When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is y(t) = y₀ sin²(ωt), where y is measured from the lower end of the unstretched spring. Then ω is:
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Question

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is y(t) = y₀ sin²(ωt), where y is measured from the lower end of the unstretched spring. Then ω is:

Options

G1 Physics · Accepted Answer

y=y₀ sin²ωt=(y₀/2)[1−cos(2ωt)]. Thus the actual SHM angular frequency is Ω=2ω and the equilibrium extension is y₀/2. For a vertical spring, equilibrium extension mg/k=y₀/2 and Ω²=k/m=2g/y₀. Hence 4ω²=2g/y₀ ⇒ ω=√[g/(2y₀)]. Correct option: 2.