Two light strings, each of length l, are fixed at points A and B on a fixed horizontal rod xy. A small bob is tied by both strings and in equilibrium, the strings are making angle 45° with the rod. If the bob is slightly displaced normal to the plane of the strings and released then period of the resulting small oscillation will be:

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$2 π \sqrt{\frac{2 \sqrt{2} l}{g}}$

$2 π \sqrt{\frac{\sqrt{2} l}{g}}$

$2 π \sqrt{\frac{l}{g}}$

$2 π \sqrt{\frac{l}{\sqrt{2} g}}$

answer is D.

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Detailed Solution

Resulting torque on the bob = mg $\frac{l}{\sqrt{2}} sinθ$

MI of bob about axis xy = $\frac{{ml}^{2}}{2}$

For small angle $θ$

$ω = \sqrt{\frac{\sqrt{2} g}{l}} \Rightarrow T = 2 π \sqrt{\frac{l}{\sqrt{2} g}}$

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