Can anyone please explain what exactly is reduced mass and how it can be used in solving problems in classical/Newtonian mechanics? I've been seeing this a lot lately and have no idea why it's being used.
$$\mu \equiv \frac{1}{\frac{1}{m_1}+\frac{1}{m_2}}.$$
The purpose of reduced mass is to turn a two-body problem into a one-body one. The classic example is two masses $m_1$ and $m_2$ interacting via their gravitational fields. The energy of the system is $$E = T + U = \frac{1}{2}m_1\dot{\mathbf{r_1}}^2 + \frac{1}{2}m_2\dot{\mathbf{r_2}}^2-\frac{GmM}{|\mathbf{r_1} - \mathbf{r_2}|},$$ where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the positions of the two masses. This equation becomes simpler if we write it in terms of the relative coordinate $\mathbf{r} \equiv \mathbf{r_1}-\mathbf{r_2}$ and the center of mass coordinate $\mathbf{R} \equiv \frac{m_1\mathbf{r_1} + m_2\mathbf{r_2}}{m_1 + m_2}$: $$E = \frac{1}{2}\mu\dot{\mathbf{r}}^2 + \frac{1}{2}M\,\dot{\mathbf{R}}^2 - \frac{Gm_1 m_2}{r}.$$ (Try showing this as an exercise.) Here, $\mu$ is the reduced mass, $\mu \equiv \frac{m_1 m_2}{m_1 + m_2}$ and $M \equiv m_1 + m_2$.
Now, because the system is isolated (i.e., there are no external forces), its COM moves at constant velocity, meaning that the COM frame is inertial. In that reference frame, $\dot{\mathbf{R}} = \mathbf{0}$, so the total energy is just $$E = \frac{1}{2}\mu\dot{\mathbf{r}}^2 - \frac{Gm_1 m_2}{r}.$$ Notice that this is exactly the same as the total energy of a single particle of mass $\mu$ and position $\mathbf{r}$, moving through the potential $U(r) = -\frac{Gm_1m_2}{r}$. So to solve the two-body problem, we just need to solve a one-body problem!
Note: It's actually much easier to solve this problem with the Lagrangian $\mathcal{L} \equiv T - U$ and the Euler-Lagrange equation, but I worked with the total energy here for the sake of simplicity.
The reduced mass also pops up in other problems. For instance, consider a system of two masses (we'll call them $m_1$ and $m_2$ again) attached to either end of an ideal spring of constant $k$. The masses are constrained to move along the $x$-axis. Denote their displacements from equilibrium by $x_1$ and $x_2$, respectively. The equations of motion are $$ \begin{cases} \ddot{x_1} = -\frac{k}{m_1}(x_1 - x_2) \\ \ddot{x_2} = \frac{k}{m_2}(x_1 - x_2) \end{cases} $$ (If you haven't seen this notation before, $\ddot{x} \equiv \frac{d^2 x}{dt^2}$.) Introducing the relative coordinate $x \equiv x_1 - x_2$ and subtracting the second equation from the first, we get $$\ddot{x_1} - \ddot{x_2} = -k\left(\frac{1}{m_1} + \frac{1}{m_2}\right)x,$$ or $$\mu\ddot{x} = -kx,$$ with the reduced mass defined as before. So the relative coordinate oscillates harmonically at angular frequency $\omega = \sqrt{\frac{k}{\mu}}$. As before, the reduced mass simplified the problem by "reducing" the number of bodies involved.
Reduced mass is a physical equation used for two (or more) objects in motion about each other, vibrating or rotating due to their own internal forces, while both objects are being subjected to an external force that translates them. It defines the "center point" behavior of the system of two objects as viewed by the external force. It allows us to solve the equations of motion of the two objects entirely by themselves using their own internal forces and individual masses. Subsequently, we can solve the equations of motion of the two objects as though they are only one object "tied together" using only the external force and the reduced mass.
An example is our solar system. It consists of many bodies (planets and the sun) in motion about themselves yet translating all together through the universe. We treat the internal rotations on their own. We treat the motion through space using the reduced mass and center of (reduced) mass coordinates.
