= Solution
With a mass source, the <continuity equation> becomes $\rho_t+\partial_i(\rho u_i)=M$. If the conservative momentum equation has no added source, repeating the elimination gives the <mass-injection term in the acoustic analogy>:
$$
\boxed{(\partial_t^2-c_0^2\Delta)\rho'=\partial_i\partial_jT_{ij}+\partial_tM.}
$$
More generally, injection can carry momentum. If its momentum source is $S_i$, the additional term is $\partial_tM-\partial_iS_i$; prescribing $M$ alone does not determine this dipole contribution. The displayed monopole formula assumes no separately imposed momentum source.
Let $\mathcal M(t)=\int M(\mathbf y,t)\,d^3y$ be the total mass rate. Convolution with the <retarded acoustic Green function>, followed by the <acoustic compact-source approximation>, gives the extra <acoustic monopole>
$$
\boxed{\rho'_M(\mathbf x,t)\sim\frac{\dot{\mathcal M}(t-r/c_0)}{4\pi c_0^2r}.}
$$
The stress quadrupole from the preceding part remains present. If required, a compact injected momentum rate $\mathcal S_i=\int S_i\,d^3y$ adds the dipole $n_i\dot{\mathcal S}_i/(4\pi c_0^3r)$.
A small pulsating bubble displaces liquid with volume flux $\dot{\mathcal V}(t)$ and acts acoustically like a mass rate $\mathcal M=\rho_0\dot{\mathcal V}$. Spherical symmetry makes this a monopole, with
$$
\boxed{\rho'\sim\frac{\rho_0}{4\pi c_0^2r}\ddot{\mathcal V}(t-r/c_0).}
$$
For volume oscillation $\mathcal V=\mathcal V_0+\Delta\mathcal V\cos\omega t$, the density amplitude is $\rho_0\omega^2|\Delta\mathcal V|/(4\pi c_0^2r)$. \b[At fixed volume-displacement amplitude, it scales as $\omega^2$.] Holding volume-flux amplitude fixed would instead give an $\omega$ scaling. The bubble must remain small compared with the wavelength, and a frequency-dependent dynamical bubble amplitude is a separate effect.
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