---
title: "Two-mass system X1/F, X2/F with F-V analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:fv-tf-m1-m2-k1b1-k2b2"
source: https://www.bctengineer.com/q/semester_3/control_systems/fv-tf-m1-m2-k1b1-k2b2
site: BCT Engineering
---

# Two-mass system X1/F, X2/F with F-V analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2072 Chaitra, 8 marks
- Video: [Translational Mechanical Systems (Solved Example)](https://www.youtube.com/watch?v=Av86LcNuiMQ) — 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-tf-m1-m2-k1b1-k2b2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

The given mechanical system has force f(t) as input and x1 and x2 as displacement outputs. Draw equivalent F-V analogous circuit and determine the transfer functions X1(s)/F(s) and X2(s)/F(s). [Figure: force f(t) applied to mass M1 (displacement x1, friction D1 with ground); M1 connected to mass M2 (displacement x2, friction D2 with ground) through spring k1 in parallel with damper b1; M2 connected to the fixed wall through damper b2 in parallel with spring k2]

## Short answer

Laplace transform of equations yields X1/F = [M2s^2+(D2+b1+b2)s+k1+k2] / Δ, X2/F = (b1s+k1) / Δ, Δ = [M1s^2+(D1+b1)s+k1][M2s^2+(D2+b1+b2)s+k1+k2] - (b1s+k1)^2. F-V: masses→L, dampers→R, springs→1/C; coupling b1,k1 share common R–C branch.

## Answer

Displacements $x_1$ ($M_1$, ground friction $D_1$, force $f$) and $x_2$ ($M_2$, ground friction $D_2$). Coupling: $k_1 \parallel b_1$; $M_2$ to wall: $k_2 \parallel b_2$.

#### Differential equations
$$
\begin{aligned}
M_1\ddot x_1 + D_1\dot x_1 + b_1(\dot x_1 - \dot x_2) + k_1(x_1 - x_2) &= f(t) \\
M_2\ddot x_2 + D_2\dot x_2 + b_2\dot x_2 + k_2x_2 + b_1(\dot x_2 - \dot x_1) + k_1(x_2 - x_1) &= 0
\end{aligned}
$$
Laplace form:
$$
\begin{aligned}
\left[M_1s^2 + (D_1 + b_1)s + k_1\right]X_1 - (b_1s + k_1)X_2 &= F \\
-(b_1s + k_1)X_1 + \left[M_2s^2 + (D_2 + b_1 + b_2)s + k_1 + k_2\right]X_2 &= 0
\end{aligned}
$$

#### Transfer functions (Cramer's rule)
$$
\Delta = \left[M_1s^2 + (D_1+b_1)s + k_1\right]\left[M_2s^2 + (D_2+b_1+b_2)s + k_1 + k_2\right] - (b_1s + k_1)^2
$$
$$
\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + (D_2 + b_1 + b_2)s + k_1 + k_2}{\Delta}, \qquad
\frac{X_2(s)}{F(s)} = \frac{b_1s + k_1}{\Delta}
$$
Expanded:
$$
\begin{aligned}
\Delta =\; & M_1M_2s^4 + \left[M_1(D_2 + b_1 + b_2) + M_2(D_1 + b_1)\right]s^3 \\
& + \left[D_1D_2 + (D_1 + D_2)b_1 + D_1b_2 + b_1b_2 + M_1(k_1 + k_2) + M_2k_1\right]s^2 \\
& + \left[(D_1 + D_2)k_1 + D_1k_2 + b_1k_2 + b_2k_1\right]s + k_1k_2
\end{aligned}
$$

#### F-V analogous circuit
$f\to e$, $M\to L$, $D,b\to R$, $k\to 1/C$, $\dot x\to i$.

| Mechanical | Electrical |
|---|---|
| $M_1$, $M_2$ | $L_1$, $L_2$ |
| $D_1$, $D_2$ | $R_{D1}$, $R_{D2}$ |
| $b_1$ (coupling) | $R_{b1}$ in common branch |
| $k_1$ (coupling) | $C_1 = 1/k_1$ in common branch |
| $b_2$, $k_2$ | $R_{b2}$, $C_2 = 1/k_2$ in mesh 2 |

```text
   L1   R_D1             L2   R_D2   R_b2
 +-LLLL-/\/\-+---------+-LLLL-/\/\--/\/\-+
 |           |                           |
(~) e      R_b1                       C2=1/k2
 |           |                           |
 |   i1    C1=1/k1          i2           |
 +-----------+---------------------------+
```
$$
\begin{aligned}
L_1\frac{di_1}{dt} + R_{D1}i_1 + R_{b1}(i_1 - i_2) + \frac{1}{C_1}\int(i_1 - i_2)dt &= e \\
L_2\frac{di_2}{dt} + (R_{D2} + R_{b2})i_2 + \frac{1}{C_2}\int i_2dt + R_{b1}(i_2 - i_1) + \frac{1}{C_1}\int(i_2 - i_1)dt &= 0
\end{aligned}
$$

## Check yourself

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

1. The numerators of X1/F and X2/F are:
   A. M2 s^2 + (D2+b1+b2)s + k1 + k2, and b1 s + k1
   B. 1 and s
   C. b1 s + k1, and M2 s^2
   D. k1, and k2
   Answer: A. X1 uses the M2-side diagonal term; X2 uses the coupling. (Key point: Numerators)

2. Delta is:
   A. (b1 s + k1)^2
   B. Product of diagonal terms minus (b1 s + k1)^2
   C. Sum of diagonals
   D. Product of diagonals
   Answer: B. Determinant of the 2x2 matrix. (Key point: Common denominator)

3. k1 and b1 between masses appear as:
   A. Only in mesh 1
   B. Voltage sources
   C. Inductors
   D. Rb1 and C1 = 1/k1 in the common branch
   Answer: D. Coupling elements are shared between meshes. (Key point: F-V mapping)

## Close variants set in other exams

- [Two-mass system with numeric values: F-V analogy and X1/F](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-tf-m1m2-k1k2k3) — Control Systems, chapter 2, asked 3×
- [F-V and F-I analogy of two-mass system, X2 output](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-tf-m1-m2-k) — Control Systems, chapter 2, asked 3×

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