Question

A system with two degrees of freedom, as shown in the figure, has masses \(m_1 = 200\ \text{kg}\) and \(m_2 = 100\ \text{kg}\) and stiffness coefficients \(k_1 = k_2 = 200\ \text{N/m}\). Then the lowest natural frequency of the system is ____________ rad/s (rounded off to one decimal place).

Hint:

For multi-DOF systems, reduce the eigenvalue problem using \(\lambda = \omega^2\) for easier computation.

Answer 1

Correct Answer: 0.7

The mass matrix and stiffness matrix are: \[ M = \begin{bmatrix} 200 & 0
0 & 100 \end{bmatrix}, \qquad K = \begin{bmatrix} 400 & -200
-200 & 200 \end{bmatrix} \] The natural frequencies are obtained from: \[ \det(K - \omega^2 M) = 0 \] Solving: \[ \omega^4 (200 \cdot 100) - \omega^2 (400 \cdot 100 + 200 \cdot 200) + (400\cdot 200 - 200^2)=0 \] \[ 20000 \omega^4 - 80000 \omega^2 + 40000 = 0 \] Divide by 20000: \[ \omega^4 - 4\omega^2 + 2 = 0 \] Let \(\lambda = \omega^2\): \[ \lambda^2 - 4\lambda + 2 = 0 \] \[ \lambda = 2 \pm \sqrt{2} \] Lowest natural frequency: \[ \omega = \sqrt{2 - \sqrt{2}} \approx 0.74\ \text{rad/s} \] Thus, the correct range is: \[ \boxed{0.7\ \text{to}\ 0.8\ \text{rad/s}} \]
Final Answer: 0.7–0.8 rad/s