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Two-mass system X2/F with F-V and F-I analogy (D3)

Short answer

X2(s)/F(s) = (D3 s + K2) / Δ, with Δ = [M1s^2+(D1+D3)s+K1+K2][M2s^2+(D2+D3)s+K2+K3] - (D3s+K2)^2. The F-V analogy maps force to voltage, mass to inductance, damper to resistance, spring to 1/capacitance; the F-I analogy maps force to current, mass to capacitance, damper to 1/resistance, spring to 1/inductance.

Find the transfer function X2(s)/F(s) for the mechanical system of figure below. Also draw the F-V and F-I analogy circuit of the system. [Figure: wall – spring K1 – mass M1 (force F(t) applied, displacement x1(t), friction D1 with ground); M1 connected to mass M2 (displacement x2(t), friction D2 with ground) through spring K2 in parallel with damper D3; M2 – spring K3 – wall]

Displacements x1x_1 (M1M_1, force FF, ground friction D1D_1) and x2x_2 (M2M_2, ground friction D2D_2). K1K_1 ties M1M_1 to the left wall, K3K_3 ties M2M_2 to the right wall, and K2∥D3K_2 \parallel D_3 couples them.

Equations of motion (Laplace)

[M1s2+(D1+D3)s+K1+K2]X1−(D3s+K2)X2=F−(D3s+K2)X1+[M2s2+(D2+D3)s+K2+K3]X2=0\begin{aligned} \left[M_1s^2 + (D_1 + D_3)s + K_1 + K_2\right]X_1 - (D_3s + K_2)X_2 &= F \\ -(D_3s + K_2)X_1 + \left[M_2s^2 + (D_2 + D_3)s + K_2 + K_3\right]X_2 &= 0 \end{aligned}

(Sum of impedances at each mass on the diagonal; coupling impedance off the diagonal.)

Transfer function

X2(s)F(s)=D3s+K2Δ\frac{X_2(s)}{F(s)} = \frac{D_3s + K_2}{\Delta} Δ=[M1s2+(D1+D3)s+K1+K2][M2s2+(D2+D3)s+K2+K3]−(D3s+K2)2\Delta = \left[M_1s^2 + (D_1+D_3)s + K_1 + K_2\right]\left[M_2s^2 + (D_2+D_3)s + K_2 + K_3\right] - (D_3s + K_2)^2

Expanded:

Δ=  M1M2s4+[D1M2+D2M1+D3(M1+M2)]s3+[D1D2+D1D3+D2D3+M1(K2+K3)+M2(K1+K2)]s2+[D1(K2+K3)+D2(K1+K2)+D3(K1+K3)]s+K1K2+K1K3+K2K3\begin{aligned} \Delta =\; & M_1M_2s^4 + \left[D_1M_2 + D_2M_1 + D_3(M_1 + M_2)\right]s^3 \\ & + \left[D_1D_2 + D_1D_3 + D_2D_3 + M_1(K_2 + K_3) + M_2(K_1 + K_2)\right]s^2 \\ & + \left[D_1(K_2 + K_3) + D_2(K_1 + K_2) + D_3(K_1 + K_3)\right]s \\ & + K_1K_2 + K_1K_3 + K_2K_3 \end{aligned}

F-V analogy (mesh)

F→eF\to e, M→LM\to L, D→RD\to R, K→1/CK\to 1/C, velocity →\to current.

  L1=M1 R=D1  C=1/K1        L2=M2 R=D2  C=1/K3
 +-LLLL-/\/\--||---+-------+-LLLL-/\/\--||---+
 |                 |                         |
(~) e           R3 = D3                      |
 |                 |                         |
 |    i1        C2 = 1/K2         i2         |
 +-----------------+-------------------------+

F-I analogy (node)

F→iF\to i, M→CM\to C, D→1/RD\to 1/R, K→1/LK\to 1/L, velocity →\to voltage.

 v1                   L2∥R3      v2
 +----+----+----+---+--LLLL--+---+----+----+
 |    |    |    |   +-/\/\/--+   |    |    |
(i)   C1   R1   L1               C2   R2   L3
 |    |    |    |                |    |    |
 +----+----+----+----------------+----+----+ gnd

(C1=M1C_1 = M_1, R1=1/D1R_1 = 1/D_1, L1=1/K1L_1 = 1/K_1; L2=1/K2L_2 = 1/K_2 in parallel with R3=1/D3R_3 = 1/D_3; C2=M2C_2 = M_2, R2=1/D2R_2 = 1/D_2, L3=1/K3L_3 = 1/K_3.) The mesh (F-V) and node (F-I) equations have exactly the same form as the two mechanical equations above.

✓ Answer checked by Claude Opus 5.5 (high effort)

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