[Figure: a uniform rod AB lying horizontally and pivoted at its midpoint O; a light spring of constant k joins each end of the rod to a rigid support, both springs horizontal and perpendicular to the rod.] Two light identical springs of constant k are attached horizontally at the two ends of a unif…

Question

[Figure: a uniform rod AB lying horizontally and pivoted at its midpoint O; a light spring of constant k joins each end of the rod to a rigid support, both springs horizontal and perpendicular to the rod.] Two light identical springs of 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 turns freely in a horizontal plane, and the other ends of the springs are fixed to rigid supports. The rod is pushed through a small angle and released. The frequency of the resulting oscillation is

Accepted Answer

(2) 1/(2π)√(6k)/m. For a small turn θ each end moves /l2θ, so each spring exerts a torque k(l²)/4θ and the total restoring torque is (kl²)/2θ. The rod's moment of inertia about O is (ml²)/(12), so ω²=(kl²/2)/(ml²/12)=(6k)/m. f=1/(2π)√(6k)/m.

Step-by-step solution

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