Question

A uniform rigid bar of mass 3 kg is hinged at point F, and supported by a spring of stiffness \( k = 100 \, {N/m} \), as shown in the figure. The natural frequency of free vibration of the system is \_\_\_\_\_\_ rad/s (answer in integer).
\includegraphics[width=0.5\linewidth]{57.png}

Hint:

For a rigid bar hinged at one end with a spring, the natural frequency is calculated using the formula \( \omega_n = \sqrt{\frac{k}{I}} \), where \( I = \frac{1}{3} m L^2 \) is the moment of inertia of the bar.

Answer 1

We are given the following:
Mass of the bar, \( m = 3 \, {kg} \),
Spring stiffness, \( k = 100 \, {N/m} \),
Length of the bar is \( L = 1 \, {m} \),
The system is hinged at point F and supported by a spring at point G. Step 1:
The moment of inertia \( I \) of a uniform rigid bar of mass \( m \) and length \( L \), hinged at one end, is given by: \[ I = \frac{1}{3} m L^2 \] Substitute the given values:
\( m = 3 \, {kg} \),
\( L = 1 \, {m} \).
\[ I = \frac{1}{3} \times 3 \times 1^2 = 1 \, {kg} \cdot {m}^2 \] Step 2:
The natural frequency \( \omega_n \) of the system is given by the formula: \[ \omega_n = \sqrt{\frac{k}{I}} \] Substitute the given values:
\( k = 100 \, {N/m} \),
\( I = 1 \, {kg} \cdot {m}^2 \).
\[ \omega_n = \sqrt{\frac{100}{1}} = \sqrt{100} = 10 \, {rad/s} \] Thus, the natural frequency of the system is: \[ \boxed{10} \, {rad/s} \]