Two-mass spring system X1/F and F-V analogy

Displacements $x_1$ ($M_1$) and $x_2$ ($M_2$, where $F$ acts). No friction ($B_1 = B_2 = 0$).

Equations of motion

$$\begin{aligned} M_1\ddot x_1 + K_1x_1 + K_2(x_1 - x_2) &= 0 \\ M_2\ddot x_2 + K_2(x_2 - x_1) &= F(t) \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + K_1 + K_2)X_1 - K_2X_2 &= 0 \\ -K_2X_1 + (M_2s^2 + K_2)X_2 &= F \end{aligned}$$

From the first equation, $X_2 = \dfrac{M_1s^2 + K_1 + K_2}{K_2}X_1$. Substituting into the second:

$$\left[\frac{(M_1s^2 + K_1 + K_2)(M_2s^2 + K_2)}{K_2} - K_2\right]X_1 = F$$

Transfer function

$$\frac{X_1(s)}{F(s)} = \frac{K_2}{(M_1s^2 + K_1 + K_2)(M_2s^2 + K_2) - K_2^2}$$ $$\frac{X_1(s)}{F(s)} = \frac{K_2}{M_1M_2s^4 + \left[M_1K_2 + M_2(K_1 + K_2)\right]s^2 + K_1K_2}$$

Only even powers of $s$ appear because there is no damping; the poles are on the $j\omega$ axis (two undamped natural frequencies).

Force-voltage analogy

$F\to e$, $M\to L$, $K\to 1/C$, velocity $\to$ current.

$$\begin{aligned} L_1\frac{di_1}{dt} + \frac{1}{C_1}\int i_1dt + \frac{1}{C_2}\int(i_1 - i_2)dt &= 0 \\ L_2\frac{di_2}{dt} + \frac{1}{C_2}\int(i_2 - i_1)dt &= e \end{aligned}$$

with $L_1 = M_1$, $L_2 = M_2$, $C_1 = 1/K_1$, $C_2 = 1/K_2$.

    L1 = M1                 L2 = M2
 +---LLLL----+-------------LLLL----+
 |           |                     |
C1=1/K1    C2=1/K2                (~) e
 |    i1     |         i2          |
 +-----------+---------------------+