A particle executes SHM along a straight line 4 cm long. When the displacement is 1 cm its velocity and acceleration are numerically equal. The time period of SHM is
A
2π s
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B
2π/√3 s
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C
2π/√5 s
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D
2π/√7 s
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Solution
The correct option is A2π/√3 s v=ω√A2−x2 a=−ω2x A=2cm ω√4−1=ω2×1 ω=√3 T=2πω=2π√3sec