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A ring of mass m can freely slide on a smooth vertical rod. The ring is symmetrically attached with two springs, as shown, each of stiffness k . Ring is displaced such that each spring makes an angle $θ$ with the horizontal. If the ring is slightly displaced vertically, then time period is............

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$\frac{2 π}{\sin θ} \sqrt{\frac{m}{2 k}}$

$2 π \sqrt{\frac{m}{k}}$

$2 π \sqrt{\frac{m \sin θ}{2 k}}$

$\frac{π}{\sin θ} \sqrt{\frac{m}{2 k}}$

answer is C.

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Detailed Solution

Let x be the displacement of the ring from equilibrium position and x' be the extension in the spring:

$Sin θ = \frac{x^{'}}{x} ∴ F_{r e s t} = 2 k x^{'} \cos (90 - θ)$

$\Rightarrow x^{'} = x \sin θ = m a & a = ω^{2} x; T = \frac{2 π}{ω}$

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