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Question:

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 rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is

12π√6kM

12π√kM

12π√24kM

12π√2kM

Solution:

Restoring torque = 2 × Fspring × L/2
Fspring = kΔx
Δx = L/2 θ
Restoring torque = 2 × k(L/2 θ) × L/2 = (kL²θ)/2
Torque = Iα = (ML²/12)α
(kL²θ)/2 = (ML²/12)α
α = (6kθ)/M
Comparing with α = -ω²θ
ω² = 6k/M
ω = √(6k/M)
f = ω/2π = (1/2π)√(6k/M)
f = (1/2π)√(6k/M)