State model of two-mass spring system, output x2

Let $x_1$ and $x_2$ be the displacements of $M_1$ and $M_2$ from their rest positions (positive to the right).

 |    K1    +----+   K2    +----+
 |--/\/\/---| M1 |--/\/\/--| M2 |--> F(t)
 |          +----+         +----+
 |           x1 ->          x2 ->

Equations of motion (no friction)

Mass $M_1$: spring $K_1$ exerts a restoring force $K_1x_1$, and spring $K_2$ exerts $K_2(x_1 - x_2)$:

$$M_1\ddot{x}_1 + K_1x_1 + K_2(x_1 - x_2) = 0$$

Mass $M_2$:

$$M_2\ddot{x}_2 + K_2(x_2 - x_1) = F(t)$$

State variables

$z_1 = x_1$, $z_2 = \dot{x}_1$, $z_3 = x_2$, $z_4 = \dot{x}_2$:

$$\begin{aligned} \dot{z}_1 &= z_2 \\ \dot{z}_2 &= -\frac{K_1 + K_2}{M_1}z_1 + \frac{K_2}{M_1}z_3 \\ \dot{z}_3 &= z_4 \\ \dot{z}_4 &= \frac{K_2}{M_2}z_1 - \frac{K_2}{M_2}z_3 + \frac{1}{M_2}F(t) \end{aligned}$$

State equation

$$\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \\ \dot{z}_3 \\ \dot{z}_4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -\dfrac{K_1+K_2}{M_1} & 0 & \dfrac{K_2}{M_1} & 0 \\ 0 & 0 & 0 & 1 \\ \dfrac{K_2}{M_2} & 0 & -\dfrac{K_2}{M_2} & 0 \end{bmatrix}\begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ z_4 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \\ \dfrac{1}{M_2} \end{bmatrix}F(t)$$

Output equation

$$y = x_2(t) = \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ z_4 \end{bmatrix}$$

With no damping, the eigenvalues of $A$ lie on the imaginary axis, so the system oscillates without decay (marginally stable).