Two-mass system equations, X2/F, F-I and F-V analogies

Displacements: $x_1$ for $M_1$ (force $f$, ground friction $B_1$) and $x_2$ for $M_2$ (on rollers, so no ground friction). $B_{12} \parallel K_{12}$ couples the masses; $B_2 \parallel K_2$ ties $M_2$ to the wall.

Differential equations

$$\begin{aligned} M_1\ddot x_1 + B_1\dot x_1 + B_{12}(\dot x_1 - \dot x_2) + K_{12}(x_1 - x_2) &= f(t) \\ M_2\ddot x_2 + B_{12}(\dot x_2 - \dot x_1) + K_{12}(x_2 - x_1) + B_2\dot x_2 + K_2x_2 &= 0 \end{aligned}$$

Transfer function $X_2(s)/F(s)$

Laplace form:

$$\begin{aligned} \left[M_1s^2 + (B_1 + B_{12})s + K_{12}\right]X_1 - (B_{12}s + K_{12})X_2 &= F \\ -(B_{12}s + K_{12})X_1 + \left[M_2s^2 + (B_{12} + B_2)s + K_{12} + K_2\right]X_2 &= 0 \end{aligned}$$

From the second equation, $X_1 = \dfrac{M_2s^2 + (B_{12}+B_2)s + K_{12} + K_2}{B_{12}s + K_{12}}X_2$. Substituting into the first:

$$\frac{X_2(s)}{F(s)} = \frac{B_{12}s + K_{12}}{\Delta}$$ $$\Delta = \left[M_1s^2 + (B_1+B_{12})s + K_{12}\right]\left[M_2s^2 + (B_{12}+B_2)s + K_{12} + K_2\right] - (B_{12}s + K_{12})^2$$

Expanded:

$$\begin{aligned} \Delta =\; & M_1M_2s^4 + \left[B_1M_2 + B_{12}(M_1 + M_2) + B_2M_1\right]s^3 \\ & + \left[B_1B_{12} + B_1B_2 + B_{12}B_2 + K_{12}(M_1 + M_2) + K_2M_1\right]s^2 \\ & + \left[B_1(K_{12} + K_2) + B_{12}K_2 + B_2K_{12}\right]s + K_{12}K_2 \end{aligned}$$

Table of analogies

Force-current analogous circuit

Node 1 ↔ $\dot x_1$, node 2 ↔ $\dot x_2$; $C_1 = M_1$, $C_2 = M_2$, $R_1 = 1/B_1$, $R_{12} = 1/B_{12}$, $L_{12} = 1/K_{12}$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.

 v1             L12∥R12     v2
 +----+----+---+--LLLL--+---+----+----+
 |    |    |   +-/\/\/--+   |    |    |
(i)   C1   R1               C2   R2   L2
 |    |    |                |    |    |
 +----+----+----------------+----+----+ gnd

$$\begin{aligned} C_1\frac{dv_1}{dt} + \frac{v_1}{R_1} + \frac{v_1 - v_2}{R_{12}} + \frac{1}{L_{12}}\int(v_1 - v_2)dt &= i \\ C_2\frac{dv_2}{dt} + \frac{v_2 - v_1}{R_{12}} + \frac{1}{L_{12}}\int(v_2 - v_1)dt + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2dt &= 0 \end{aligned}$$

Force-voltage analogous circuit

Mesh 1 ↔ $\dot x_1$, mesh 2 ↔ $\dot x_2$; $L_1 = M_1$, $L_2 = M_2$, $R_1 = B_1$, $R_{12} = B_{12}$, $C_{12} = 1/K_{12}$, $R_2 = B_2$, $C_2 = 1/K_2$.

   L1    R1               L2    R2    C2
 +-LLLL-/\/\-+----------+-LLLL-/\/\---||--+
 |           |                            |
(~) e      R12                            |
 |           |                            |
 |   i1     C12            i2             |
 +-----------+----------------------------+

$$\begin{aligned} L_1\frac{di_1}{dt} + R_1i_1 + R_{12}(i_1 - i_2) + \frac{1}{C_{12}}\int(i_1 - i_2)dt &= e \\ L_2\frac{di_2}{dt} + R_{12}(i_2 - i_1) + \frac{1}{C_{12}}\int(i_2 - i_1)dt + R_2i_2 + \frac{1}{C_2}\int i_2dt &= 0 \end{aligned}$$

Both sets have the same form as the mechanical equations, confirming the analogies.