Past exam of the elasticity and plasticity course of the University of Oxford / 2007 / 2 / i

Decomposing a plane displacement into components parallel and perpendicular to its wavevector gives compressional and shear speedsThe incident polarization is longitudinal. Its reflected longitudinal wave reverses the vertical wavevector component while retaining its horizontal component. The converted shear wave has wavevector $k_s(\sin \beta ,-\cos \beta )$ perpendicular to polarization $(\cos \beta ,\sin \beta )$.

Every term must have the same phase along the rigid boundary. ThusChoose $\cos \beta >0$ so the reflected shear wave propagates downward into $y<0$. At $y=0$, both displacement components must vanish for every boundary phase:Solving this two-by-two system yieldsFor a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha <\pi$; choosing $\beta \in (-\pi /2,\pi /2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha \le \pi /2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no critical angle for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At normal incidence $\alpha =\pi /2$, $\beta =0$, $R_1=1$ and $R_2=0$, cancelling the incident displacement with the reversed longitudinal polarization. This is compressional-wave reflection from a rigid elastic boundary.