Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 40C / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv

i

Solution

Use the acoustic velocity potential convention

u = \nabla ϕ, p^{'} = - ρ_{0} ϕ_{t} .

For a spherically symmetric outgoing wave with time dependence

e^{iω t}

and

k = ω / c_{0}

, write

ϕ (r) = C \frac{e ^{- ik (r - a)}}{r} .

Linearizing the boundary condition at the mean surface

r = a

gives

ϕ^{'} (a) = iω ϵ .

Since

ϕ^{'} (a) = - \frac{C}{a ^{2}} (1 + ika),

we find

C = - \frac{iω ϵ a ^{2}}{1 + ika} .

Thus the outgoing acoustic field of a pulsating sphere is

ϕ (r, t) = Re [- \frac{iω ϵ a ^{2}}{1 + iωa / c _{0}} \frac{e ^{iω t - i (ω / c_{0}) (r - a)}}{r}] .

ii

Solution

For real harmonic quantities

f (t) = Re (f e^{iω t}), g (t) = Re (g e^{iω t}),

define the period average by

⟨ f g ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} f (t) g (t) d t = \frac{1}{2} Re (f g^{*}) .

This is the time average of harmonic power.

At

r = a

, put

x = ωa / c_{0}

. The velocity and pressure amplitudes are

u_{r} (a) = iω ϵ, p (a) = - i ρ_{0} ω ϕ (a) = - \frac{ρ _{0} ω ^{2} ϵ a}{1 + i x} .

Hence the mean outward work rate per unit area is

⟨ p^{'} u_{r} ⟩ = \frac{1}{2} Re (p u_{r}^{*}) = \frac{ρ _{0} ω ^{4} ϵ ^{2} a ^{2}}{2 c _{0} ( 1 + x ^{2} )} .

Multiplication by the area

4 π a^{2}

yields

⟨ P ⟩ = \frac{2 π ρ _{0} ω ^{4} ϵ ^{2} a ^{4} / c _{0}}{1 + ω ^{2} a ^{2} / c _{0}^{2}} = 2 π a^{2} ρ_{0} ω^{2} ϵ^{2} c_{0} \frac{ω ^{2} a ^{2}}{c _{0}^{2} + ω ^{2} a ^{2}} .

This is the mean acoustic power of a pulsating sphere.

iii

Solution

For a harmonic field, the time-averaged energy densities at

r = a

are

⟨ K ⟩ = \frac{ρ _{0}}{4} ∣ u_{r} ∣^{2} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4},

⟨ W ⟩ = \frac{∣ p ∣ ^{2}}{4 ρ _{0} c _{0}^{2}} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4} \frac{x ^{2}}{1 + x ^{2}} .

From part (ii),

∣ ⟨ I_{r} ⟩ ∣ = \frac{ρ _{0} ω ^{2} ϵ ^{2} c _{0}}{2} \frac{x ^{2}}{1 + x ^{2}} .

Consequently

\frac{c _{0} ⟨ K + W ⟩}{∣ ⟨ I _{r} ⟩ ∣} = \frac{1 + 2 x ^{2}}{2 x ^{2}} = 1 + \frac{c _{0}^{2}}{2 ω ^{2} a ^{2}} .

This is the surface acoustic energy-to-flux ratio of a pulsating sphere.

iv

Solution

If

c_{0} ≪ ωa

, then

x ≫ 1

. The sphere is acoustically large, and

⟨ K ⟩ \sim ⟨ W ⟩, ∣ ⟨ I_{r} ⟩ ∣ \sim c_{0} ⟨ K + W ⟩ .

Pressure and velocity are nearly in phase, so the near-surface field behaves locally like a progressive radiating wave. The power approaches

2 π a^{2} ρ_{0} ω^{2} ϵ^{2} c_{0} .

If

c_{0} ≫ ωa

, then

x ≪ 1

. The sphere is acoustically compact, and

\frac{⟨ W ⟩}{⟨ K ⟩} \sim x^{2} ≪ 1, \frac{c _{0} ⟨ K + W ⟩}{∣ ⟨ I _{r} ⟩ ∣} \sim \frac{1}{2 x ^{2}} ≫ 1.

The surface field is mainly reactive kinetic energy: pressure and velocity are nearly in quadrature, energy is stored and returned during each cycle, and only a small mean acoustic energy flux is radiated. The mean power scales as

⟨ P ⟩ \sim \frac{2 π ρ _{0} ω ^{4} ϵ ^{2} a ^{4}}{c _{0}} .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version