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A point mass m is suspended from free end of rod of length $l,$ mass m. Then the time period for small amplitude of oscillations will be:

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$2 π \sqrt{\frac{4 l}{3 g}}$

$2 π \sqrt{\frac{8 l}{9 g}}$

$2 π \sqrt{\frac{l}{g}}$

$2 π \sqrt{\frac{8 l}{15 g}}$

answer is C.

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Detailed Solution

$T = 2 π \sqrt{\frac{I}{m g d}} = 2 π \sqrt{\frac{\frac{4}{3} m l^{2}}{2 m g .3 \frac{l}{4}}} = 2 π \sqrt{\frac{8 l}{9 g}}$

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