EasyPhysics
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A mass m is suspended from a vertical spring and produces an extension of 0.1 m and comes in equilibrium. If the mass is then pulled down slightly from its equilibrium position and released, the time period of its oscillation will be (g = 10 m/s^2)
A
π/√5 s
B
π/5 s
C
π/2 s
D
2π/√3 s
Explanation
At equilibrium, mg = kx. So m/k = x/g. Time period T = 2π√(m/k) = 2π√(x/g). Given x = 0.1 and g = 10, T = 2π√(0.1/10) = 2π√(1/100) = 2π/10 = π/5 s.
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