 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/40a/a/solution

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

u = \nabla ϕ, p = - ρ_{0} \frac{\partial ϕ}{\partial t} .

Put

k = \frac{ω}{c _{0}}, α = k R .

For a spherically symmetric harmonic solution, write

ϕ (r, t) = F (r) sin ω t .

\nabla^{2} F = \frac{1}{r} \frac{d ^{2}}{d r ^{2}} (r F)

\frac{d ^{2}}{d r ^{2}} (r F) + k^{2} (r F) = 0.

The rigid inner sphere requires

u_{r} (R, t) = F^{'} (R) sin ω t = 0.

A form satisfying this condition automatically is

F (r) = \frac{C}{r} [cos k (r - R) + \frac{1}{k R} sin k (r - R)] .

Indeed, if the bracket is

h (r)

, then

h (R) = 1

and

h^{'} (R) = 1/ R

, so

(h / r)^{'} = 0

R

At the outer sphere,

ε p_{0} cos ω t = - ρ_{0} ω F (2 R) cos ω t,

F (2 R) = - \frac{ε p _{0}}{ρ _{0} ω} .

Since

F (2 R) = \frac{C}{2 R} (cos α + \frac{s i n α}{α}),

we obtain

C = - \frac{2 Rε p _{0}}{ρ _{0} ω ( c o s α + α ^{- 1} s i n α )} .

Therefore

ϕ (r, t) = - \frac{2 Rε p _{0}}{ρ _{0} ω r} \frac{cos k ( r - R ) + ( k R ) ^{- 1} sin k ( r - R )}{cos α + α ^{- 1} sin α} sin ω t .

The assumption

α \leq π /2

keeps the displayed denominator positive.

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