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OPhO 2024, Теория — Задача 2. Bouncy Bubble

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2024 Online Physics Olympiad: Invitational Contest — Theoretical Examination · 25 т.

Условие

In this problem, we will investigate the interaction between fast oscillations and gradual changes in a physical system.

2.1

A large volume of incompressible, non-viscous liquid with density $\rho$ is kept at temperature $T_c$ and pressure $P_c$ . A spherical bubble consisting of $N$ particles of ideal gas with temperature $T_0>T_c$ is introduced into the liquid. Neglect surface tension and any heat transfer between the liquid and the gas.

a. Find the equilibrium radius $R_0$ of the bubble.

The bubble’s radius is perturbed slightly from equilibrium and its oscillations are observed; the gas remains near thermal equilibrium at all times. Assume that the motion of the liquid is laminar and radial, and that the density of the gas is negligible compared to $\rho$ . You may express future answers in terms of $R_0$ .

b. Find the frequency $\omega$ of the bubble’s small oscillations.

2.2

Now, assume that the interface between the gas and the liquid has thermal conductance per unit area $\kappa$ . Then, because of heat loss, the bubble will shrink over time, approaching a final radius $R_f$ (which you may use in future answers). The shrinkage is slow enough that the kinetic energy of the liquid can be neglected.

c. If the bubble starts at radius $R_0$ , find the approximate time $\tau$ until it shrinks to radius $(R_0+R_f)/2$ . Express your answer to the lowest order in the quantity $\alpha=R_0/R_f-1$ .

d. Next, the bubble starts off oscillating around $R_0$ with amplitude $R_0\delta_0$ , where $\delta_0\ll1$ ; assume that the oscillations are much faster than the shrinkage. Find the time-averaged final radius $R_f'$ of the bubble, to the lowest order in $\delta_0$ . Qualitatively explain the reason for any difference between $R_f'$ and $R_f$ .

e. Given the situation in part (d), find the approximate time $\tau'$ until the bubble’s time-averaged radius shrinks to $(R_0+R_f')/2$ , to the lowest orders in $\alpha'=R_0/R_f'-1$ and $\delta_0$ .

Published organizer clarifications — separately quoted from the 2024 OPhO Clarifications document.

T2 - Bouncy Bubble

The ideal gas is monatomic. “To the lowest order” means the lowest-order nonzero correction.

(a) $T_0$ is the temperature at the bubble’s equilibrium radius.

(c) $\kappa$ is the thermal conductance per unit area of the bubble’s surface. The bubble starts at radius $R_0$ . You should express future answers in terms of $R_f$ instead of $T_c$ or $P_c$ (when possible).

(d) The oscillations begin around radius $R_0$ .

(e) When expanding to the lowest order in $\alpha'$ and $\delta_0$ , you may neglect the cross term. You may not use $R_f'$ in your answer.

Решение

Bouncy Bubble – Solution

Total: 25.0 pts

a.

At equilibrium, the pressure in the bubble is $P_c$ . Using the Ideal Gas Law:

P_cV=P_c\cdot\frac43\pi R_0^3=NkT_0\Rightarrow R_0=\sqrt[3]{\frac{3NkT_0}{4\pi P_c}}

Grading Scheme: 1.0 pts total.

0.5 pts for use of the Ideal Gas Law.
0.5 pts for a correct expression for $R_0$ .

b.

Let the bubble have radius $R_0+x$ . Setting $U=0$ at equilibrium, the work-energy theorem implies:

U=\int_0^x4\pi R_0^2(P_c-P)\,da

Using $PV^{5/3}=\mathrm{const}$ :

\begin{aligned}U&=\int_0^x4\pi R_0^2\left(P_c-\frac{P_c(\frac43\pi R_0^3)^{5/3}}{(\frac43\pi(R_0+a)^3)^{5/3}}\right)\,da\\&=\int_0^x4\pi R_0^2P_c\left(1-\left(1+\frac a{R_0}\right)^{-5}\right)\,da\\&\approx\int_0^x20\pi P_cR_0a\,da\\&=10\pi P_cR_0x^2\end{aligned}

The kinetic energy comes from the motion of the liquid. If the bubble’s radius is changing at rate $\dot x$ , the liquid at radius $r$ must have speed $\frac{R_0^2}{r^2}\dot x$ by incompressibility. Thus:

K=\int_R^\infty\frac12(4\pi\rho r^2\,dr)\left(\frac{R_0^2}{r^2}\dot x\right)^2=2\pi\rho R_0^4\dot x^2\int_R^\infty\frac1{r^2}\,dr=2\pi\rho R_0^3\dot x^2

We find $\omega=\frac1{R_0}\sqrt{\frac{5P_c}\rho}$ .

