---
title: "Undamped dynamic vibration absorber"
subject: "Theory of Mechanism and Machine II"
chapter: 7
key: "semester_7/theory_of_mechanism_and_machine_ii:dynamic-vibration-absorber-design"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/dynamic-vibration-absorber-design
site: BCT Engineering
---

# Undamped dynamic vibration absorber

Theory of Mechanism and Machine II · Chapter 7: Vibration of Two Degree of Freedom Systems

- Asked in: year not given, 4+5 marks
- Video: [Dynamic vibration absorber part 1](https://www.youtube.com/watch?v=dbQQOjAuiss) — Mechanical Engineering by Ashish Purohit
- 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/dynamic-vibration-absorber-design
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-7

## Question

(a) Explain the principle of an undamped dynamic vibration absorber and show that the amplitude of the main mass is zero when the absorber is tuned to the exciting frequency.
(b) A machine of mass 100 kg is mounted on springs and is excited by a harmonic force at 1500 rpm, which is also its natural speed. Design an absorber of mass 10 kg to remove the resonance. Find the absorber spring stiffness and the two new natural frequencies of the combined system.

## Short answer

An undamped absorber tuned to the excitation frequency causes zero main‑mass amplitude by cancelling the exciting force. Absorber spring stiffness is 246.7 kN/m; the two new natural speeds are approximately 1281 rpm and 1756 rpm.

## Answer

#### (a) Principle
A small auxiliary spring-mass system ($m_2$, $k_2$) is attached to the main system ($m_1$, $k_1$), which is excited by $F_0\sin\omega t$. When $\sqrt{k_2/m_2} = \omega$, the absorber's force on the main mass cancels the exciting force and the main mass stays still. The absorber mass vibrates instead.

```text
  F0 sin(wt)
     |
   [ m1 ]--k2--[ m2 ]
     |
     k1
   ///
```
**Equations of motion:**
$$
m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 = F_0\sin\omega t, \qquad m_2\ddot{x}_2 + k_2(x_2 - x_1) = 0
$$

Put $x = X\sin\omega t$:

$$
\begin{bmatrix} k_1 + k_2 - m_1\omega^2 & -k_2 \\ -k_2 & k_2 - m_2\omega^2 \end{bmatrix}\begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} F_0 \\ 0 \end{Bmatrix}
$$

$$
X_1 = \frac{(k_2 - m_2\omega^2)F_0}{(k_1+k_2-m_1\omega^2)(k_2 - m_2\omega^2) - k_2^2}, \qquad X_2 = \frac{k_2F_0}{(\ldots)}
$$

$X_1 = 0$ when $k_2 - m_2\omega^2 = 0$, i.e. $\omega = \sqrt{k_2/m_2}$. Then $X_2 = -\dfrac{F_0}{k_2}$, so the absorber spring force $k_2X_2 = -F_0$ exactly cancels the exciting force. The absorber works only at the one tuned frequency; away from it, the combined system has two resonances.

#### (b) Design
Operating frequency: $\Omega = \dfrac{2\pi \times 1500}{60} = 157.08$ rad/s. The main system is at resonance, so $k_1 = m_1\Omega^2 = 100 \times 157.08^2 = 2467401$ N/m.

**Tuning:** $k_2 = m_2\Omega^2 = 10 \times 157.08^2 = 246740\ \text{N/m} = 246.7\ \text{kN/m}$.

**New natural frequencies:** the mass ratio is $\mu = m_2/m_1 = 0.1$. The absorber and the main system have the same natural frequency $\Omega$. With $r = \omega/\Omega$, the frequency equation is

$$
r^4 - (2+\mu)r^2 + 1 = 0 \;\Rightarrow\; r^2 = \frac{2.1 \pm \sqrt{2.1^2 - 4}}{2} = 0.7298,\ 1.3702
$$

$$
\omega_{n1} = \Omega\sqrt{0.7298} = 134.19\ \text{rad/s}\ (1281\ \text{rpm}), \qquad \omega_{n2} = \Omega\sqrt{1.3702} = 183.87\ \text{rad/s}\ (1756\ \text{rpm})
$$

The two new resonant speeds lie about 15% below and 17% above the operating speed, so the machine runs between them, safely away from both.

**Answer:** Absorber stiffness $= 246.7$ kN/m; new natural speeds $\approx 1281$ rpm and $1756$ rpm.

## Check yourself

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

1. Why does the main mass stand still when the absorber is tuned (sqrt(k2/m2) = w)?
   A. The absorber spring force k2 X2 = -F0 exactly cancels the exciting force
   B. The absorber adds damping to the main mass
   C. The absorber mass is much bigger than the main mass
   D. The main spring becomes infinitely stiff
   Answer: A. At the tuned frequency X2 = -F0/k2, so the absorber spring pushes back on m1 with a force equal and opposite to F0. (Key point: Principle of absorber)

2. From X1 = (k2 - m2 w^2) F0 / (...), X1 = 0 when:
   A. k1 = m1 w^2
   B. k2 = m2 w^2
   C. k1 + k2 = m1 w^2
   D. m1 = m2
   Answer: B. The numerator vanishes when k2 - m2 w^2 = 0, i.e. w = sqrt(k2/m2). (Key point: Condition for zero amplitude)

3. Machine excited at 1500 rpm (157.08 rad/s); absorber mass 10 kg. The absorber stiffness is:
   A. 24.7 kN/m
   B. 2467 kN/m
   C. 246.7 kN/m
   D. 1570.8 kN/m
   Answer: C. k2 = m2 w^2 = 10 x 157.08^2 = 246 740 N/m = 246.7 kN/m. (Key point: Absorber stiffness design)

4. After adding the absorber the combined system has:
   A. a single natural frequency at the operating speed
   B. no natural frequencies
   C. two natural frequencies, on either side of the operating speed
   D. two natural frequencies both above the operating speed
   Answer: C. The two new resonances (about 1281 and 1756 rpm) lie below and above 1500 rpm; the absorber works only near its tuned frequency. (Key point: Limitation of absorber)

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