---
title: "Free body diagram and X1/F1 of two-mass system"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:fbd-tf-m1-m2-k12-b12"
source: https://www.bctengineer.com/q/semester_3/control_systems/fbd-tf-m1-m2-k12-b12
site: BCT Engineering
---

# Free body diagram and X1/F1 of two-mass system

Control Systems · Chapter 2: Component Modeling

- Asked in: 2066 Bhadra (old course), 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/fbd-tf-m1-m2-k12-b12
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

For the mechanical system shown in fig. 1(b), draw free body diagram, write complete differential equations and identify the transfer function X1(s)/F1(s). [Figure: wall – spring K2 – mass M2 (displacement x2(t), friction B2 with ground); M2 connected to mass M1 (displacement x1(t), friction B1 with ground) through spring K12 in parallel with damper B12; force f1(t) applied to M1]

## Short answer

Differential equations: M1x1''+B1x1'+B12(x1'-x2')+K12(x1-x2)=f1, M2x2''+B2x2'+K2x2+B12(x2'-x1')+K12(x2-x1)=0. Transfer function X1(s)/F1(s)=[M2s²+(B2+B12)s+K12+K2] / { [M1s²+(B1+B12)s+K12][M2s²+(B2+B12)s+K12+K2] - (B12s+K12)² }.

## Answer

Displacements $x_1$ ($M_1$, friction $B_1$, force $f_1$) and $x_2$ ($M_2$, friction $B_2$, spring $K_2$ to the wall). $K_{12} \parallel B_{12}$ couples them.

#### Free body diagrams
```text
 Opposing forces (to the left)        Applied
 M1x1'', B1x1'
 K12(x1-x2), B12(x1'-x2')   ┌────┐
        ◄───────────────────┤ M1 ├────► f1(t)
                            └────┘
 M2x2'', B2x2', K2x2
 K12(x2-x1), B12(x2'-x1')   ┌────┐
        ◄───────────────────┤ M2 │     (none)
                            └────┘
```
(For each mass, the applied force acts in the positive direction and all reaction forces oppose it.)

#### Differential equations
$$
\begin{aligned}
M_1\ddot x_1 + B_1\dot x_1 + B_{12}(\dot x_1 - \dot x_2) + K_{12}(x_1 - x_2) &= f_1(t) \\
M_2\ddot x_2 + B_2\dot x_2 + K_2x_2 + B_{12}(\dot x_2 - \dot x_1) + K_{12}(x_2 - x_1) &= 0
\end{aligned}
$$

#### Laplace form
$$
\begin{aligned}
\left[M_1s^2 + (B_1 + B_{12})s + K_{12}\right]X_1 - (B_{12}s + K_{12})X_2 &= F_1 \\
-(B_{12}s + K_{12})X_1 + \left[M_2s^2 + (B_2 + B_{12})s + K_{12} + K_2\right]X_2 &= 0
\end{aligned}
$$

#### Transfer function
By Cramer's rule:
$$
\frac{X_1(s)}{F_1(s)} = \frac{M_2s^2 + (B_2 + B_{12})s + K_{12} + K_2}{\Delta}
$$
$$
\Delta = \left[M_1s^2 + (B_1+B_{12})s + K_{12}\right]\left[M_2s^2 + (B_2+B_{12})s + K_{12} + K_2\right] - (B_{12}s + K_{12})^2
$$
Expanded:
$$
\begin{aligned}
\Delta =\; & M_1M_2s^4 + \left[B_1M_2 + B_2M_1 + B_{12}(M_1 + M_2)\right]s^3 \\
& + \left[B_1B_2 + B_1B_{12} + B_2B_{12} + K_{12}(M_1 + M_2) + K_2M_1\right]s^2 \\
& + \left[B_1(K_{12} + K_2) + B_2K_{12} + B_{12}K_2\right]s + K_{12}K_2
\end{aligned}
$$
Check: at $s = 0$, $X_1/F_1 = (K_{12} + K_2)/(K_{12}K_2) = 1/K_{12} + 1/K_2$, the compliance of the two springs in series.

## Check yourself

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

1. What forces oppose motion of M1 in the free body diagram?
   A. M1 x1'', B1 x1', K12(x1 - x2) and B12(x1' - x2')
   B. Only K12 x1
   C. M1 x1'' and K2 x1
   D. B2 x2' and K2 x2
   Answer: A. M1 has inertia, ground friction B1 and the coupling spring and damper forces; f1 is the applied force. (Key point: Free body diagram forces)

2. Which equation describes M2?
   A. M2 x2'' + B2 x2' + K2 x2 + B12(x2' - x1') + K12(x2 - x1) = 0
   B. M2 x2'' + B2 x2' + K2 x2 = f1
   C. M2 x2'' + K12(x2 - x1) = 0
   D. M2 x2'' + B12 x2' + K2 x2 = f1
   Answer: A. M2 has no applied force, and is tied to the wall by K2 and to M1 through K12 and B12. (Key point: Equation for M2)

3. What is the numerator of X1(s)/F1(s)?
   A. B12 s + K12
   B. M2 s^2 + (B2 + B12) s + K12 + K2
   C. M1 s^2 + (B1 + B12) s + K12
   D. K12 K2
   Answer: B. F1 acts on M1, so by Cramer's rule the numerator is M2's diagonal term. (Key point: Numerator)

4. At s = 0, X1/F1 equals
   A. 1/K12
   B. 1/K2
   C. 1/K12 + 1/K2
   D. K12 + K2
   Answer: C. The two springs are in series for a static force, so compliances add. (Key point: Steady-state check)

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