Two-mass system X2/F with F-V analogy (B1, K3 link)

Displacements: $x_1$ ($M_1$), $x_2$ ($M_2$) and $x_3$ = junction between damper $B_1$ (on the $M_1$ side) and spring $K_3$ (on the $M_2$ side).

Equations of motion (Laplace)

$$\begin{aligned} (M_1s^2 + B_{v1}s + K_1)X_1 + B_1s(X_1 - X_3) &= F \\ B_1s(X_3 - X_1) + K_3(X_3 - X_2) &= 0 \\ \left[M_2s^2 + (B_{v2} + B_2)s + K_2\right]X_2 + K_3(X_2 - X_3) &= 0 \end{aligned}$$

Eliminating the junction

$B_1$ in series with $K_3$:

$$K_{eq}(s) = \frac{B_1sK_3}{B_1s + K_3}$$

With $a = M_1s^2 + B_{v1}s + K_1$ and $b = M_2s^2 + (B_{v2} + B_2)s + K_2$:

$$\frac{X_2}{F} = \frac{K_{eq}}{(a + K_{eq})(b + K_{eq}) - K_{eq}^2} = \frac{K_{eq}}{ab + K_{eq}(a + b)}$$

Transfer function

Multiplying by $(B_1s + K_3)$:

$$\frac{X_2(s)}{F(s)} = \frac{B_1K_3s}{(B_1s + K_3)(M_1s^2 + B_{v1}s + K_1)\left[M_2s^2 + (B_{v2} + B_2)s + K_2\right] + B_1K_3s\left[(M_1 + M_2)s^2 + (B_{v1} + B_{v2} + B_2)s + K_1 + K_2\right]}$$

The denominator is 5th order (leading term $B_1M_1M_2s^5$, constant term $K_1K_2K_3$). The zero at $s = 0$ shows that a steady force cannot move $M_2$, because the damper $B_1$ transmits no static force.

F-V analogous circuit

$F\to e$, $M\to L$, $B\to R$, $K\to 1/C$, velocity $\to$ mesh current. $B_1$ and $K_3$ are in series (same force), so in the F-V circuit they become the parallel pair $R = B_1 \parallel C = 1/K_3$, shared by mesh 1 ($M_1$) and mesh 2 ($M_2$); the loop inside the pair is mesh 3 (junction $x_3$).

  L1  R=Bv1 C=1/K1      L2  R=Bv2+B2 C=1/K2
 +LLLL-/\/\--||---+-----LLLL---/\/\----||--+
 |                |                        |
(~) e       +-----+-----+                  |
 |    i1    |           |      i2          |
 |        R=B1   i3   C=1/K3               |
 |          |           |                  |
 |          +-----+-----+                  |
 |                |                        |
 +----------------+------------------------+

Mesh equations:

$$\begin{aligned} L_1\frac{di_1}{dt} + R_{v1}i_1 + \frac{1}{C_1}\int i_1dt + R_1(i_1 - i_3) &= e \\ R_1(i_3 - i_1) + \frac{1}{C_3}\int(i_3 - i_2)dt &= 0 \\ L_2\frac{di_2}{dt} + (R_{v2} + R_2)i_2 + \frac{1}{C_2}\int i_2dt + \frac{1}{C_3}\int(i_2 - i_3)dt &= 0 \end{aligned}$$

($R_1 = B_1$ is shared by meshes 1 and 3; $C_3 = 1/K_3$ by meshes 3 and 2.)