Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m$. the rod is pivoted at its centre $‘O’$ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is
JEE MAIN 2019, Diffcult
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Torque on rod at displacement $\theta$ from mean position $\theta$ is very small. $x=\frac{L}{2} \theta$
$\tau=2 k x \frac{L}{2}=2 k \frac{L^{2}}{4} \theta=\frac{k L^{2}}{2} \theta$
Now, $\tau=1 \alpha$
$\frac{\mathrm{kL}^{2}}{2} \theta=\frac{\mathrm{mL}^{2}}{12} \alpha \quad ; \quad \alpha=\frac{6 \mathrm{k}}{\mathrm{m}} \theta$
$\tau=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}}$
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