Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 40A (Waves)

Table of contents
- a 40A
  - Solution a
- b 40A
  - Solution b
- c 40A
  - Solution c

a

Solution

Use the acoustic velocity potential convention

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 .

The wave equation and

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

give the radial Helmholtz equation

\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

at

R

.

At the outer sphere,

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

so

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.

b

Solution

At every radius, the pressure perturbation has the form

p (r, t) = - ρ_{0} ω F (r) cos ω t,

whereas the radial velocity is

u_{r} (r, t) = F^{'} (r) sin ω t .

Define the average over one forcing period

T = 2 π / ω

by

⟨ I_{r} ⟩ = \frac{1}{T} \int_{0}^{T} p (r, t) u_{r} (r, t) d t .

Then

⟨ I_{r} ⟩ = - ρ_{0} ω F (r) F^{'} (r) \frac{1}{T} \int_{0}^{T} cos ω t sin ω t d t = 0 .

The same is true for the net flux through every concentric sphere.

Pressure and velocity are in temporal quadrature, so this is a reactive acoustic energy flux. The forcing establishes a standing wave in the closed, lossless annulus: compressional and kinetic energy exchange during each cycle, but no energy is transported outward on average. This is the pressure-forced standing acoustic wave in a spherical annulus.

c

Solution

At larger frequencies the denominator

D (α) = cos α + \frac{s i n α}{α}

can vanish. Its zeros satisfy

tan α = - α

and are the normal-mode frequencies for a pressure node at the outer sphere and zero radial velocity at the inner sphere. Near one of these radial acoustic resonances, the ideal undamped harmonic response becomes very large; exactly at resonance the linear lossless initial-value problem develops secular growth rather than a bounded periodic solution.

In a physical system, fluid viscosity, thermal losses, acoustic radiation, and damping in the elastic shell keep the amplitude finite and shift its phase. A nonzero pressure-velocity phase component then supplies positive mean power. At sufficiently large amplitude the linear acoustic approximation can also fail, leading to nonlinear frequency shifts or shock formation.

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