---
title: "Vertical two-mass system transfer function X1/F"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-vertical-m1-m2-k1-b1"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-vertical-m1-m2-k1-b1
site: BCT Engineering
---

# Vertical two-mass system transfer function X1/F

Control Systems · Chapter 2: Component Modeling

- Asked in: 2065 Shrawan (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/tf-vertical-m1-m2-k1-b1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Derive the transfer function X1(s)/F(s) for the system shown below. [Figure: vertical system; mass M1 (displacement x1(t)) hangs from a fixed support through spring K1; mass M2 (displacement x2(t)) hangs below M1 through damper B1; damper B2 connects M2 to the fixed support above; force F(t) acts downward on M2]

## Short answer

Static equilibrium offset removed. Equations: M1x1''+K1x1+B1(x1'-x2')=0, M2x2''+B2x2'+B1(x2'-x1')=F. Transfer function X1(s)/F(s)=B1 / [M1M2s³+(B1(M1+M2)+B2M1)s²+(B1B2+K1M2)s+K1(B1+B2)].

## Answer

Measure $x_1$, $x_2$ downward from the static equilibrium position, so gravity is balanced by the initial spring stretch and drops out of the equations.

#### Forces
- $M_1$: spring $K_1$ (to support) gives $K_1x_1$; damper $B_1$ (to $M_2$) gives $B_1(\dot x_1 - \dot x_2)$; inertia $M_1\ddot x_1$.
- $M_2$: applied $F(t)$ downward; damper $B_2$ (to support) gives $B_2\dot x_2$; $B_1$ gives $B_1(\dot x_2 - \dot x_1)$; inertia $M_2\ddot x_2$.

#### Differential equations
$$
\begin{aligned}
M_1\ddot x_1 + K_1x_1 + B_1(\dot x_1 - \dot x_2) &= 0 \\
M_2\ddot x_2 + B_2\dot x_2 + B_1(\dot x_2 - \dot x_1) &= F(t)
\end{aligned}
$$

#### Laplace form
$$
\begin{aligned}
(M_1s^2 + B_1s + K_1)X_1 - B_1sX_2 &= 0 \\
-B_1sX_1 + \left[M_2s^2 + (B_1 + B_2)s\right]X_2 &= F
\end{aligned}
$$

#### Solving
From the first equation, $X_2 = \dfrac{M_1s^2 + B_1s + K_1}{B_1s}X_1$. Substituting:
$$
\left[\frac{(M_1s^2 + B_1s + K_1)\,s(M_2s + B_1 + B_2)}{B_1s} - B_1s\right]X_1 = F
$$
$$
\frac{X_1(s)}{F(s)} = \frac{B_1s}{s(M_1s^2 + B_1s + K_1)(M_2s + B_1 + B_2) - B_1^2s^2}
$$
Cancelling $s$:
$$
\frac{X_1(s)}{F(s)} = \frac{B_1}{(M_1s^2 + B_1s + K_1)(M_2s + B_1 + B_2) - B_1^2s}
$$

#### Final form
$$
\frac{X_1(s)}{F(s)} = \frac{B_1}{M_1M_2s^3 + \left[B_1(M_1 + M_2) + B_2M_1\right]s^2 + (B_1B_2 + K_1M_2)s + K_1(B_1 + B_2)}
$$
Check: under a constant force, $M_2$ finally moves at a steady velocity $v_2 = F/(B_1 + B_2)$ (it is held only by dampers), and $M_1$ settles where $K_1x_1 = B_1v_2$, i.e. $x_1 = \dfrac{B_1F}{K_1(B_1 + B_2)}$, which is the value of the transfer function at $s = 0$.

## Check yourself

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

1. Which equation describes M2?
   A. M2 x2'' + B2 x2' + B1(x2' - x1') = F
   B. M2 x2'' + K1 x2 = F
   C. M2 x2'' + B1 x2' = 0
   D. M2 x2'' + B2 x2' = F
   Answer: A. M2 has F, damper B2 to the support, and damper B1 to M1. (Key point: Equation for M2)

2. What is the numerator of X1/F?
   A. K1
   B. B1
   C. M2 s^2
   D. B1 + B2
   Answer: B. F acts on M2, which drives M1 through damper B1, so the numerator is B1 after cancelling s. (Key point: Transfer function)

3. Under constant F, what is the final x1?
   A. F/K1
   B. B1 F / (K1 (B1 + B2))
   C. F/(B1 + B2)
   D. B2 F / K1
   Answer: B. M2 reaches velocity F/(B1+B2), and M1 settles where K1 x1 = B1 v2. (Key point: Steady-state value)

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