= Solution
Write the density and pressure as $\rho_0+\rho'$ and $p_0+p'$, and let the small fluid velocity be the gradient of an <acoustic velocity potential>, $\mathbf u=\nabla\phi$. The <linear homentropic acoustic equations> are
$$
\rho'_t+\rho_0\nabla^2\phi=0,
\qquad
\rho_0\nabla\phi_t=-\nabla p',
\qquad
p'=c_0^2\rho',
$$
where the <adiabatic sound speed> of a perfect gas is $c_0^2=\gamma p_0/\rho_0$. Absorbing a function of time into $\phi$, the momentum equation gives
$$
\boxed{p'=-\rho_0\phi_t.}
$$
Eliminating $p'$ and $\rho'$ gives the acoustic <wave equation>
$$
\boxed{\phi_{tt}=c_0^2\nabla^2\phi.}
$$
Let the shell displacement be $a(t)-a_0=\operatorname{Re}(\eta e^{-i\omega t})$ and put $k=\omega/c_0$, $\theta=ka_0$. The spherically symmetric solution regular at the centre is
$$
\phi(r,t)=\operatorname{Re}\left(A\frac{\sin kr}{r}e^{-i\omega t}\right).
$$
The <kinematic boundary condition> $\phi_r(a_0,t)=\dot a(t)$ gives
$$
A\frac{\sin\theta}{a_0^2}(\theta\cot\theta-1)
=-i\omega\eta.
$$
Since $p'=-\rho_0\phi_t$, the pressure amplitude on the shell is therefore
$$
\widehat p(a_0)=
\frac{\rho_0\omega^2a_0}{\theta\cot\theta-1}\eta.
$$
The radial <Newton's second law> per unit area is
$$
m\ddot a=p'(a_0,t)-\kappa(a-a_0),
$$
so its complex amplitude satisfies
$$
(\kappa-m\omega^2)\eta=\widehat p(a_0).
$$
For a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(mc_0^2)$ yield the frequency equation for <spherically symmetric vibration of a gas-filled elastic shell>:
$$
\boxed{\theta^2\left(1+\frac{\alpha}{\theta\cot\theta-1}\right)
=\frac{\kappa a_0^2}{mc_0^2},
\qquad
\alpha=\frac{\rho_0a_0}{m}.}
$$
The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta\cot\theta-1$ runs from $+\infty$ to $-\infty$ on $(n\pi,(n+1)\pi)$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.
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