Torque on the rod about O,
τ=kx.2l×2
Iα=k.2l.θ.2l×2
Iα=k2l2.θ
∴ω2=M12l2k2l2=M6k
f=2π1M6k
JEE Main 2019 — Physics Waves & Oscillations
2019mcqhard
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:
Official previous-year question
Held on 12 Jan 2019 · Verified 6 Jul 2026.
Options
- A
2π1m3k
- B
2π1mk
- C
2π1m6k
- D
2π1m2k
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