Mass inside a mass: transfer function and F-I analogy

Let $x_1$ = displacement of the outer frame $M_1$ and $x_2$ = displacement of the inner mass $M_2$. Assumption: no friction (none is shown) between $M_1$ and ground or between $M_2$ and $M_1$.

Free body equations

• $M_1$: force $f$ acts; spring $k_2$ to the wall resists by $k_2x_1$; spring $k_1$ pulls by $k_1(x_1 - x_2)$.
• $M_2$: only spring $k_1$ acts, $k_1(x_2 - x_1)$.

$$\begin{aligned} M_1\ddot x_1 + k_1(x_1 - x_2) + k_2x_1 &= f \\ M_2\ddot x_2 + k_1(x_2 - x_1) &= 0 \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + k_1 + k_2)X_1 - k_1X_2 &= F \\ -k_1X_1 + (M_2s^2 + k_1)X_2 &= 0 \end{aligned}$$

From the second equation, $X_1 = \dfrac{M_2s^2 + k_1}{k_1}X_2$. Substituting:

$$\left[(M_1s^2 + k_1 + k_2)(M_2s^2 + k_1) - k_1^2\right]X_2 = k_1F$$

Transfer function

$$\frac{X_2(s)}{F(s)} = \frac{k_1}{M_1M_2s^4 + \left[M_1k_1 + M_2(k_1 + k_2)\right]s^2 + k_1k_2}$$

With no damping the poles lie on the $j\omega$ axis (undamped oscillation at two natural frequencies).

F-I analogous circuit

$f\to i$, $M\to C$, $k\to 1/L$, velocity $\to$ node voltage. Note that $M_2$ is referred to the ground reference (absolute velocity), so its capacitor goes to ground.

 v1           L1    v2
 +----+----+--LLLL--+
 |    |    |        |
(i)   C1   L2       C2
 |    |    |        |
 +----+----+--------+ gnd

($C_1 = M_1$, $L_2 = 1/k_2$, $L_1 = 1/k_1$, $C_2 = M_2$.)

$$\begin{aligned} C_1\frac{dv_1}{dt} + \frac{1}{L_2}\int v_1dt + \frac{1}{L_1}\int(v_1 - v_2)dt &= i \\ C_2\frac{dv_2}{dt} + \frac{1}{L_1}\int(v_2 - v_1)dt &= 0 \end{aligned}$$