---
title: "F-V and F-I analogy of mass-spring-damper system"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:fv-fi-m-k1-k2-b2"
source: https://www.bctengineer.com/q/semester_3/control_systems/fv-fi-m-k1-k2-b2
site: BCT Engineering
---

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

Control Systems · Chapter 2: Component Modeling

- Asked in: 2080 Bhadra, 4 marks
- Video: [Analogous Systems](https://www.youtube.com/watch?v=RUwQkXroaig) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/fv-fi-m-k1-k2-b2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

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]

## 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.

## Answer

Take $x_1$ (mass) and $x_2$ (junction of $K_1$, $K_2$, $B_2$) as the two displacements. The junction has no mass.

#### Equations of motion
$$
\begin{aligned}
M\ddot{x}_1 + K_1(x_1 - x_2) &= f \\
K_1(x_2 - x_1) + B_2\dot{x}_2 + K_2 x_2 &= 0
\end{aligned}
$$

#### F-V analogy (mesh/loop)
$f \to e$, $M \to L$, $B \to R$, $K \to 1/C$, $\dot x \to i$. Each displacement gives one mesh; $K_1$ is common to both.
$$
\begin{aligned}
L\frac{di_1}{dt} + \frac{1}{C_1}\int (i_1 - i_2)\,dt &= e \\
\frac{1}{C_1}\int (i_2 - i_1)\,dt + R_2 i_2 + \frac{1}{C_2}\int i_2\,dt &= 0
\end{aligned}
$$
with $L = M$, $C_1 = 1/K_1$, $R_2 = B_2$, $C_2 = 1/K_2$.
```text
      L = M               R2 = B2
  +---LLLL----+---------/\/\/----+
  |           |                  |
 (~) e      C1=1/K1           C2=1/K2
  |    i1     |       i2         |
  +-----------+------------------+
```

#### F-I analogy (node)
$f \to i$, $M \to C$, $B \to 1/R$, $K \to 1/L$, $\dot x \to v$. Each displacement gives one node; $K_1$ joins the nodes.
$$
\begin{aligned}
C\frac{dv_1}{dt} + \frac{1}{L_1}\int (v_1 - v_2)\,dt &= i \\
\frac{1}{L_1}\int (v_2 - v_1)\,dt + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2\,dt &= 0
\end{aligned}
$$
with $C = M$, $L_1 = 1/K_1$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.
```text
    v1          L1=1/K1         v2
  +---+--------LLLL-------+--------+
  |   |                   |        |
 (i) C=M              R2=1/B2  L2=1/K2
  |   |                   |        |
  +---+-------------------+--------+  ground
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the second equation of motion equal to zero on the right side?
   A. No external force acts on the junction, which has no mass
   B. The damper B2 cancels the force
   C. Spring K1 is ideal
   D. The wall absorbs all force
   Answer: A. The junction of K1, K2, B2 is massless and has no applied force, so the forces on it sum to zero. (Key point: Massless junction equation)

2. In the F-V circuit, how does K1 appear?
   A. As a current source
   B. As a resistor in mesh 2
   C. Only in mesh 1 as an inductor
   D. As C1 = 1/K1 in the branch common to both meshes
   Answer: D. Each displacement gives one mesh, and the spring between them is the shared branch C1 = 1/K1. (Key point: F-V circuit structure)

3. In the F-I analogy, which mapping is correct?
   A. M -> C, B -> 1/R (conductance), K -> 1/L, velocity -> node voltage
   B. M -> C, B -> R, K -> L
   C. M -> R, B -> 1/C, K -> 1/L
   D. M -> L, B -> R, K -> C
   Answer: A. Force-current: force is current, mass is capacitance, damper is conductance 1/R, spring is 1/L, and each displacement is a node. (Key point: F-I element mapping)

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