Vertical two-mass system transfer function X1/F

Measure $x_1$, $x_2$ downward from the static equilibrium position, so gravity is balanced by the initial spring stretch and drops out of the equations.

Forces

• $M_1$: spring $K_1$ (to support) gives $K_1x_1$; damper $B_1$ (to $M_2$) gives $B_1(\dot x_1 - \dot x_2)$; inertia $M_1\ddot x_1$.
• $M_2$: applied $F(t)$ downward; damper $B_2$ (to support) gives $B_2\dot x_2$; $B_1$ gives $B_1(\dot x_2 - \dot x_1)$; inertia $M_2\ddot x_2$.

Differential equations

$$\begin{aligned} M_1\ddot x_1 + K_1x_1 + B_1(\dot x_1 - \dot x_2) &= 0 \\ M_2\ddot x_2 + B_2\dot x_2 + B_1(\dot x_2 - \dot x_1) &= F(t) \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + B_1s + K_1)X_1 - B_1sX_2 &= 0 \\ -B_1sX_1 + \left[M_2s^2 + (B_1 + B_2)s\right]X_2 &= F \end{aligned}$$

Solving

From the first equation, $X_2 = \dfrac{M_1s^2 + B_1s + K_1}{B_1s}X_1$. Substituting:

$$\left[\frac{(M_1s^2 + B_1s + K_1)\,s(M_2s + B_1 + B_2)}{B_1s} - B_1s\right]X_1 = F$$ $$\frac{X_1(s)}{F(s)} = \frac{B_1s}{s(M_1s^2 + B_1s + K_1)(M_2s + B_1 + B_2) - B_1^2s^2}$$

Cancelling $s$:

$$\frac{X_1(s)}{F(s)} = \frac{B_1}{(M_1s^2 + B_1s + K_1)(M_2s + B_1 + B_2) - B_1^2s}$$

Final form

$$\frac{X_1(s)}{F(s)} = \frac{B_1}{M_1M_2s^3 + \left[B_1(M_1 + M_2) + B_2M_1\right]s^2 + (B_1B_2 + K_1M_2)s + K_1(B_1 + B_2)}$$

Check: under a constant force, $M_2$ finally moves at a steady velocity $v_2 = F/(B_1 + B_2)$ (it is held only by dampers), and $M_1$ settles where $K_1x_1 = B_1v_2$, i.e. $x_1 = \dfrac{B_1F}{K_1(B_1 + B_2)}$, which is the value of the transfer function at $s = 0$.