question_answer

A rod of uniform diameter is suspended from one of its ends in vertical plane. The mass of the rod is 'm' and length T, the natural frequency of this rod in Hz for small amplitude is:

A) \[\frac{1}{2\pi }\sqrt{\frac{g}{l}}\]                     

B) \[\frac{1}{2\pi }\sqrt{\frac{g}{3l}}\]

C) \[\frac{1}{2\pi }\sqrt{\frac{2g}{3l}}\]                 

D) \[\frac{1}{2\pi }\sqrt{\frac{3g}{2l}}\]

Correct Answer: D

Solution :

The equation of motion is (figure)             \[\frac{m{{l}^{2}}}{3}\,\ddot{\theta }\,\left( \frac{mgl}{2} \right)\,\theta =0\]             \[\ddot{\theta }+\frac{3}{2}\,\frac{g}{l}\theta =0\]             \[{{f}_{n}}=\frac{1}{2\pi }\,\sqrt{\frac{3g}{2l}}\]