---
title: "Free body diagram and X1/F, X2/F of two-mass system"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:fbd-m1-m2-k1k2-b1b2"
source: https://www.bctengineer.com/q/semester_3/control_systems/fbd-m1-m2-k1k2-b1b2
site: BCT Engineering
---

# Free body diagram and X1/F, X2/F of two-mass system

Control Systems · Chapter 2: Component Modeling

- Asked in: 2070 Chaitra (old course), 10 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-m1-m2-k1k2-b1b2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Draw the free body diagram, write the differential equations and find the mentioned transfer functions X2(s)/F(s) and X1(s)/F(s) of the system below. [Figure: wall – spring k1 – mass M1 (displacement x1, friction B1 with ground) – spring k2 – mass M2 (displacement x2, friction B2 with ground); force F(t) applied to M2]

## Short answer

The differential equations are M1 x1'' + B1 x1' + k1 x1 + k2(x1 - x2) = 0 and M2 x2'' + B2 x2' + k2(x2 - x1) = F. The transfer functions are X2(s)/F(s) = (M1 s^2 + B1 s + k1 + k2) / Δ and X1(s)/F(s) = k2 / Δ, where Δ = (M1 s^2 + B1 s + k1 + k2)(M2 s^2 + B2 s + k2) - k2^2.

## Answer

Displacements $x_1$ ($M_1$) and $x_2$ ($M_2$), both positive to the right; $F(t)$ acts on $M_2$.

#### Free body diagrams
```text
            M1                          M2
 k1x1    ┌──────┐  k2(x2-x1)  k2(x2-x1) ┌──────┐
 ◄───────┤      ├──────►      ◄─────────┤      ├──► F(t)
 M1x1''  │      │                M2x2'' │      │
 ◄───────┤  M1  │                ◄──────┤  M2  │
 B1x1'   │      │                B2x2'  │      │
 ◄───────┤      │                ◄──────┤      │
         └──────┘                       └──────┘
```
- On $M_1$: inertia $M_1\ddot x_1$, friction $B_1\dot x_1$, spring $k_1x_1$ (all opposing motion); the spring $k_2$ pulls it to the right with $k_2(x_2 - x_1)$, i.e. a force $k_2(x_1 - x_2)$ opposing the motion.
- On $M_2$: applied $F$; opposing $M_2\ddot x_2$, $B_2\dot x_2$, $k_2(x_2 - x_1)$.

#### Differential equations
$$
\begin{aligned}
M_1\ddot x_1 + B_1\dot x_1 + k_1x_1 + k_2(x_1 - x_2) &= 0 \\
M_2\ddot x_2 + B_2\dot x_2 + k_2(x_2 - x_1) &= F(t)
\end{aligned}
$$

#### Laplace form
$$
\begin{aligned}
(M_1s^2 + B_1s + k_1 + k_2)X_1 - k_2X_2 &= 0 \\
-k_2X_1 + (M_2s^2 + B_2s + k_2)X_2 &= F
\end{aligned}
$$
Determinant:
$$
\Delta = (M_1s^2 + B_1s + k_1 + k_2)(M_2s^2 + B_2s + k_2) - k_2^2
$$
Expanded:
$$
\begin{aligned}
\Delta =\; & M_1M_2s^4 + (B_1M_2 + B_2M_1)s^3 + \left[B_1B_2 + M_1k_2 + M_2(k_1 + k_2)\right]s^2 \\
& + \left[B_1k_2 + B_2(k_1 + k_2)\right]s + k_1k_2
\end{aligned}
$$

#### Transfer functions (Cramer's rule)
$$
\frac{X_2(s)}{F(s)} = \frac{M_1s^2 + B_1s + k_1 + k_2}{\Delta}
$$
$$
\frac{X_1(s)}{F(s)} = \frac{k_2}{\Delta}
$$
Check: at steady state ($s\to 0$) with a constant force, $X_1/F = 1/k_1$ and $X_2/F = (k_1+k_2)/(k_1k_2) = 1/k_1 + 1/k_2$, which is correct for two springs in series.

## Check yourself

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

1. Where does F(t) act and what is the resulting equation for M2?
   A. On M1; M2 x2'' + B2 x2' + k2(x2 - x1) = 0
   B. On M2; M2 x2'' + B2 x2' + k2(x2 - x1) = F(t)
   C. On M2; M2 x2'' + B2 x2' + k2 x2 = F(t)
   D. On both masses equally
   Answer: B. F acts on M2, and the spring k2 force depends on the relative displacement x2 - x1. (Key point: Force on M2)

2. Which equation describes M1?
   A. M1 x1'' + B1 x1' + k1 x1 = F
   B. M1 x1'' + B1 x1' + k1 x1 + k2(x1 - x2) = 0
   C. M1 x1'' + k1 x1 = 0
   D. M1 x1'' + B1 x1' + k2(x1 - x2) = 0
   Answer: B. M1 feels spring k1 to the wall and k2 to M2, and has no applied force. (Key point: Equation for M1)

3. What is the numerator of X2(s)/F(s)?
   A. k2
   B. M1 s^2 + B1 s + k1 + k2
   C. M2 s^2 + B2 s + k2
   D. k1 k2
   Answer: B. By Cramer's rule the numerator is the diagonal term of the equation that does not contain F, namely M1's term. (Key point: X2/F)

4. Why is the numerator of X1(s)/F(s) just k2?
   A. F acts on M1
   B. M1 is driven only through spring k2 from M2
   C. k1 is zero
   D. Friction is neglected
   Answer: B. F acts on M2, and M1 is reached only through the coupling spring k2. (Key point: X1/F)

5. At s -> 0 with constant force, what is X2/F?
   A. 1/k1
   B. 1/k2
   C. 1/k1 + 1/k2
   D. k1 + k2
   Answer: C. Two springs in series add their compliances, so X2/F = 1/k1 + 1/k2. (Key point: Steady-state check)

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