Matrix iteration for the first mode

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 $\lambda$.

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 $\{1,\ 1.548,\ 1.745\}$.