State space model of vertical two-mass system

Let $y_1$ be the downward displacement of $M_1$ and $y = y_2$ that of $M_2$, both measured from the static equilibrium position, so gravity and the static spring force cancel and do not appear.

   //////////  support
       |
       K  (spring)
       |
   +-------+
   |  M1   | <-- u(t), y1
   +-------+
       |
       B  (damper)
       |
   +-------+
   |  M2   |  y(t) = y2
   +-------+

Equations of motion

For $M_1$: spring force $Ky_1$ and damper force $B(\dot{y}_1 - \dot{y}_2)$ oppose its motion:

$$M_1\ddot{y}_1 + B(\dot{y}_1 - \dot{y}_2) + Ky_1 = u(t)$$

For $M_2$: only the damper acts on it:

$$M_2\ddot{y}_2 + B(\dot{y}_2 - \dot{y}_1) = 0$$

State variables

$$x_1 = y_1,\quad x_2 = \dot{y}_1,\quad x_3 = y_2,\quad x_4 = \dot{y}_2$$ $$\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -\frac{K}{M_1}x_1 - \frac{B}{M_1}x_2 + \frac{B}{M_1}x_4 + \frac{1}{M_1}u \\ \dot{x}_3 &= x_4 \\ \dot{x}_4 &= \frac{B}{M_2}x_2 - \frac{B}{M_2}x_4 \end{aligned}$$

State-space representation

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \\ \dot{x}_4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -\dfrac{K}{M_1} & -\dfrac{B}{M_1} & 0 & \dfrac{B}{M_1} \\ 0 & 0 & 0 & 1 \\ 0 & \dfrac{B}{M_2} & 0 & -\dfrac{B}{M_2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} + \begin{bmatrix} 0 \\ \dfrac{1}{M_1} \\ 0 \\ 0 \end{bmatrix}u(t)$$ $$y(t) = \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$$

$y_2$ itself is needed as a state only because it is the output; the dynamics depend on $y_1$, $\dot{y}_1$ and $\dot{y}_2$.