Question

Thin rod of mass $M$ and length $L$, is pivoted from one of its ends oscillates with frequency:

• A. $\frac{1}{2\pi }\sqrt{\frac{2g}{3L}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{3g}{2L}}$
• C. $2\pi \sqrt{\frac{2L}{3g}}$
• D. $\frac{1}{\pi }\sqrt{\frac{g}{L}}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{3g}{2L}}$

For rigid body oscillating about horizontal axis passing a point is given by

$T=2\pi \sqrt{\frac{I}{mgR}}$ where $R=L/2$ i.e distance of the point of pivot to the centre of gravity and $I$ is moment of inertia.

$I=1/3m{L}^2$

After putting the values of $R$ and $I$ we have

$T=2\pi \sqrt{\frac{2L}{3g}}$

Hence frequency is $f=\frac{1}{2\pi }\sqrt{\frac{3g}{2L}}$