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Question

The time period of oscillation of a particle that executes S.H.M. is 1.2sec. The time starting from extreme position, its velocity will be half of its velocity at mean position is
  1. 0.1 s
  2. 0.2 s
  3. 0.6 s
  4. 0.4 s

A
0.1 s
B
0.2 s
C
0.4 s
D
0.6 s
Solution
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Time is zero at extreme position
T=1.2sec
v=ωA2x2;vmax=ωA
v=vmax2=ωA2=ω A2x2
A24=A2x2;x=32A
x=A cos ωt
32A=A cos ωt
ωt=π6

t=π6×ω
t=/π6×1.22π/=0.1sec

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