Grading Scheme: 5.0 pts total

1.0 pts for using the work-energy theorem to find an integral for $U$ in terms of $P_c-P$
1.0 pts for using the adiabatic condition to linearize $P_c-P$ .
0.5 pts for a correct expression for $U$ .
1.0 pts for using incompressibility to derive the velocity profile of the liquid, $v(r)\propto1/r^2$ .
0.5 pts for a correct expression for $K$ .
1.0 pts for a correct expression for $\omega$ .

c.

Because the bubble begins at equilibrium and shrinks slowly, it will remain near equilibrium. Thus, we will always have $T=\frac{4\pi P_cr^3}{3Nk}$ . Because the external pressure is constant:

\dot H=\dot Q=-4\pi r^2\kappa(T-T_c)

Using $H=\frac52NkT$ , we have:

\dot H=\frac52Nk\dot T=\frac52Nk\frac{dT}{dr}\dot r=10\pi P_cr^2\dot r\Rightarrow\dot r=-\frac{2\kappa}{5P_c}(T-T_c)=-\frac{8\pi\kappa}{15Nk}\left(r^3-\frac{3NkT_c}{4\pi P_c}\right)

Note that $R_f^3=\frac{3NkT_c}{4\pi P_c}$ . Separating variables:

\int_{R_0}^{(R_0+R_f)/2}\frac1{r^3-R_f^3}\,dr=\int_0^\tau-\frac{8\pi\kappa}{15Nk}\,dt\Rightarrow\frac1{R_f^2}\int_{1+\alpha/2}^{1+\alpha}\frac1{x^3-1}\,dx=\frac{8\pi\kappa}{15Nk}\tau

\int_{1+\alpha/2}^{1+\alpha}\frac1{x^3-1}\,dx\approx\int_{1+\alpha/2}^{1+\alpha}\left(\frac1{3(x-1)}-\frac13\right)\,dx=\frac{\ln(2)}3-\frac\alpha6

Thus, $\tau=\frac{5\ln(2)Nk}{8\pi\kappa R_f^2}\left(1-\frac\alpha{2\ln(2)}\right)$ .

Grading Scheme: 5.0 pts total.

0.5 pts for a correct expression for $\dot Q$ in terms of $T-T_c$
1.0 pts for equating $\dot Q$ to $\dot H$ or $C_p\dot T$ due to constant external pressure.
1.5 pts for substituting the expression for $T$ and obtaining a correct differential equation in $r$ .
1.0 pts for separating variables and approximating the integrand as shown.
1.0 pts for a correct final answer, or 0.5 pts for an answer with a correct leading term.

d.

Let the oscillating radius be $r(1+\delta\cos(\omega t))$ . We claim that the total energy $E$ of the oscillations is proportional to $\omega$ , giving $\delta=\frac{R_0^2}{r^2}\delta_0$ . To see this, we consider the adiabatic invariant, $I$ : the area enclosed by the system’s motion in phase space. For any simple harmonic oscillator, $I=2\pi E/\omega$ . Crucially, $I$ remains approximately constant under slow external changes, giving $E\propto\omega$ as desired. You can read more here.

Using $TV^{2/3}=\mathrm{const}$ , the oscillating temperature is $\frac T{(1+\delta\cos(\omega t))^2}$ . Thus, we have:

\begin{aligned}\dot Q&=-4\pi\kappa r^2(1+\delta\cos(\omega t))^2\left(\frac T{(1+\delta\cos(\omega t))^2}-T_c\right)\\&=-4\pi\kappa r^2\left(T-(1+2\delta\cos(\omega t)+\delta^2\cos^2(\omega t))T_c\right)\end{aligned}

\begin{aligned}\Rightarrow\langle\dot Q\rangle&=-4\pi\kappa r^2\left(T-\left(1+\frac{\delta^2}2\right)T_c\right)\\&=-4\pi\kappa r^2T_c\left(\frac{r^3}{R_f^3}-1-\frac{R_0^4\delta_0^2}{2r^4}\right)\end{aligned}

