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 is

Step-by-step solution

Given amplitude A = 4 cm and displacement x = 3 cm. The magnitude of velocity of a particle in S.H.M. is given by v = A^2 - x^2 . The magnitude of acceleration is given by a = ^2 x . According to the given condition, v = a at x = 3 cm. A^2 - x^2 = ^2 x 4^2 - 3^2 = (3) 16 - 9 = 3 7 = 3 = 7 3 rad/s The time period T is given by T = 2 . T = 2 7 3 = 6 7 s Answer: 6 7 s