 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/38a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

z

direction,

u = w (x, y, t) e_{z} .

Then

\nabla \cdot u = 0

, and the elastic-wave equation reduces to

w_{tt} = c_{S}^{2} (w_{xx} + w_{yy}), c_{S} = \frac{μ}{ρ} .

For

w = Y (y) e^{i (k x - ω t)}

, the rigid boundary gives

Y (0) = 0

, while vanishing shear traction at the free surface gives

Y^{'} (h) = 0

. Thus

Y_{n} (y) = A sin (q_{n} y), q_{n} h = (n + \frac{1}{2}) π, n = 0, 1, 2, \dots .

ω_{n}^{2} = c_{S}^{2} (k^{2} + q_{n}^{2}), q_{n} = \frac{( n + 1/2 ) π}{h} .

 New to topics? Read the docs here!