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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π4 S

π2 S

π10 S

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

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