Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 40A (Waves)

Take the divergence of the elastic equation. With $\theta =\nabla \cdot \mathbf{u}$,Thus dilatational P-waves have speed

Taking the curl and writing $\mathbf{\omega }=\nabla \times \mathbf{u}$ eliminates the gradient term and givesThus rotational S-waves have speed

Let $\eta =\hat{\mathbf{k}}\cdot \mathbf{x}-ct$. Thenand substitution gives

For a P-wave, $c=c_P$ and the varying part of $\mathbf{f}$ is longitudinal:For an S-wave, $c=c_S$ and $\mathbf{f}$ is transverse:

All wavevectors lie in the $xz$ plane. The incident P-wave travels upward toward the interface. In the lower medium there are a reflected P-wave, polarized parallel to its reflected wavevector, and a reflected SV-wave, polarized in the $xz$ plane perpendicular to its wavevector. In the upper medium there are transmitted P and SV waves with the corresponding longitudinal and transverse polarizations. No SH wave is generated by this in-plane incident field.

Frequency and tangential phase matching give Snell's law: every wave has the same $\omega$ and the same tangential wavenumber $k_x$, while its normal wavenumber has the sign appropriate to reflection or transmission.

For a perfectly bonded interface, displacement and traction are continuous:in the two-dimensional problem. In three dimensions one also includes $[u_y]=[{\sigma }_{yz}]=0$. HereThese four in-plane conditions determine the four reflected and transmitted amplitudes.

Let $\omega =k{c}_P$, and let $\psi$ be the transmission angle. Since ${\hat{c}}_P=2c_P$, tangential phase matching givesThe assumption of no evanescence requires $\sin \varphi \le 1/2$.

With a common suppressed factor $e^{-i\omega t}$, takeFor an inviscid liquid, continuity of normal displacement and normal stress giveswhere $Z=\lambda /c_P$ and $\hat{Z}=\hat{\lambda }/{\hat{c}}_P$. Solving,

The condition $|R|=|T|$ givesThe sign that could instead produce $Z\cos \psi =(\hat{Z}+2Z)\cos \varphi$ leads to ${\sin }^2\varphi >1$ and is inadmissible. HenceUsing ${\cos }^2\psi =1-4{\sin }^2\varphi$ and squaring gives