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A uniform pole of length l = 2 L is laid on smooth horizontal table as shown in figure. The mass of pole is M and it is connected to a frictionless axis at O. A spring with force constant k is connected to the other end. The pole is displaced by a small angle θ₀ from equilibrium position and released such that it performs small oscillations. Then

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$ω_{0} = \sqrt{\frac{M}{3 k}}$

$ω_{0} = \sqrt{\frac{k}{3 M}}$

$ω_{0} = \sqrt{\frac{3 k}{M}}$

$ω_{0} = \sqrt{\frac{k}{2 M}}$

answer is C.

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Detailed Solution

$I = \frac{1}{3} M {(2 L)}^{2} = \frac{4}{3} {ML}^{2}$

Force applied by the spring is F = – kx

⇒ F = – k (2Lθ)

(θ is the angular displacement from the equilibrium position). Further

$𝜏 = |\vec{r} \times \vec{F}| = 4 L^{2} ksinθ = - 4 L^{2} kθ$

Also, $𝜏 = Iα = I \ddot{θ} = - 4 L^{2} kθ$

$\Rightarrow \ddot{θ} + \frac{3 k}{M} θ = 0 \Rightarrow ω_{0} = \sqrt{\frac{3 k}{M}}$

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