A rod of length L is supported by two ideal strings of length l such that the system hangs in a vertical plane.
Case 1: Rod is kept horizontal, displaced slightly perpendicular to the plane and allowed to oscillate.
Case 2: Rod is given a small twist about central axis and then allowed to oscillate.
Let $T_{1} {,T}_{2}$ be the periods of oscillations in the two respective cases.

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$T_{1} = 2 π \sqrt{\frac{l}{2 g}}$

$T_{2} = 2 π \sqrt{\frac{l}{3 g}}$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

$T_{2} = 2 π \sqrt{\frac{L}{6 g}}$

answer is A, D.

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Detailed Solution

Case 1 : Similar to general case of simple pendulum.

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

Case 2 : Let $θ$ be the angle that the rod rotate about its central axis and $ϕ$ be the angle made by string with the vertical.

$\begin{array}{l} In vertical, 2 T \cos ϕ = m g - - - (1) \\ In horizontal plane, - (2 T \sin ϕ) \frac{L}{2} = I α \\ \Rightarrow - (2 T \sin ϕ) \frac{L}{2} = \frac{m L^{2}}{12} \frac{d^{2} θ}{d t^{2}} \\ \Rightarrow - (T ϕ) = \frac{m L}{12} \frac{d^{2} θ}{d t^{2}} \\ \Rightarrow - \frac{m g ϕ}{2} = \frac{m L}{12} \frac{d^{2} θ}{d t^{2}} \\ \Rightarrow \frac{d^{2} θ}{d t^{2}} = - (\frac{6 g}{L}) ϕ \\ α = - (\frac{3 g}{l}) θ (∴ \frac{L}{2} θ = l ϕ) \\ c o m p a r i n g w i t h α = - ω^{2} θ \end{array}$

$\Rightarrow T_{2} = 2 π \sqrt{\frac{l}{3 g}}$

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