Question

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length \(\ell\) and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is :

• A. \(\frac{1}{2\pi}\sqrt{\frac{3k}{m}}\)
• B. \(\frac{1}{2\pi}\sqrt{\frac{6k}{m}}\)
• C. \(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
• D. \(\frac{1}{2\pi}\sqrt{\frac{2k}{m}}\)

Hint:

Each spring gives a restoring torque \(k\frac{\ell}{2}\theta\cdot\frac{\ell}{2}\); equate the total torque to \(I\alpha\) with \(I = \frac{m\ell^2}{12}\).

Answer 1

The Correct Option is B

Let the rod rotate through a small angle \(\theta\) about O. Each end moves a distance \(\frac{\ell}{2}\theta\) along x, so each spring is deformed by \(\frac{\ell}{2}\theta\). The spring on A is attached to the right wall and the spring on B to the left wall, so as the rod turns both springs push the rod back towards equilibrium.
Force by each spring: \(F = k\frac{\ell}{2}\theta\), acting at lever arm \(\frac{\ell}{2}\).
Total restoring torque \(\tau = 2\cdot k\frac{\ell}{2}\theta\cdot\frac{\ell}{2} = \frac{k\ell^2}{2}\theta\).
Moment of inertia of the rod about its centre: \(I = \frac{m\ell^2}{12}\).
\(I\ddot\theta = -\frac{k\ell^2}{2}\theta \Rightarrow \omega^2 = \frac{k\ell^2/2}{m\ell^2/12} = \frac{6k}{m}\).
\(f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}\).
Answer: (2) \(\frac{1}{2\pi}\sqrt{\frac{6k}{m}}\)