Single-mass system transfer function with F-V and F-I

Displacements: $x_1$ (free end where $f$ acts) and $x_2$ (mass $M_2$). The free end has no mass.

Equations of motion

At the massless end, the spring force equals the applied force:

$$K_1(x_1 - x_2) = f(t)$$

For $M_2$:

$$M_2\ddot x_2 + B_2\dot x_2 + K_2x_2 + K_1(x_2 - x_1) = 0$$

Transfer functions

Laplace form:

$$\begin{aligned} K_1X_1 - K_1X_2 &= F \\ -K_1X_1 + (M_2s^2 + B_2s + K_1 + K_2)X_2 &= 0 \end{aligned}$$

Adding the two equations: $(M_2s^2 + B_2s + K_2)X_2 = F$, so

$$\frac{X_2(s)}{F(s)} = \frac{1}{M_2s^2 + B_2s + K_2}$$

and since $X_1 = X_2 + F/K_1$:

$$\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + B_2s + K_1 + K_2}{K_1(M_2s^2 + B_2s + K_2)}$$

The spring $K_1$ only passes the force on; it does not change the dynamics of $M_2$.

F-V analogy

$f\to e$, $M\to L$, $B\to R$, $K\to 1/C$. Mesh 1 (velocity $\dot x_1$) and mesh 2 ($\dot x_2$) share $C_1 = 1/K_1$.

                   L2=M2   R2=B2   C2=1/K2
 +--------+--------LLLL----/\/\-----||---+
 |        |                              |
(~) e   C1=1/K1                          |
 |   i1   |              i2              |
 +--------+------------------------------+

$$\frac{1}{C_1}\int(i_1 - i_2)dt = e, \qquad L_2\frac{di_2}{dt} + R_2i_2 + \frac{1}{C_2}\int i_2dt + \frac{1}{C_1}\int(i_2 - i_1)dt = 0$$

F-I analogy

$f\to i$, $M\to C$, $B\to 1/R$, $K\to 1/L$.

 v1  L1   v2
 +--LLLL--+----+----+
 |        |    |    |
(i)       C2   R2   L2
 |        |    |    |
 +--------+----+----+ gnd

($L_1 = 1/K_1$, $C_2 = M_2$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.)

$$\frac{1}{L_1}\int(v_1 - v_2)dt = i, \qquad C_2\frac{dv_2}{dt} + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2dt + \frac{1}{L_1}\int(v_2 - v_1)dt = 0$$