The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

40 Views|Posted 5 months ago
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5 months ago

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

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Physics Oscillations 2025

Physics Oscillations 2025

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