Consider the wave equation in the x-direction: $$ \rho\frac{\partial^2u}{\partial t^2}=\frac{\sigma_{xx}}{\partial x}+\frac{\sigma_{xy}}{\partial x}+\frac{\sigma_{xz}}{\partial x} \tag{1} $$ where $\rho$ is the density of the medium and $\sigma_{xx},\sigma_{xy},\sigma_{xz}$ are stresses acting on the face perpendicular to the x-axis, and point to the x, y and z directions, respectively.
This wave equation is essential equivalent to Newton's second law of motion: $$ m \times a \tag{2} = F_1 $$ where $m$ and $a$ are mass and acceleration of a point mass, and $F_1$ is the force giving rise to the acceleration. Here I changed the order of the sides of the equation to make the terms in (2) correspond with the terms in (1).
Now, to account for the source generating the wave, my textbook says we add a source term to the RHS of equation (1): $$ \rho\frac{\partial^2u}{\partial t^2}=\frac{\sigma_{xx}}{\partial x}+\frac{\sigma_{xy}}{\partial x}+\frac{\sigma_{xz}}{\partial x} +\rho F_2 \tag{3} $$ Here, I understand that $F_2$ is the source that gives rise to the wave (please correct me if that's incorrect).
My question is how adding $\rho F_2$ represents the source generating the wave? All I see if that we just added an arbitrary term to the RHS of equation (2) and if we compare equation (3) to Newton's second law, it is equivalent to: $$ m \times a = F_1 + F_2 $$ which seems redundant to me.
