 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/39c/b/iii/solution

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

The fluid velocity at the piston equals its velocity,

u_{p} (t) = \frac{d}{d t} [- ϵ t (1 - t)] = - ϵ (1 - 2 t) .

Using

c - c_{0} = (γ - 1) u /2

in part (ii) gives

\frac{δ p}{p _{0}} = γ \frac{u}{c _{0}} + \frac{γ ( γ + 1 )}{4} \frac{u ^{2}}{c _{0}^{2}} + O (ϵ^{3}) .

Over

0 \leq t \leq 1

⟨ u_{p} ⟩ = 0, ⟨ u_{p}^{2} ⟩ = ϵ^{2} \int_{0}^{1} (1 - 2 t)^{2} d t = \frac{ϵ ^{2}}{3} .

Thus

⟨ \frac{δ p}{p _{0}} ⟩ = \frac{γ ( γ + 1 )}{12} \frac{ϵ ^{2}}{c _{0}^{2}} + O (ϵ^{3}) .

The linear pressure oscillation has zero mean, while the quadratic compressibility produces a positive mean pressure, the finite-amplitude acoustic analogue of radiation pressure.

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