---
title: "Matrix iteration for the first mode"
subject: "Theory of Mechanism and Machine II"
chapter: 9
key: "semester_7/theory_of_mechanism_and_machine_ii:matrix-iteration-fundamental-frequency"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/matrix-iteration-fundamental-frequency
site: BCT Engineering
---

# Matrix iteration for the first mode

Theory of Mechanism and Machine II · Chapter 9: Vibration of Multi Degree of Freedom Systems

- Asked in: year not given, 8 marks
- Video: [Matrix Method • Numerical Problem On Matrix Method • Multi Degree Freedom System • In Hindi](https://www.youtube.com/watch?v=oPpW2DliCZc) — Learn Mech Tech
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
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- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-9

## Question

Three masses $m_1 = 4$ kg, $m_2 = 2$ kg and $m_3 = 1$ kg are connected in series to a fixed wall by three identical springs of stiffness 2000 N/m (wall - spring - $m_1$ - spring - $m_2$ - spring - $m_3$). Using the flexibility matrix and matrix iteration, find the fundamental natural frequency and the first mode shape.

## Short answer

Flexibility matrix a_ij = min(i,j)/k leads to dynamic matrix [D]=[a][M]. Matrix iteration with normalisation converges to λ₁ = 4.42×10⁻³ s², giving ω₁ ≈ 15.04 rad/s and mode shape {1, 1.548, 1.745}.

## Answer

**Data:** $m_1 = 4$, $m_2 = 2$, $m_3 = 1$ kg; $k = 2000$ N/m for each spring.

#### Flexibility and dynamic matrices
For this chain, the flexibility influence coefficient is $a_{ij} = \dfrac{\min(i,j)}{k}$:

$$
[a] = \frac{1}{2000}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix}\ \text{m/N}
$$

The free vibration equation $\{X\} = \omega^2[a][M]\{X\}$ is rewritten with the **dynamic matrix** $[D] = [a][M]$:

$$
[D]\{X\} = \lambda\{X\}, \qquad \lambda = \frac{1}{\omega^2}
$$

$$
[D] = \frac{1}{2000}\begin{bmatrix} 4 & 2 & 1 \\ 4 & 4 & 2 \\ 4 & 4 & 3 \end{bmatrix}
$$

#### Iteration
Start with the trial vector $\{1, 1, 1\}^T$, multiply by $[D]$ and normalise by dividing by the first element. The first element of $[D]\{X\}$ divided by the first element of $\{X\}$ is the estimate of $\lambda$.

| Iter. | Trial $\{X\}$ | $[D]\{X\}\times10^3$ | $\lambda\times10^3$ (s²) |
|---|---|---|---|
| 1 | [1.0000, 1.0000, 1.0000] | [3.5000, 5.0000, 5.5000] | 3.5000 |
| 2 | [1.0000, 1.4286, 1.5714] | [4.2143, 6.4286, 7.2143] | 4.2143 |
| 3 | [1.0000, 1.5254, 1.7119] | [4.3814, 6.7627, 7.6186] | 4.3814 |
| 4 | [1.0000, 1.5435, 1.7389] | [4.4130, 6.8259, 7.6954] | 4.4130 |
| 5 | [1.0000, 1.5468, 1.7438] | [4.4187, 6.8374, 7.7093] | 4.4187 |
| 6 | [1.0000, 1.5474, 1.7447] | [4.4197, 6.8395, 7.7118] | 4.4197 |
| 7 | [1.0000, 1.5475, 1.7449] | [4.4199, 6.8398, 7.7123] | 4.4199 |
| 8 | [1.0000, 1.5475, 1.7449] | [4.4199, 6.8399, 7.7123] | 4.4199 |

The mode shape and $\lambda$ have converged. From the last cycle,

$$
\lambda_1 = 4.4199\times10^{-3}\ \text{s}^2, \qquad \omega_1 = \frac{1}{\sqrt{\lambda_1}} = 15.04\ \text{rad/s}, \qquad f_1 = \frac{\omega_1}{2\pi} = 2.394\ \text{Hz}
$$

The first mode shape (normalised to $X_1 = 1$) is $\{1,\ 1.548,\ 1.745\}^T$. All three masses move in phase, with the largest motion at the free end.

(An exact eigenvalue solution gives $\omega_1 = 15.04$ rad/s, which confirms the result.)

**Answer:** $\omega_1 \approx 15.04$ rad/s ($f_1 \approx 2.394$ Hz); mode shape $\{1,\ 1.548,\ 1.745\}$.

## Check yourself

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

1. The dynamic matrix used in matrix iteration is:
   A. [D] = [k][M]
   B. [D] = [a][M], with [D]{X} = lambda{X}
   C. [D] = [M][k]^-1 only for equal masses
   D. [D] = [a] + [M]
   Answer: B. From {X} = w^2 [a][M]{X}, [D] = [a][M] and lambda = 1/w^2. (Key point: Dynamic matrix)

2. For a chain of identical springs fixed at the wall, a_ij equals:
   A. max(i,j) k
   B. min(i,j)/k
   C. 1/(i j k)
   D. k/(i+j)
   Answer: B. A unit load at j stretches springs up to min(i,j), so the deflection at i is min(i,j)/k. (Key point: Flexibility coefficients of the chain)

3. Repeated multiplication by [D] and normalising converges to:
   A. the highest mode, with smallest lambda
   B. the lowest mode, with largest lambda, giving the fundamental frequency
   C. a random mode
   D. zero vector
   Answer: B. The largest eigenvalue lambda = 1/w^2 corresponds to the lowest frequency, so the fundamental mode dominates. (Key point: Iteration converges to)

4. With lambda1 = 4.4199x10^-3 s^2, the fundamental frequency is:
   A. w1 = 15.04 rad/s
   B. w1 = 66.5 rad/s
   C. w1 = 0.0044 rad/s
   D. w1 = 4.42 rad/s
   Answer: A. w1 = 1/sqrt(lambda1) = 1/sqrt(0.0044199) = 15.04 rad/s. (Key point: Result)

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