At equilibrium, we must have $\langle\dot Q\rangle=0$ . Thus, assuming that $R_f'=R_f(1+c)$ :

(1+c)^3-1-\frac{R_0^4\delta_0^2}{2R_f^4(1+c)^4}\approx3c-\frac{R_0^4\delta_0^2}{2R_f^4}=0\Rightarrow R_f'=R_f\left(1+\frac{R_0^4\delta_0^2}{6R_f^4}\right)

The fundamental reason why $R_f'>R_f$ is that the rate of heat loss is a concave function under adiabatic perturbations of $r$ . Thus, the rate of heat loss under the fast, roughly adiabatic oscillations averages out to less than it does with no oscillation, so the equilibrium radius is slightly larger.

Grading Scheme: 9.0 pts total.

3.0 pts for use of the adiabatic invariant or similar to derive $\delta(r)$ .
1.0 pts for obtaining $\delta(r)\propto1/r^2$ .
1.0 pts for finding an expression for the oscillating temperature using the adiabatic condition.
1.5 pts for substituting the oscillating radius and temperature into $\dot Q$ and time-averaging
0.5 pts for setting $\langle\dot Q\rangle=0$ at equilibrium.
1.0 pts for a correct final answer.
1.0 pts for mentioning the concavity of $\dot Q$ as the cause of the discrepancy between $R_f$ and $R_f'$ , or 0.5 points for only mentioning nonlinearity.

e.

Now, we must consider the change in energy of the oscillations in addition to the change in $H$ . Thus:

\dot H+\dot E=\dot H+\frac{dE}{dr}\dot r=\langle\dot Q\rangle

Note that we do not time-average $\dot H$ because it is a state variable (when the system crosses through equilibrium, $H$ is uniquely determined by $r$ ). Then, using $E=10\pi P_cr^3\delta^2=10\pi P_c\frac{R_0^4\delta_0^2}r$ :

10\pi P_cr^2\dot r-10\pi P_c\frac{R_0^4\delta_0^2}{r^2}\dot r=-4\pi\kappa r^2T_c\left(\frac{r^3}{R_f^3}-1-\frac{R_0^4\delta_0^2}{2r^4}\right)

\Rightarrow\dot r=-\frac{8\pi\kappa}{15Nk}\left(r^3-R_f^3\left(1+\frac{R_0^4\delta_0^2}{2r^4}\right)\right)\bigg/\left(1-\frac{R_0^4\delta_0^2}{r^4}\right)\approx-\frac{8\pi\kappa}{15Nk}\frac{r^3-R_f'^3}{1-\delta_0^2}

Here, we take $\frac{R_0^4}{r^4}\approx1$ because we don’t care about the cross term between $\delta_0$ and $\alpha'$ . Thus, we have:

\tau'=\frac{15Nk(1-\delta_0^2)}{8\pi\kappa}\int_{R_0}^{(R_0+R_f')/2}\frac1{r^3-R_f'^3}\,dr\approx\frac{5\ln(2)Nk}{8\pi\kappa R_f'^2}\left(1-\frac{\alpha'}{2\ln(2)}\right)(1-\delta_0^2)

Using $R_f'\approx R_f\left(1+\frac{\delta_0^2}6\right)$ , we find $\tau'=\frac{5\ln(2)Nk}{8\pi\kappa R_f^2}\left(1-\frac{\alpha'}{2\ln(2)}\right)\left(1-\frac{4\delta_0^2}3\right)$ .

Grading Scheme: 5.0 pts total.

1.0 pts for substituting the time-averaged $\dot Q$ into the differential equation; 0.5 pts if time-averaging $\dot H$ or $C_p\dot T$ as well.
2.0 pts for realizing that the energy of the oscillations changes and adding this term into the differential equation.
1.0 pts for obtaining a differential equation analogous to the equation given for part (c), with the substitution of $R_f'$ for $R_f$ , and using the same techniques to obtain $\tau'$ .
1.0 pts for a correct final answer, or 0.5 pts for a final answer with the right correction for $\delta_0$ .

Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf