 Solution

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

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

Let

η = \hat{k} \cdot x - c t

. Then

\nabla^{2} u = f^{''} (η), \nabla (\nabla \cdot u) = \hat{k} (\hat{k} \cdot f^{''} (η)),

and substitution gives

ρ c^{2} f^{''} = (λ + μ) \hat{k} (\hat{k} \cdot f^{''}) + μ f^{''} .

c = c_{P}

and the varying part of

f

is longitudinal:

f (η) = \hat{k} F (η), θ = F^{'} (η), ω = 0.

For an S-wave,

c = c_{S}

and

f

is transverse:

\hat{k} \cdot f = 0, θ = 0, ω = \hat{k} \times f^{'} (η) .

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