Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 38A (Waves)

Take a shear-horizontal wave polarized in the $z$ direction,Then $\nabla \cdot \mathbf{u}=0$, and the elastic-wave equation reduces toFor $w=Y(y)e^{i(kx-\omega t)}$, the rigid boundary gives $Y(0)=0$, while vanishing shear traction at the free surface gives $Y^{\prime }(h)=0$. ThusThe dispersion relation of each guided shear-horizontal mode is

The cutoff occurs at $k=0$:For $k>0$, the phase velocity and group velocity areTheir product is $c_{p,n}{c}_{g,n}=c_S^2$. On a plot against positive $k$, $c_{p,n}$ decreases from $+\infty$ to $c_S$, while $c_{g,n}$ increases from zero to $c_S$; larger mode number shifts both transitions to larger wavenumber.

Use the real modeThe kinetic energy density and isotropic linear-elastic energy density areAverage over one period and integrate across the layer. Since ${\int }_0^h{\sin }^2(q_{n}y)dy={\int }_0^h{\cos }^2(q_{n}y)dy=h/2$, the mean energy per unit horizontal area isThe $x$-directed elastic-energy flux is $S_x=-{\sigma }_{zx}{w}_t=-\mu w_x{w}_t$. Its corresponding average isUsing ${\omega }^2=c_S^2(k^2+q_n^2)$ givesand therefore