 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/38d/b/solution

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

The linearized kinematic condition at the undisturbed surface is

\partial_{t} η = \partial_{z} ϕ_{z = 0} .

For the given complex amplitudes this says

- iω A = k B, B = - \frac{iω}{k} A .

u = (ϕ_{x}, 0, ϕ_{z})

. Its nonzero complex strain amplitudes are

e_{xx} = - k^{2} B e^{k z}, e_{zz} = k^{2} B e^{k z}, e_{x z} = e_{z x} = i k^{2} B e^{k z} .

Using the stated rule for period averages,

⟨ e : e ⟩ = 2 k^{4} ∣ B ∣^{2} e^{2 k z} .

Thus the mean dissipation per unit horizontal area is

D = 2 μ \int_{- \infty}^{0} ⟨ e : e ⟩ d z = 2 μ k^{3} ∣ B ∣^{2} = 2 μ k ω^{2} ∣ A ∣^{2} = 2 μg k^{2} ∣ A ∣^{2} .

E = ρ g ∣ A ∣^{2} /2

consequently obeys

d E / d t = - D

ν = μ / ρ

\frac{d ∣ A ∣}{d t} = - 2 ν k^{2} ∣ A ∣, ∣ A (t) ∣ = ∣ A (0) ∣ e^{- 2 ν k^{2} t} .

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