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I have attached the question image.

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You can use hook's law $F=-kx$ . If $\tau$ is the torque on board , then force at the spring, $F=\frac{\tau}{L}$ and $\tau=I\alpha$ , where $I=\frac{mL^2}{3}$ , M.I of rod about one of its end . U can replace x by $x=L\theta$ and appliying to hook's law yield $I\dfrac{d^2\theta}{dt^2}=-L^2k\theta$ this gives $\omega=\sqrt{\frac{L^2k}{I}}$. Therefore, time period, $T=2\pi\sqrt{\frac{I}{L^2k}}$ substituting your values, you can arrive at a simple value (I am not going to do it here).

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