A small block of mass m is connected to one end of a massless rod. A spring of force constant k is attached to the other end of the rod. The rod is pivoted at P. The block is displaced slightly downward and left free to oscillate. For small angular displacement $(θ)$ of the block, which of the following options are correct

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Restoring torque is $k a^{2} θ$

angular acceleration is $\frac{k a^{2} θ}{m b^{}}$

Time period of oscillation is $2 π \sqrt{\frac{b^{2} m}{a^{2} k}}$

Time period of oscillation is $π \sqrt{\frac{b^{2} m}{a^{2} k}}$

answer is A, C.

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Detailed Solution

$A \to (q); B \to (r); C \to (p); D \to (s)$
For small angle $θ$ ,the restoring torque
$τ = - (k a θ) \times a = K a^{2} (- θ)$
$∴ a = \frac{τ}{1} = \frac{k a^{2}}{m b^{2}} (- θ)$
On comparing with , $a = - ω^{2} θ, w e g e t$
$ω = \sqrt{\frac{k a^{2}}{m b^{2}}} a n d T = 2 π \frac{b}{a} \sqrt{\frac{m}{k}}$
Energy of oscillation, $E = \frac{1}{2} k A^{2} = \frac{1}{2} k {(a θ)}^{2}$