---
title: "Three-mass chain frequencies and modes"
subject: "Theory of Mechanism and Machine II"
chapter: 8
key: "semester_7/theory_of_mechanism_and_machine_ii:three-dof-eigenvalues-mode-shapes"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/three-dof-eigenvalues-mode-shapes
site: BCT Engineering
---

# Three-mass chain frequencies and modes

Theory of Mechanism and Machine II · Chapter 8: Vibration of Multi Degree of Freedom Systems

- Asked in: year not given, 8 marks
- 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/three-dof-eigenvalues-mode-shapes
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-8

## Question

Three equal masses of 10 kg are connected in series to a fixed wall by three identical springs of stiffness 100 kN/m each (wall - spring - $m_1$ - spring - $m_2$ - spring - $m_3$, the last mass free). Set up the eigenvalue problem, find the natural frequencies and the mode shapes (normalised to $X_1 = 1$), and verify the orthogonality of the first two modes with respect to the mass matrix.

## Short answer

Eigenvalue problem from [K]{X}=ω²[M]{X}. Natural frequencies: 44.5, 124.7, 180.2 rad/s. Mode shapes with X1=1: {1, 1.802, 2.247}, {1, 0.445, -0.802}, {1, -1.247, 0.555}. First two modes are orthogonal with respect to the mass matrix.

## Answer

**Data:** $m_1 = m_2 = m_3 = 10$ kg, $k = 100\,000$ N/m.

#### Mass and stiffness matrices
$$
[M] = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}, \qquad [K] = 10^5\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}\ \text{N/m}
$$

#### Eigenvalue problem
$$
\left([K] - \omega^2[M]\right)\{X\} = \{0\}
$$

Let $\lambda = \dfrac{m\omega^2}{k} = \dfrac{10\omega^2}{10^5}$. The characteristic equation $\det\left([K] - \omega^2[M]\right) = 0$ becomes

$$
\begin{vmatrix} 2-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda \end{vmatrix} = 0
$$

$$
(2-\lambda)\left[(2-\lambda)(1-\lambda) - 1\right] - (1-\lambda) = 0 \;\Rightarrow\; \lambda^3 - 5\lambda^2 + 6\lambda - 1 = 0
$$

Solving the cubic (Newton or trial):

$$
\lambda_1 = 0.1981,\quad \lambda_2 = 1.5550,\quad \lambda_3 = 3.2470
$$

#### Natural frequencies
$$
\omega = \sqrt{\lambda\,\frac{k}{m}} = \sqrt{10000\lambda}
$$

| Mode | $\lambda$ | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|---|
| 1 | 0.1981 | 44.5 | 7.08 |
| 2 | 1.5550 | 124.7 | 19.85 |
| 3 | 3.2470 | 180.2 | 28.68 |

#### Mode shapes
From the first row, $X_2 = (2-\lambda)X_1$; from the second row, $X_3 = (2-\lambda)X_2 - X_1$.

| Mode | $X_1$ | $X_2$ | $X_3$ |
|---|---|---|---|
| 1 | 1 | 1.802 | 2.247 |
| 2 | 1 | 0.445 | -0.802 |
| 3 | 1 | -1.247 | 0.555 |

```text
 Mode 1: all in phase, no node
 Mode 2: one node (between m2 and m3)
 Mode 3: two nodes
```

#### Orthogonality check (modes 1 and 2)
$$
\{X\}_1^T[M]\{X\}_2 = 10\left[(1)(1) + (1.802)(0.445) + (2.247)(-0.802)\right] = 0.000 \approx 0\ \ \checkmark
$$

**Answer:** $\omega_1 = 44.5$, $\omega_2 = 124.7$, $\omega_3 = 180.2$ rad/s; mode shapes as tabulated; orthogonality verified.

## Check yourself

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

1. For wall-k-m1-k-m2-k-m3 (last mass free), the stiffness matrix (in units of k) is:
   A. [2 -1 0; -1 2 -1; 0 -1 2]
   B. [1 -1 0; -1 2 -1; 0 -1 1]
   C. [2 -1 0; -1 2 -1; 0 -1 1]
   D. [2 1 0; 1 2 1; 0 1 1]
   Answer: C. The last mass is attached to only one spring so its diagonal is k; others are 2k with -k off-diagonals. (Key point: Stiffness matrix)

2. The natural frequencies come from:
   A. [K]{X} = {0}
   B. det([K] - w^2 [M]) = 0
   C. det([M]) = 0
   D. [K][M] = w^2 [I]
   Answer: B. Free vibration gives ([K] - w^2 [M]){X} = 0 and non-trivial solutions need the determinant to vanish. (Key point: Eigenvalue problem)

3. With lambda = m w^2/k, the lowest root lambda1 = 0.1981 gives omega1 of about:
   A. 44.5 rad/s
   B. 14.1 rad/s
   C. 198 rad/s
   D. 7.08 rad/s
   Answer: A. w = sqrt(10000 x 0.1981) = 44.5 rad/s (7.08 Hz is the frequency in Hz). (Key point: Frequency scaling)

4. How many nodes does the second mode have?
   A. none
   B. one, between m2 and m3
   C. two
   D. three
   Answer: B. Mode 2 is (1, 0.445, -0.802): one sign change between m2 and m3, so one node; mode 3 has two. (Key point: Mode shape nodes)

5. Orthogonality of modes 1 and 2 is verified by showing:
   A. {X}1^T [M] {X}2 = 0
   B. {X}1 = {X}2
   C. {X}1^T {X}1 = 1
   D. the frequencies are equal
   Answer: A. Different modes satisfy {X}r^T [M] {X}s = 0; here 1 + 0.802 - 1.802 = 0. (Key point: Orthogonality check)

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