Variation of velocity (V) with distance from equilibrium position (x) of a particle executing SHM along a straight line is shown in the diagram. Its time period is
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a
π4 S
b
π2 S
c
π10 S
d
π5 S
answer is D.
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Detailed Solution
Vmax=Aω, Amplitude A = 0.1 m0.1 x ω = 1⇒ω = 10 rad/sTime period T=2πω=2π10 sec=π5 sec