 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/40a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Take the divergence of the elastic equation. With

θ = \nabla \cdot u

ρ θ_{tt} = (λ + 2 μ) \nabla^{2} θ .

Thus dilatational P-waves have speed

c_{P} = \frac{λ + 2 μ}{ρ} .

Taking the curl and writing

ω = \nabla \times u

eliminates the gradient term and gives

ρ ω_{tt} = μ \nabla^{2} ω .

Thus rotational S-waves have speed

c_{S} = \frac{μ}{ρ} .

 New to topics? Read the docs here!