A particle executes linear S.H.M. with amplitude 4 cm . The magnitude of velocity and acceleration is equal when it is at 3 cm from mean position. Time period of oscillation is

Step-by-step solution

The velocity and acceleration magnitudes in simple harmonic motion are given by v = A^2 - x^2 and a = ^2 x respectively. Setting |v| = |a| at displacement x = 3 cm with amplitude A = 4 cm yields: 16 - 9 = ^2 3 7 = 3 ^2 Dividing both sides by (nonzero for oscillation): 7 = 3 Solving gives = 7 3 rad/s. The period T = 2 becomes: T = 2 7 3 = 6 7 seconds This result corresponds to option B .