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The string of a simple pendulum is replaced by a uniform rod of length L and mass M while bob has a mass m. it is allowed to make small oscillations. Its time period is

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$2 π \sqrt{(\frac{2 M}{3 m}) \frac{L}{g}}$

$2 π \sqrt{\frac{2 (M + 3 m) L}{3 (M + 2 m) g}}$

$2 π \sqrt{(\frac{M + m}{m + 3 m}) \frac{L}{g}}$

answer is B.

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Detailed Solution

$I_{s} α = M g \frac{L}{2} \sin θ + m g L \sin θ$

$α = \frac{g (\frac{M}{2} + m)}{(\frac{M}{3} + m) L} θ$

$∴ ω^{2} = \frac{3 g}{2 L} (\frac{M + 2 m}{M + 3 m})$

$T = 2 π \sqrt{\frac{2}{3} \frac{(M + 3 m) L}{(M + 2 m) g}}$

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