---
title: "Three-mass system X2/F with F-V and F-I analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-fi-three-mass"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-three-mass
site: BCT Engineering
---

# Three-mass system X2/F with F-V and F-I analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2074 Asoj, 6 marks
- Video: [2.Force Current and Force Voltage Analogy solved problem with system equation](https://www.youtube.com/watch?v=jgCfYudsIFY) — ELECTRICAL HUB
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-three-mass
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

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 (displacement X1); M1 connected to mass M2 (displacement X2) through damper D1 in parallel with spring K2; M2 connected to mass M3 (displacement X3) through spring K3; force F(t) applied to M3; ground friction Fv1 = Fv2 = Fv3 = 0]

## Short answer

Write equations for three masses, Laplace, solve with Cramer's rule: X2/F = K3 (M1 s^2 + D1 s + K1+K2) / Δ, with Δ the 6th-order determinant from the impedance matrix. F-V: force→voltage, masses→L, dampers→R, springs→1/C. F-I: force→current, masses→C, dampers→1/R, springs→1/L.

## Answer

Displacements $X_1, X_2, X_3$ of $M_1, M_2, M_3$; no ground friction ($f_{v1} = f_{v2} = f_{v3} = 0$).

#### Equations of motion (Laplace)
$$
\begin{aligned}
(M_1s^2 + D_1s + K_1 + K_2)X_1 - (D_1s + K_2)X_2 &= 0 \\
-(D_1s + K_2)X_1 + (M_2s^2 + D_1s + K_2 + K_3)X_2 - K_3X_3 &= 0 \\
-K_3X_2 + (M_3s^2 + K_3)X_3 &= F
\end{aligned}
$$

#### Transfer function by Cramer's rule
$$
\frac{X_2(s)}{F(s)} = \frac{K_3(M_1s^2 + D_1s + K_1 + K_2)}{\Delta}
$$
$$
\begin{aligned}
\Delta =\; & (M_1s^2 + D_1s + K_1 + K_2)\left[(M_2s^2 + D_1s + K_2 + K_3)(M_3s^2 + K_3) - K_3^2\right] \\
& - (D_1s + K_2)^2(M_3s^2 + K_3)
\end{aligned}
$$
($\Delta$ is a 6th-order polynomial; its constant term is $K_1K_2K_3$.)

#### F-V analogy (mesh)
$F\to e$, $M\to L$, $D\to R$, $K\to 1/C$, velocity $\to$ mesh current. Elements between two masses are shared by their meshes.
```text
  L1=M1       L2=M2          L3=M3
 +-LLLL-+-----LLLL----+------LLLL----+
 |      |             |              |
C1     R1=D1        C3=1/K3         (~) e
=1/K1   |             |              |
 |     C2=1/K2        |              |
 | i1   |      i2     |      i3      |
 +------+-------------+--------------+
```
Mesh 1: $L_1, C_1$ plus shared ($R_1 + C_2$); mesh 2: $L_2$ plus shared branches; mesh 3: $L_3$, source $e$, shared $C_3$.

#### F-I analogy (node)
$F\to i$, $M\to C$, $D\to 1/R$, $K\to 1/L$, velocity $\to$ node voltage.
```text
 v1         L2∥R1      v2  L3   v3
 +----+---+--LLLL--+---+--LLLL--+----+
 |    |   +-/\/\/--+   |        |    |
 C1   L1               C2       C3  (i)
 |    |                |        |    |
 +----+----------------+--------+----+ gnd
```
($C_1 = M_1$, $C_2 = M_2$, $C_3 = M_3$, $L_1 = 1/K_1$, $L_3 = 1/K_3$.) $L_2 = 1/K_2$ and $R_1 = 1/D_1$ are in parallel between nodes 1 and 2.

## Check yourself

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

1. Where is F applied and what appears on the right of the M3 equation?
   A. F on M2
   B. F on M3, so right side is F
   C. F on M1
   D. No force
   Answer: B. Force acts on M3, so the third equation equals F; the others equal zero. (Key point: Equations structure)

2. The numerator of X2/F is:
   A. M1 s^2 K3
   B. D1 s + K2
   C. K3
   D. K3 (M1 s^2 + D1 s + K1 + K2)
   Answer: D. Cramer's rule cofactor path from M3 to M2 gives K3 times the M1 block. (Key point: Numerator)

3. The order of the denominator is:
   A. 3
   B. 4
   C. 5
   D. 6
   Answer: D. Three masses with second-order terms each give a 6th-order polynomial. (Key point: Order)

4. In the F-I circuit, D1 and K2 between M1 and M2 become:
   A. R1 = 1/D1 in parallel with L2 = 1/K2 between nodes 1 and 2
   B. Both to ground
   C. Series between nodes
   D. Capacitors
   Answer: A. Parallel mechanical elements remain parallel between the two nodes in the F-I analogy. (Key point: F-I coupling)

## Close variants set in other exams

- [Two-mass system X2/F with F-V and F-I analogy (D3)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-fi-m1-m2-2) — Control Systems, chapter 2, asked 3×
- [Two-mass system X2/F with F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-k1k2-b1b2b3) — Control Systems, chapter 2, asked 3×

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