Two-mass system transfer function and F-V analogy

Assumptions: $f_{c1}$, $f_{c2}$ are viscous friction coefficients of $m_1$, $m_2$ with the ground; $x_3$ is the displacement of the massless junction between $K_2$ and $B_1$ ($K_2$ on the $m_1$ side); zero initial conditions.

Equations of motion (Laplace form)

Mass $m_1$:

$$F(s) = (m_1s^2 + f_{c1}s + K_1)X_1 + K_2(X_1 - X_3)$$

Junction: $\;K_2(X_3 - X_1) + B_1s(X_3 - X_2) = 0$

Mass $m_2$:

$$0 = (m_2s^2 + f_{c2}s + B_2s)X_2 + B_1s(X_2 - X_3)$$

Solving

$K_2$ and $B_1$ in series carry the same force, so they act as one coupling element

$$Z_c(s) = \frac{K_2B_1s}{K_2 + B_1s}$$

Let $Z_1 = m_1s^2 + f_{c1}s + K_1$ and $Z_2 = m_2s^2 + (f_{c2}+B_2)s$. Then

$$\begin{bmatrix} Z_1 + Z_c & -Z_c \\ -Z_c & Z_2 + Z_c \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} F \\ 0 \end{bmatrix}$$ $$\frac{X_2}{F} = \frac{Z_c}{(Z_1+Z_c)(Z_2+Z_c) - Z_c^2} = \frac{Z_c}{Z_1Z_2 + Z_c(Z_1 + Z_2)}$$

Substituting $Z_c$, multiplying by $(K_2 + B_1s)$ and cancelling the common factor $s$:

$$\frac{X_2(s)}{F(s)} = \frac{K_2B_1}{(K_2 + B_1s)(m_1s^2 + f_{c1}s + K_1)(m_2s + f_{c2} + B_2) + K_2B_1\left[(m_1+m_2)s^2 + (f_{c1}+f_{c2}+B_2)s + K_1\right]}$$

The denominator is a fourth-order polynomial; its leading term is $B_1m_1m_2s^4$ and its constant term is $K_1K_2(B_1 + B_2 + f_{c2})$.

Force–voltage (F-V) analogy

Force → voltage, mass → $L$, friction/damper → $R$, spring → $C = 1/K$, velocity → loop current. Mechanical series ($K_2$, $B_1$) → electrical parallel ($C_2 \parallel R_{B1}$), shared by both loops.

  +--[L1]--[Rf1]--[C1]--+--[L2]--[Rf2]--[RB2]--+
  |                     |                      |
 (F)  i1 ->          +--+--+          <- i2    |
  |                  C2   RB1                  |
  |                  +--+--+                   |
  |                     |                      |
  +---------------------+----------------------+
 L1=m1, Rf1=fc1, C1=1/K1, C2=1/K2, RB1=B1
 L2=m2, Rf2=fc2, RB2=B2

Loop equations, with $i = \dot x$:

$$\begin{aligned} F &= L_1\frac{di_1}{dt} + R_{f1}i_1 + \frac{1}{C_1}\int i_1dt + v_p \\ 0 &= L_2\frac{di_2}{dt} + (R_{f2} + R_{B2})i_2 - v_p \end{aligned}$$

where $v_p$ is the voltage across $C_2 \parallel R_{B1}$ carrying $(i_1 - i_2)$ in total; these match the mechanical equations.