Question

A rigid massless rod pinned at one end has a mass $m$ attached to its other end. The rod is supported by a linear spring of stiffness $k$ as shown in the figure. The natural frequency of this system is: 

• A. $\displaystyle \frac{1}{2\pi}\sqrt{\frac{kL^{2}}{4m(L^{2}+H^{2})}}$
• B. $\displaystyle \frac{1}{2\pi}\sqrt{\frac{kL^{2}}{m(L^{2}+H^{2})}}$
• C. $\displaystyle \frac{1}{2\pi}\sqrt{\frac{4kL^{2}}{m(L^{2}+H^{2})}}$
• D. $\displaystyle \frac{1}{2\pi}\sqrt{\frac{k(L^{2}+H^{2})}{4mL^{2}}}$

Hint:

For rotational systems: use $K_\theta = k d^2$ where $d$ is the perpendicular distance from pivot to spring, and $I = mr^{2}$ for a point mass.

Answer 1

The Correct Option is A

Step 1: Determine rotational stiffness about the hinge.
The spring is located at the midpoint of the rod, i.e., at a distance $L/2$ from the pivot. A small rotation $\theta$ causes the midpoint to move vertically by \[ \delta = \frac{L}{2}\theta. \] Spring force: \[ F = k\delta = k\left(\frac{L}{2}\right)\theta. \] Moment of this force about the hinge: \[ M = F\left(\frac{L}{2}\right) = k\left(\frac{L}{2}\right)^2 \theta = k\frac{L^{2}}{4}\theta. \] Thus, rotational spring stiffness is \[ K_\theta = k\frac{L^{2}}{4}. \]

Step 2: Compute rotational inertia of the mass.
The mass $m$ is located at distance \[ r = \sqrt{L^{2} + H^{2}} \] from the hinge. So moment of inertia is \[ I = mr^{2} = m(L^{2} + H^{2}). \]

Step 3: Write natural frequency equation for rotation.
\[ \omega = \sqrt{\frac{K_\theta}{I}} = \sqrt{\frac{kL^{2}/4}{m(L^{2}+H^{2})}} = \sqrt{\frac{kL^{2}}{4m(L^{2}+H^{2})}}. \]

Step 4: Convert to frequency.
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{kL^{2}}{4m(L^{2}+H^{2})}}. \]
Thus, the correct option is (A).