Before shock formation, the piston launches a right-moving simple wave into the undisturbed gas. The opposite invariant is fixed at its ambient value:

R_{-} = u - \frac{2 c}{γ - 1} = - \frac{2 c _{0}}{γ - 1} .

Therefore the simple-wave relation is

c = c_{0} + \frac{γ - 1}{2} u .

(2)

ii

 0  0

Solution

 0  0

The motion is isentropic before a shock forms, and the perfect-gas relations give

\frac{p}{p _{0}} = (\frac{c}{c _{0}})^{2 γ / (γ - 1)} .

(1)

Writing

δ = δc / c_{0}

and applying the generalized binomial expansion,

\frac{δ p}{p _{0}} = \frac{2 γ}{γ - 1} δ + \frac{γ ( γ + 1 )}{( γ - 1 ) ^{2}} δ^{2} + O (δ^{3}) .

(2)

iii

 0  0

Solution

 0  0

The fluid velocity at the piston equals its velocity,

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

(1)

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}) .

(2)

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} .

(3)

Thus

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

(4)

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

 Ancestors (10)

 View article source

 Discussion (0)

New discussion

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

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 39C / b

 Discussion (0)  Subscribe (1)

 Discussion (0)