Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 79 3 iv Solution Created 2026-10-03 Updated 2026-10-14
Let be either endpoint at which is prescribed, and put , and . Integrating the constant-coefficient velocity–traction system givesTaking the ratio yields the uniform-layer SH impedance update and its accompanying velocity:The initial value sets an arbitrary normalization; specifying impedance alone cannot specify the overall wave wave amplitude. Positive propagates data from , while negative propagates a specified load from , so the same expressions solve both requested initial-value problems.
Expanding sine and cosine givesAbsorbing the reference-point phases into the two constants gives the requested forward/backward exponential superposition. Equivalently, wherever the ratios are defined,The exceptional values give the pure directional solutions. If a denominator in the impedance formula vanishes, propagate the finite pair through that point; this is an impedance-chart pole, not a failure of the physical solution. The upper-half-plane assumptions ensure .
Reflection from a layered SH impedance 2026-10-14
Set the lower half-space load to using the causal vertical wavenumber branch, which excludes a wave arriving from depth infinity. Apply the uniform-layer SH impedance update upwards through every layer, keeping particle velocity and traction continuous. For upper half-space impedance , incident velocity and reflected velocity , the top values are and . Their ratio gives the displayed reflection coefficient, incorporating every underlying interface and multiple reflection.