 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/40c/a/solution

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

ρ_{0} \frac{\partial ^{2}}{\partial t ^{2}} (\nabla \cdot u) = (λ + 2 μ) \nabla^{2} (\nabla \cdot u) .

Thus the dilatation is a wave of speed

c_{P} = \frac{λ + 2 μ}{ρ _{0}} .

Taking the curl, and using

\nabla \times \nabla f = 0

, gives

ρ_{0} \frac{\partial ^{2}}{\partial t ^{2}} (\nabla \times u) = μ \nabla^{2} (\nabla \times u),

so the rotation propagates at

c_{S} = \frac{μ}{ρ _{0}} .

These are respectively the speeds of P-waves and S-waves.

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