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Question: Answered & Verified by Expert
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 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 is:

PhysicsOscillationsJEE MainJEE Main 2019 (12 Jan Shift 1)
Options:
  • A 12π3km
  • B 12πkm
  • C 12π6km
  • D 12π2km
Solution:
2356 Upvotes Verified Answer
The correct answer is: 12π6km
Torque on the rod about O,



τ=kx.l2×2

α​=k.l2.θ.l2×2

α​=kl22.θ

 ω2=kl22Ml212=6kM

f=12π6kM

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