---
title: "Free damped vibration: three cases"
subject: "Theory of Mechanism and Machine II"
chapter: 6
key: "semester_7/theory_of_mechanism_and_machine_ii:damped-free-vibration-cases-derivation"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/damped-free-vibration-cases-derivation
site: BCT Engineering
---

# Free damped vibration: three cases

Theory of Mechanism and Machine II · Chapter 6: Vibration of Single Degree of Freedom Systems

- Asked in: year not given, 8 marks
- Video: [Damped Free Vibrations with Viscous Damping-Theory (Equation of motion) [DOM]](https://www.youtube.com/watch?v=YpiSZxDj7ws) — Education Lessons
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/damped-free-vibration-cases-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-6

## Question

Derive the equation of motion of a spring-mass-viscous damper system in free vibration. Discuss the under-damped, critically damped and over-damped cases with the help of sketches of the response.

## Short answer

Equation: m x'' + c x' + k x = 0. With ω_n=√(k/m), c_c=2√(km), ζ=c/c_c. Under-damped (ζ<1): oscillatory decay x = X e^{-ζ ω_n t} sin(ω_d t+φ). Critically damped (ζ=1): x = (A+Bt)e^{-ω_n t}, fastest non-oscillatory return. Over-damped (ζ>1): two real negative roots, slow non-oscillatory return.

## Answer

#### Equation of motion
For mass $m$, spring $k$ and viscous damper $c$, with displacement $x$ from equilibrium:

$$
m\ddot{x} + c\dot{x} + kx = 0
$$

Assume $x = Ae^{st}$:

$$
ms^2 + cs + k = 0 \;\Rightarrow\; s_{1,2} = -\frac{c}{2m} \pm \sqrt{\left(\frac{c}{2m}\right)^2 - \frac{k}{m}}
$$

The **critical damping coefficient** is the value of $c$ that makes the term under the square root zero: $c_c = 2\sqrt{km} = 2m\omega_n$. The **damping ratio** is $\zeta = c/c_c$. Then

$$
s_{1,2} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)
$$

#### Case 1: under-damped ($\zeta < 1$)
The roots are complex: $s = -\zeta\omega_n \pm i\omega_d$, with damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$.

$$
x(t) = Xe^{-\zeta\omega_n t}\sin(\omega_d t + \phi)
$$

The motion is oscillatory with an exponentially decaying amplitude.

#### Case 2: critically damped ($\zeta = 1$)
Equal real roots $s = -\omega_n$:

$$
x(t) = (A + Bt)\,e^{-\omega_n t}
$$

There is no oscillation; the mass returns to equilibrium in the shortest possible time.

#### Case 3: over-damped ($\zeta > 1$)
Two distinct negative real roots:

$$
x(t) = Ae^{(-\zeta + \sqrt{\zeta^2-1})\omega_n t} + Be^{(-\zeta - \sqrt{\zeta^2-1})\omega_n t}
$$

There is no oscillation; the return to equilibrium is slower than in the critical case.

```text
 x
 x0*
 | \\ \
 |  \\  '.__
 |   \ '.    '----.______________  over-damped (slow return)
 |    \  '-.___
 |     \       '------.________    critically damped (fastest, no overshoot)
 +------\-----------.------.----------> t
 |       \        /   \__/         under-damped (oscillates, decays)
 |        '.____.'
```

| Case | $\zeta$ | Roots | Motion |
|---|---|---|---|
| Under-damped | $<1$ | complex pair | decaying oscillation |
| Critical | $=1$ | equal real | non-oscillatory, fastest return |
| Over-damped | $>1$ | two real | non-oscillatory, slow return |

## Check yourself

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

1. Assuming x = A e^(st) in m x'' + c x' + k x = 0, what equation gives s?
   A. m s^2 + c s + k = 0
   B. m s + c + k = 0
   C. s^2 + c s - k = 0
   D. m s^2 - c s + k = 0
   Answer: A. Substituting x = A e^(st) gives (m s^2 + c s + k) A e^(st) = 0, so the quadratic m s^2 + c s + k = 0 must hold. (Key point: Characteristic equation)

2. The critical damping coefficient is the value of c that makes the root of the characteristic equation:
   A. complex with a large imaginary part
   B. equal to zero damping ratio
   C. positive and real
   D. repeated (the square-root term zero), c_c = 2 sqrt(km)
   Answer: D. At c = c_c = 2 sqrt(km) the discriminant vanishes, giving equal real roots; zeta = c/c_c. (Key point: Critical damping coefficient)

3. For zeta < 1 the motion is:
   A. non-oscillatory, returning to equilibrium in the shortest time
   B. oscillation with exponentially decaying amplitude at damped frequency wd = wn sqrt(1 - zeta^2)
   C. oscillation with constant amplitude at wn
   D. a slow creep back to equilibrium with no oscillation
   Answer: B. Complex roots -zeta wn ± i wd give x = X e^(-zeta wn t) sin(wd t + phi), a decaying oscillation. (Key point: Under-damped response)

4. Compared with the over-damped case, the critically damped system:
   A. oscillates once before settling
   B. never returns to equilibrium
   C. returns to equilibrium in the shortest time without oscillating
   D. returns more slowly because roots are equal
   Answer: C. Critical damping (zeta = 1) gives x = (A + Bt) e^(-wn t): no oscillation and the fastest return; over-damping is slower. (Key point: Critical vs over-damped)

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