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Simple Harmonic Oscillation

A thin unifor...

Question

A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillations of small amplitude is

A

$2 π \sqrt{\frac{l}{g}}$

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B

$2 π \sqrt{\frac{2 l}{3 g}}$

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C

$2 π \sqrt{\frac{3 l}{2 g}}$

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D

$2 π \sqrt{\frac{l}{2 g}}$

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Solution

The correct option is B
$2 π \sqrt{\frac{2 l}{3 g}}$

$T = 2 π \sqrt{\frac{I}{C}} = 2 π \sqrt{\frac{\frac{m l^{2}}{3}}{m g \frac{l}{2}}}$

Here C represents the position of centre of mass

$T = 2 π \sqrt{\frac{2 l}{3 g}}$

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