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A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

Asked in NEET 2021 · Single spring

(1) 0.0628 s
(2) 6.28 s
(3) 3.14 s
(4) 0.628 s

Answer: (4) 0.628 s

Step-by-step solution

k=F/x=(10)/(0.05)=200 N/m.

T=2π√m/k=2π√2/(200)=(2π)/(10).

T=0.628 s.

Why the other options are wrong

(1) A power of ten low; it would need a spring constant of 2×10⁴ N/m, a hundred times the 200 N/m the data give.
(2) Reads 5 cm as 5 m: k=(10)/5=2 N/m gives T=2π√2/2=2π s.
(3) No single slip gives π s; it would need m/k=1/4 s², that is k=8 N/m.

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