IIT-JEE 2009 · Paper 2 · Physics Q42

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PhysicsSingle correct+3 / -1Question 42 of 57

A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$ . The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is

Top view: a rod pivoted at its centre between two parallel walls; its upper end is joined by a spring to the left wall and its lower end by a spring to the right wall, the springs perpendicular to the rod.

\dfrac1{2\pi}\sqrt{\dfrac{2k}M}

\dfrac1{2\pi}\sqrt{\dfrac kM}

\dfrac1{2\pi}\sqrt{\dfrac{6k}M}

\dfrac1{2\pi}\sqrt{\dfrac{24k}M}

Answer and worked solution

Official answer: C

Restoring torque. Turning through θ stretches each spring by $\frac L2\theta$ , so each exerts $k\frac L2\theta$ with lever arm $\frac L2$ : total torque $\dfrac{kL^2}2\theta$ .

Frequency. $I=\dfrac{ML^2}{12}$ , so
$\omega^2=\frac{kL^2/2}{ML^2/12}=\frac{6k}M,\qquad f=\frac1{2\pi}\sqrt{\frac{6k}M}.$

(C)

Topics:Simple and compound pendulum Torque and rotational dynamics

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IIT-JEE 2009 · Paper 2, Physics question 42