Two oscillating masses: motion, centre of mass, energy

Let $x_1$ and $x_2$ be the displacements of $m_1$ and $m_2$ from the positions where both springs are unstretched, measured to the right. The plane is smooth, so only the spring forces and $F$ act horizontally. Extension of spring $k_1$ is $x_1$; extension of spring $k_2$ is $(x_2 - x_1)$.

(a) Equations of motion

Spring $k_1$ pulls $m_1$ back with $k_1x_1$; spring $k_2$ pulls $m_1$ forward with $k_2(x_2-x_1)$:

$$m_1\ddot x_1 = -k_1x_1 + k_2(x_2 - x_1)$$

On $m_2$, spring $k_2$ pulls backward and $F$ pushes forward:

$$m_2\ddot x_2 = -k_2(x_2 - x_1) + F$$

(b) Motion of the centre of mass

$$x_G = \frac{m_1x_1 + m_2x_2}{m_1+m_2}$$

Adding the two equations, the internal spring force $k_2(x_2-x_1)$ cancels:

$$(m_1+m_2)\,\ddot x_G = F - k_1x_1$$

So the centre of mass moves as if the total mass were acted on by the external forces only: $F$ and the wall-spring force $k_1x_1$.

(c) Energy of the system

$$T = \tfrac12 m_1\dot x_1^2 + \tfrac12 m_2\dot x_2^2$$ $$V = \tfrac12 k_1x_1^2 + \tfrac12 k_2(x_2 - x_1)^2$$

Equivalently, with total mass $M = m_1+m_2$, $T = \tfrac12 M\dot x_G^2 + \tfrac12 m_1(\dot x_1-\dot x_G)^2 + \tfrac12 m_2(\dot x_2-\dot x_G)^2$ (kinetic energy of the centre of mass plus kinetic energy relative to it). The work done by $F$ equals the change in $T+V$.