)$, the instantaneous energy flux of a transverse string wave is $P(x,t)=-T{y}_x{y}_t=T{A}^{2}k\omega {\sin }^2(kx-\omega t)$. Averaging over one period gives $⟨{\sin }^2⟩=1/2$. Since $c=\omega /k$ and $T=\rho c^2$,The displayed power is a cycle average, not the time-dependent instantaneous power. Here $\rho$ denotes linear mass density, rather than the volume density used in the surface-wave calculation.

The wave speed is $c=\sqrt{T/\rho }=50\mathrm{m}{\mathrm{s}}^{-1}$, so the frequency is $f=c/\lambda =100\mathrm{Hz}$ and $\omega =200\pi {\mathrm{s}}^{-1}$. Substituting the amplitude in metres,Power scales with the square of amplitude; converting millimetres to metres before squaring is essential.

Write the complex displacements as $y_1=(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x})e^{-i\omega t}$ and $y_2=C{e}^{i{k}_{2}x-i\omega t}$. At a massless join, continuity of displacement and transverse force givesEliminating $B$ gives the displacement transmission coefficient and reflection coefficient,The angular frequency is shared across a stationary interface, whereas the wavelength changes. The equal-tension assumption is used in the slope condition.

At the same tension, $k\propto \sqrt{\rho }$, so $k_2/k_1=3$, $t=1/2$ and $r=-1/2$. The transmitted power fraction is not $t^2$, because the two strings have different wave impedances. Using the energy flux of a transverse string wave,The reflected fraction is $\mathcal{R}=|r{|}^2=1/4$, so $\mathcal{R}+\mathcal{T}=1$. The transmitted displacement amplitude is $2.5\mathrm{mm}$ and its wavelength is $0.5/3\mathrm{m}$. The negative $r$ indicates a reflected phase reversal; it does not mean negative reflected energy.