JEE PYQ: Oscillations Question 31
Question 31 - 2019 (09 Jan 2019 Shift 1)
Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system. The restoring torque is $\tau = k\theta$. If the rod is rotated by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
(1) $\frac{3k\theta_0^2}{l}$
(2) $\frac{2k\theta_0^2}{l}$
(3) $\frac{k\theta_0^2}{l}$
(4) $\frac{k\theta_0^2}{2l}$
Show Answer
Answer: (3)
Solution
$\omega = \sqrt{\frac{k}{I}}$. Tension at mean position $= m\omega^2 \frac{l}{3} \cdot \theta_0^2 = \frac{k\theta_0^2}{l}$.
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