Two-mass system X/F with F-V and F-I analogy

Assumptions: the masses have no friction with the ground; $x_1$ is the displacement of $M_1$, $x$ the displacement of $M_2$, and $x_3$ the displacement of the junction between damper $B$ and spring $K_3$ (B on the $M_1$ side).

Equations of motion (Laplace, zero initial conditions)

Mass $M_1$:

$$F(s) = (M_1s^2 + K_1)X_1 + K_2(X_1 - X) + Bs(X_1 - X_3)$$

Junction ($x_3$, massless): $\;Bs(X_3 - X_1) + K_3(X_3 - X) = 0$

Mass $M_2$: $\;M_2s^2X + K_2(X - X_1) + K_3(X - X_3) = 0$

Solving

The series $B$–$K_3$ branch acts as an equivalent spring $\dfrac{BsK_3}{Bs+K_3}$. Together with $K_2$ in parallel, the coupling between the masses is

$$K_c(s) = K_2 + \frac{BK_3s}{Bs+K_3} = \frac{N(s)}{Bs+K_3}, \quad N(s) = B(K_2+K_3)s + K_2K_3$$

The two mass equations become $(M_1s^2+K_1+K_c)X_1 - K_cX = F$ and $-K_cX_1 + (M_2s^2+K_c)X = 0$. So

$$\frac{X(s)}{F(s)} = \frac{K_c}{(M_1s^2+K_1+K_c)(M_2s^2+K_c) - K_c^2} = \frac{K_c}{(M_1s^2+K_1)(M_2s^2+K_c) + K_cM_2s^2}$$

Multiplying top and bottom by $(Bs+K_3)$ and expanding:

$$\frac{X(s)}{F(s)} = \frac{B(K_2+K_3)s + K_2K_3}{\Delta(s)}$$ $$\begin{aligned} \Delta(s) ={}& BM_1M_2s^5 + K_3M_1M_2s^4 + B\left[K_1M_2 + (K_2+K_3)(M_1+M_2)\right]s^3 \\ &+ K_3\left[K_1M_2 + K_2(M_1+M_2)\right]s^2 + BK_1(K_2+K_3)s + K_1K_2K_3 \end{aligned}$$

Force–voltage (F-V) analogy

Force → voltage, mass → inductance $L = M$, damper → resistance $R = B$, spring → capacitance $C = 1/K$, velocity → loop current. Mechanical parallel → electrical series and vice versa, so the coupling becomes $C_2$ in series with ($R \parallel C_3$).

  +--[L1]--[C1]--+--[L2]--+
  |              |        |
 (f) i1->       C2   <-i2 |
  |              |        |
  |           +--+--+     |
  |           R     C3    |
  |           +--+--+     |
  |              |        |
  +--------------+--------+
 L1=M1, C1=1/K1, C2=1/K2, R=B, C3=1/K3, L2=M2

Force–current (F-I) analogy

Force → current source, mass → capacitance $C = M$, damper → conductance ($R = 1/B$), spring → inductance $L = 1/K$, velocity → node voltage. Each displacement point becomes a node.

     v1                v3             v2
  +---o---+---[R]------o-----[L3]---+---o
  |   |   |                         |   |
  |   |   +----------[L2]-----------+   |
  |   |   |                             |
 (f)  C1  L1                            C2
  |   |   |                             |
  +---+---+-----------------------------+
            ground (reference)
 C1=M1, L1=1/K1, L2=1/K2, R=1/B, L3=1/K3, C2=M2

Node equation at $v_1$, for example: $f = M_1\dot v_1 + K_1\!\int v_1dt + K_2\!\int(v_1-v_2)dt + B(v_1-v_3)$, which matches the force equation of $M_1$ with $v = \dot x$.