JEE PYQ: Oscillations Question 43
Question 43 - 2019 (12 Jan 2019 Shift 2)
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 two springs are fixed to rigid supports. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:
(1) $\frac{1}{2\pi}\sqrt{\frac{3k}{m}}$
(2) $\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$
(3) $\frac{1}{2\pi}\sqrt{\frac{6k}{m}}$
(4) $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$
Show Answer
Answer: (3)
Solution
Restoring torque: $\tau = -\frac{Kl^2}{2}\theta$. $I = \frac{ml^2}{12}$. $f = \frac{1}{2\pi}\sqrt{\frac{6K}{m}}$.
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