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JEE Main201912 Jan 2019Morning ShiftPhysicsOscillationsActual

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod A B of length l and mass m . The rod is pivoted at its center ' 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

Options

  1. A1 2 π 3 k m
  2. B1 2 π k m
  3. C1 2 π 6 k m
  4. D1 2 π 2 k m

Correct answer

C. 1 2 π 6 k m

Step-by-step solution

Torque on the rod about O , τ = k x . l 2 × 2 I α​ = k . l 2 . θ . l 2 × 2 I α​ = k l 2 2 . θ ∴ ω 2 = k l 2 2 M l 2 12 = 6 k M f = 1 2 π 6 k M

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