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 Incoming links: Acoustic energy conservation

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 40B / a / Solution Created 2026-09-24 Updated 2026-09-29

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Write the density and pressure as

ρ = ρ_{0} + ρ^{'}

and

p = p_{0} + p^{'}

, and let

u

be the small fluid velocity. The linear homentropic acoustic equations are

ρ_{t}^{'} + ρ_{0} \nabla \cdot u = 0, ρ_{0} u_{t} = - \nabla p^{'}, p^{'} = c_{0}^{2} ρ^{'},

(1)

where the adiabatic sound speed is

c_{0}^{2} = (\frac{\partial p}{\partial ρ})_{s} .

(2)

If the flow is irrotational, write

u = \nabla ϕ

. The momentum equation integrates spatially to

p^{'} = - ρ_{0} ϕ_{t},

(3)

a purely time-dependent integration term having been absorbed into

ϕ

. Substitution into continuity gives the wave equation

ϕ_{tt} - c_{0}^{2} \nabla^{2} ϕ = 0,

(4)

so the wave speed is

c_{0}

To obtain energy conservation, dot the momentum equation with

u

and use the product rule:

\frac{\partial}{\partial t} (\frac{1}{2} ρ_{0} ∣ u ∣^{2}) = - \nabla \cdot (p^{'} u) + p^{'} \nabla \cdot u .

(5)

The continuity and pressure-density relations give

p^{'} \nabla \cdot u = - \frac{p ^{'} p _{t}^{'}}{ρ _{0} c _{0}^{2}} = - \frac{\partial}{\partial t} (\frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}) .

(6)

Therefore the acoustic energy conservation law is

\frac{\partial E}{\partial t} + \nabla \cdot I = 0,

(7)

with acoustic energy density and intensity

E = \frac{1}{2} ρ_{0} ∣ u ∣^{2} + \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = p^{'} u .

(8)

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 40C / a / Solution Created 2026-09-24 Updated 2026-09-25

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Let

ρ^{'}

and

p^{'}

be the density and pressure perturbations and let

u

be the velocity perturbation. The linear acoustics equations about a uniform quiescent state are

ρ_{t}^{'} + ρ_{0} \nabla \cdot u = 0, ρ_{0} u_{t} = - \nabla p^{'}, p^{'} = c_{0}^{2} ρ^{'} .

(1)

The last relation is the homentropic pressure perturbation law.

Taking the scalar product of the momentum equation with

u

gives

\frac{\partial}{\partial t} (\frac{1}{2} ρ_{0} ∣ u ∣^{2}) = - u \cdot \nabla p^{'} = - \nabla \cdot (p^{'} u) + p^{'} \nabla \cdot u .

(2)

The continuity and pressure relations imply

p^{'} \nabla \cdot u = - \frac{p ^{'} p _{t}^{'}}{ρ _{0} c _{0}^{2}} = - \frac{\partial}{\partial t} (\frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}) .

(3)

Therefore the acoustic energy conservation law is

\frac{\partial}{\partial t} (K + W) + \nabla \cdot I = 0,

(4)

with

K = \frac{1}{2} ρ_{0} ∣ u ∣^{2}, W = \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = p^{'} u .

(5)

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Pressure-forced standing acoustic wave in a spherical annulus Created 2026-09-24 Updated 2026-09-24

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For

R < r < 2 R

, a rigid inner sphere and harmonic pressure

p^{'} (2 R, t) = ε p_{0} cos ω t

(1)

produce a standing radial field. With

k = ω / c_{0}

and

α = k R

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

(2)

The pressure and radial velocity are in temporal quadrature, so the period-averaged acoustic intensity vanishes.

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