A particle is performing simple harmonic motion. If its velocities are v1 and v2 at the displacements from the mean position arc y1 and y2 respectively, then its time period is
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a
2πy21+y22v12+v22
b
2πv22-v21y21-y22
c
2πy21-y22v22-v21
d
2πv21+v22y21+y22
answer is C.
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Detailed Solution
Velocity of a particle in SHM v = ωA2-y2v1 = ωA2-y12 ⇒v21 = ω2A2-ω2y21...(i)and v2 = ωA2-y22 ⇒ v22 = ω2A2-ω2y22-----(ii)from equations (i) and (ii), we getv22-v21 = ω2(y12-y22)ω = v22-v21y21-y22 ⇒ T = 2πω = 2πy21-y22v22-v21