---
title: "Two-mass system X2/F with F-V analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-m1-m2-k1k2-b1b2b3"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-m1-m2-k1k2-b1b2b3
site: BCT Engineering
---

# Two-mass system X2/F with F-V analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2075 Asoj, 6+2 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-m1-m2-k1k2-b1b2b3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

For the mechanical system shown below find the transfer function X2(s)/F(s). Draw the force voltage analogy. [Figure: wall – spring k1 – mass M2 (displacement x2, friction B2 with ground); M2 connected to mass M1 (displacement x1) through damper B1 in parallel with spring k2; force F applied to M1; M1 has friction B3 with ground]

## Short answer

X2(s)/F(s) = (B1 s + k2) / Δ, with Δ = [M1 s^2+(B1+B3)s+k2][M2 s^2+(B1+B2)s+k1+k2] - (B1 s+k2)^2. The F-V analogy: force→voltage, velocity→current, masses→inductors, dampers→resistors, springs→capacitors; B1 and k2 become series R and C in the shared mesh branch.

## Answer

Displacements: $x_1$ for $M_1$ (force $F$ applied, ground friction $B_3$) and $x_2$ for $M_2$ (ground friction $B_2$, spring $k_1$ to wall). $B_1 \parallel k_2$ couples the masses.

#### Equations of motion
$$
\begin{aligned}
M_1\ddot x_1 + B_3\dot x_1 + B_1(\dot x_1 - \dot x_2) + k_2(x_1 - x_2) &= F \\
M_2\ddot x_2 + B_2\dot x_2 + k_1x_2 + B_1(\dot x_2 - \dot x_1) + k_2(x_2 - x_1) &= 0
\end{aligned}
$$
Laplace form:
$$
\begin{aligned}
\left[M_1s^2 + (B_1 + B_3)s + k_2\right]X_1 - (B_1s + k_2)X_2 &= F \\
-(B_1s + k_2)X_1 + \left[M_2s^2 + (B_1 + B_2)s + k_1 + k_2\right]X_2 &= 0
\end{aligned}
$$

#### Transfer function
$$
\frac{X_2(s)}{F(s)} = \frac{B_1s + k_2}{\Delta}
$$
$$
\Delta = \left[M_1s^2 + (B_1+B_3)s + k_2\right]\left[M_2s^2 + (B_1+B_2)s + k_1 + k_2\right] - (B_1s + k_2)^2
$$
Expanded:
$$
\begin{aligned}
\Delta =\; & M_1M_2s^4 + \left[B_1(M_1 + M_2) + B_2M_1 + B_3M_2\right]s^3 \\
& + \left[B_1B_2 + B_1B_3 + B_2B_3 + M_1(k_1 + k_2) + M_2k_2\right]s^2 \\
& + \left[B_1k_1 + B_2k_2 + B_3(k_1 + k_2)\right]s + k_1k_2
\end{aligned}
$$

#### F-V analogy
$F\to e$, $M\to L$, $B\to R$, $k\to 1/C$, $\dot x \to i$. Mesh 1 ($i_1$) for $M_1$, mesh 2 ($i_2$) for $M_2$; the coupling $B_1 \parallel k_2$ becomes the common branch $R_1$ in series with $C_2$ (parallel mechanical elements carry the same velocity difference, so they become series elements in the shared branch).
```text
    L1=M1  R3=B3            L2=M2   R2=B2
 +--LLLL--/\/\/--+----------LLLL---/\/\/--+
 |               |                        |
(~) e         R1 = B1                  C1 = 1/k1
 |               |                        |
 |    i1      C2 = 1/k2        i2         |
 +---------------+------------------------+
```
$$
\begin{aligned}
L_1\frac{di_1}{dt} + R_3i_1 + R_1(i_1 - i_2) + \frac{1}{C_2}\int(i_1 - i_2)dt &= e \\
L_2\frac{di_2}{dt} + R_2i_2 + \frac{1}{C_1}\int i_2dt + R_1(i_2 - i_1) + \frac{1}{C_2}\int(i_2 - i_1)dt &= 0
\end{aligned}
$$

## Check yourself

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

1. Which term appears in M2's equation but not M1's?
   A. k1 x2 (spring to wall)
   B. k2 only
   C. B3 x1'
   D. F
   Answer: A. M2 is attached to the wall by k1; M1 has the force F and ground friction B3. (Key point: Equations of motion)

2. The numerator of X2/F is:
   A. B1 s + k2 (the coupling)
   B. k1 k2
   C. M1 s^2
   D. B2 s
   Answer: A. Force passes through M1 then the coupling B1 || k2 to reach M2. (Key point: Transfer function numerator)

3. Parallel B1 and k2 between masses become, in the F-V circuit:
   A. Two inductors
   B. R1 parallel with C2
   C. Both in mesh 1 only
   D. R1 in series with C2 in the common branch
   Answer: D. Parallel mechanical elements share the velocity difference, so they become series elements in the shared branch. (Key point: Parallel coupling in F-V)

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

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