Three-mass chain frequencies and modes

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 shapes

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

 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.