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

Short answer

Force acts on M1 directly. Coupled equations give X2(s)/F(s) = B1/Δ', with Δ' = M1M2s^3+[ B1(M1+M2)+B2M1]s^2+(B1B2+K2M1)s+B1K2. F-V circuit: three loops; K1 as capacitor between loop0 and loop1, B1 as resistor between loop1 and loop2. F-I analogue: current source, nodes; K1→L=1/K1, B1→R=1/B1.

Find transfer function and develop F-V and F-I analogous circuits of the following figure. [Figure: mass M2 is a hollow frame resting on the ground (viscous friction B2 between M2 and ground), connected to the left wall through spring K2; mass M1 is a plate sliding inside M2 with viscous friction B1 between M1 and M2; force F(t) pulls M1 through spring K1]

Displacements: x1x_1 (plate M1M_1), x2x_2 (frame M2M_2). The force FF acts on the free end of spring K1K_1; since that end has no mass, the spring transmits the full force, K1(x0−x1)=FK_1(x_0 - x_1) = F, so FF effectively acts on M1M_1. (Here x0x_0 is the free-end displacement.)

Equations of motion

  • M1M_1: pulled by FF; viscous friction B1B_1 relative to M2M_2.
  • M2M_2: friction B1B_1 from M1M_1, ground friction B2B_2, spring K2K_2 to the wall.
M1x¨1+B1(x˙1−x˙2)=FM2x¨2+B1(x˙2−x˙1)+B2x˙2+K2x2=0\begin{aligned} M_1\ddot x_1 + B_1(\dot x_1 - \dot x_2) &= F \\ M_2\ddot x_2 + B_1(\dot x_2 - \dot x_1) + B_2\dot x_2 + K_2x_2 &= 0 \end{aligned}

Laplace form:

(M1s2+B1s)X1−B1sX2=F−B1sX1+[M2s2+(B1+B2)s+K2]X2=0\begin{aligned} (M_1s^2 + B_1s)X_1 - B_1sX_2 &= F \\ -B_1sX_1 + \left[M_2s^2 + (B_1 + B_2)s + K_2\right]X_2 &= 0 \end{aligned}

Transfer functions

Determinant (after cancelling a common ss):

Δ′=(M1s+B1)[M2s2+(B1+B2)s+K2]−B12s=M1M2s3+[B1(M1+M2)+B2M1]s2+(B1B2+K2M1)s+B1K2\Delta' = (M_1s + B_1)\left[M_2s^2 + (B_1 + B_2)s + K_2\right] - B_1^2s = M_1M_2s^3 + \left[B_1(M_1 + M_2) + B_2M_1\right]s^2 + (B_1B_2 + K_2M_1)s + B_1K_2 X2(s)F(s)=B1Δ′\frac{X_2(s)}{F(s)} = \frac{B_1}{\Delta'} X1(s)F(s)=M2s2+(B1+B2)s+K2s Δ′\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + (B_1 + B_2)s + K_2}{s\,\Delta'}

and for the point where the force is applied, X0(s)=X1(s)+F(s)/K1X_0(s) = X_1(s) + F(s)/K_1.

F-V analogous circuit

F→eF\to e, M→LM\to L, B→RB\to R, K→1/CK\to 1/C. Mesh i0i_0 = free-end velocity, i1i_1 = velocity of M1M_1, i2i_2 = velocity of M2M_2. K1K_1 is shared by meshes 0 and 1; B1B_1 (relative friction) by meshes 1 and 2.

              L1=M1             L2=M2  R2=B2  C2=1/K2
 +------+-----LLLL----+---------LLLL---/\/\----||--+
 |      |             |                            |
(~) e  C=1/K1       R1=B1                          |
 |  i0  |      i1     |            i2              |
 +------+-------------+----------------------------+

F-I analogous circuit

F→iF\to i, M→CM\to C, B→1/RB\to 1/R, K→1/LK\to 1/L. Node 0 = free end, node 1 = M1M_1, node 2 = M2M_2; C1=M1C_1 = M_1, C2=M2C_2 = M_2, R2=1/B2R_2 = 1/B_2, L2=1/K2L_2 = 1/K_2.

  v0    L=1/K1    v1   R1=1/B1    v2
 +---+---LLLL---+---+---/\/\/---+---+---+---+
 |                  |               |   |   |
(i)                C1              C2  R2  L2
 |                  |               |   |   |
 +------------------+---------------+---+---+ gnd

✓ Answer checked by Claude Opus 5.5 (high effort)

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