A wire is bent at an angle $θ$ . A rod of mass m can slide along the bended wire without friction as shown in figure. A soap film is maintained in the frame kept in a vertical position and the rod is in equilibrium as shown in the figure. If rod is displaced slightly in vertical direction, then the time period of small oscillation of the rod is

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$2 π \sqrt{\frac{l}{g \tan θ}}$

$2 π \sqrt{\frac{l}{g}}$

$2 π \sqrt{\frac{l \cos θ}{g}}$

$2 π \sqrt{\frac{l}{g \cos θ}}$

answer is A.

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Detailed Solution

Let S be the surface tension of the soap film. For equilibrium of rod

$\begin{array}{l} 2 (l + y) \tan \frac{θ}{2} \\ m g = {(F_{surface})}_{1} \\ m g = (2 l \tan \frac{θ}{2}) S \times 2; m g = 4 S l \tan \frac{θ}{2} \end{array}$

If the rod is displaced from its mean position by small displacement y, then restoring force on the rod is

$\begin{array}{l} F_{rest} = - [{(F_{surface})}_{2} - m g] - {(F_{surface})}_{1} \\ = - [4 S (l + y) \tan \frac{θ}{2} - m g] - [4 S \tan \frac{θ}{2} y] \\ a = - \frac{4 S \tan \frac{θ}{2}}{m} y = \frac{- 4 S \tan \frac{θ}{2}}{(\frac{4 S l \tan \frac{θ}{2}}{g})} y \\ \frac{d^{2} y}{d t^{2}} = - (\frac{g}{l}) y ∴ T = 2 π \sqrt{\frac{l}{g}} \end{array}$