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F-V and F-I analogy of mass-spring-damper system

Short answer

F-V analogy: force → voltage, mass → inductor, damper → resistor, spring → capacitor; mesh circuit has L=M with C1=1/K1, R2=B2, C2=1/K2. F-I analogy: force → current, mass → capacitor, damper → conductance (1/R), spring → inductor; node circuit has C=M, L1=1/K1, R2=1/B2, L2=1/K2.

Develop F-V and F-I analogy circuit of the mechanical system shown below. [Figure: force f applied to mass M (displacement X1, on rollers on ground); M connected through spring K1 to a point of displacement X2; from that point, damper B2 in parallel with spring K2 connect to the fixed wall]

Take x1x_1 (mass) and x2x_2 (junction of K1K_1, K2K_2, B2B_2) as the two displacements. The junction has no mass.

Equations of motion

Mx¨1+K1(x1−x2)=fK1(x2−x1)+B2x˙2+K2x2=0\begin{aligned} M\ddot{x}_1 + K_1(x_1 - x_2) &= f \\ K_1(x_2 - x_1) + B_2\dot{x}_2 + K_2 x_2 &= 0 \end{aligned}

F-V analogy (mesh/loop)

f→ef \to e, M→LM \to L, B→RB \to R, K→1/CK \to 1/C, x˙→i\dot x \to i. Each displacement gives one mesh; K1K_1 is common to both.

Ldi1dt+1C1∫(i1−i2) dt=e1C1∫(i2−i1) dt+R2i2+1C2∫i2 dt=0\begin{aligned} L\frac{di_1}{dt} + \frac{1}{C_1}\int (i_1 - i_2)\,dt &= e \\ \frac{1}{C_1}\int (i_2 - i_1)\,dt + R_2 i_2 + \frac{1}{C_2}\int i_2\,dt &= 0 \end{aligned}

with L=ML = M, C1=1/K1C_1 = 1/K_1, R2=B2R_2 = B_2, C2=1/K2C_2 = 1/K_2.

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

F-I analogy (node)

f→if \to i, M→CM \to C, B→1/RB \to 1/R, K→1/LK \to 1/L, x˙→v\dot x \to v. Each displacement gives one node; K1K_1 joins the nodes.

Cdv1dt+1L1∫(v1−v2) dt=i1L1∫(v2−v1) dt+v2R2+1L2∫v2 dt=0\begin{aligned} C\frac{dv_1}{dt} + \frac{1}{L_1}\int (v_1 - v_2)\,dt &= i \\ \frac{1}{L_1}\int (v_2 - v_1)\,dt + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2\,dt &= 0 \end{aligned}

with C=MC = M, L1=1/K1L_1 = 1/K_1, R2=1/B2R_2 = 1/B_2, L2=1/K2L_2 = 1/K_2.

    v1          L1=1/K1         v2
  +---+--------LLLL-------+--------+
  |   |                   |        |
 (i) C=M              R2=1/B2  L2=1/K2
  |   |                   |        |
  +---+-------------------+--------+  ground

✓ Answer checked by Claude Opus 5.5 (high effort)

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