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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:More generally, injection can carry momentum. If its momentum source is $S_i$, the additional term is ${\partial }_{t}M-{\partial }_i{S}_i$; prescribing $M$ alone does not determine this dipole contribution. The displayed monopole formula assumes no separately imposed momentum source.

Let $\mathcal{M}(t)=\int M(\mathbf{y},t)d^{3}y$ be the total mass rate. Convolution with the retarded acoustic Green function, followed by the acoustic compact-source approximation, gives the extra acoustic monopoleThe stress quadrupole from the preceding part remains present. If required, a compact injected momentum rate ${\mathcal{S}}_i=\int S_i{d}^{3}y$ adds the dipole $n_i{\dot{\mathcal{S}}}_i/(4\pi c_0^{3}r)$.

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, withFor volume oscillation $\mathcal{V}={\mathcal{V}}_0+\Delta \mathcal{V}\cos \omega t$, the density amplitude is ${\rho }_0{\omega }^2|\Delta \mathcal{V}|/(4\pi c_0^{2}r)$. 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.