Question

A long uniform rod of length L and mass M is free to rotate in a vertical plane about a horizontal axis through its one end 'O'. A spring of force constant k is connected vertically between one end of the rod and ground. When the rod is in equilibrium it is parallel to the ground.

What is the period of small oscillations that result when the rod is rotated slightly and released?

• A. $T=2\pi \sqrt{\frac{M}{3k}}$
• B. $T=2\pi \sqrt{\frac{M}{k}}$
• C. $T=2\pi \sqrt{\frac{2M}{k}}$
• D. $T=2\pi \sqrt{\frac{2M}{3k}}$

Answer 1

The correct option is A

$T=2\pi \sqrt{\frac{M}{3k}}$

(a) Restoring torque about 'O' due to elastic force of the spring

$\tau =-FL=-kyL(F=ky)\tau =-k{L}^2\theta (asy=L\theta )\tau =I\alpha =\frac{1}{3}M{L}^2\frac{d^2\theta }{d{t}^2}\frac{1}{3}M{L}^2\frac{d^2\theta }{d{t}^2}=-k{L}^2\theta \frac{d^2\theta }{d{t}^2}=-\frac{3k}{M}\theta \omega =\sqrt{\frac{3k}{M}}\Rightarrow T=2\pi \sqrt{\frac{M}{3k}}$