Two-mass system between walls with F-I analogy

Displacements: $x_1$ ($M_1$), $x_2$ ($M_2$) and $x_3$ = junction between damper $B_2$ (on the $M_1$ side) and spring $k_2$ (on the $M_2$ side).

Differential equations

$$\begin{aligned} M_1\ddot x_1 + B_1\dot x_1 + k_1x_1 + B_2(\dot x_1 - \dot x_3) &= F(t) \\ B_2(\dot x_3 - \dot x_1) + k_2(x_3 - x_2) &= 0 \\ M_2\ddot x_2 + B_3\dot x_2 + k_3x_2 + k_2(x_2 - x_3) &= 0 \end{aligned}$$

Eliminating the junction

The series $B_2$–$k_2$ combination acts like one element:

$$K_{eq}(s) = \frac{B_2s\cdot k_2}{B_2s + k_2}$$

With $a = M_1s^2 + B_1s + k_1$ and $b = M_2s^2 + B_3s + k_3$:

$$\begin{aligned} (a + K_{eq})X_1 - K_{eq}X_2 &= F \\ -K_{eq}X_1 + (b + K_{eq})X_2 &= 0 \end{aligned}$$

Transfer function

$$\frac{X_2}{F} = \frac{K_{eq}}{ab + K_{eq}(a + b)}$$

Multiplying numerator and denominator by $(B_2s + k_2)$:

$$\frac{X_2(s)}{F(s)} = \frac{B_2k_2\,s}{(B_2s + k_2)(M_1s^2 + B_1s + k_1)(M_2s^2 + B_3s + k_3) + B_2k_2s\left[(M_1+M_2)s^2 + (B_1+B_3)s + k_1 + k_3\right]}$$

The denominator is 5th order (leading term $B_2M_1M_2s^5$, constant term $k_1k_2k_3$).

Force-current analogy

$F\to i$, $M\to C$, $B\to 1/R$, $k\to 1/L$, velocity $\to$ node voltage.

  v1                v3                v2
 +--+--+--+--/\/\/--+--LLLL--+--+--+--+
 |  |  |  |  1/B2      1/k2  |  |  |
(i) C1 R  L                  C2 R  L
 | =M1 1/ 1/                =M2 1/ 1/
 |  |  B1 k1                 |  B3 k3
 +--+--+--+------------------+--+--+ gnd

Node equation at node 1: $C_1\dfrac{dv_1}{dt} + B_1v_1 + k_1\displaystyle\int v_1dt + B_2(v_1 - v_3) = i$ (conductances written directly), and similarly for nodes 3 and 2.