Two particles A and B execute S.H.M.s of periods T and 5T 2 . If they start from the mean position, then the phase difference between them when the particle A completes two oscillations will be (in radian)

Step-by-step solution

Time taken by particle A to complete two oscillations is t = 2T . The phase of a particle executing S.H.M. starting from the mean position at time t is given by = t = 2 T' t , where T' is the time period. For particle A, the phase at t = 2T is: _A = 2 T 2T = 4 For particle B, the time period is T_B = 5T 2 . The phase at t = 2T is: _B = 2 5T 2 2T = 4 5T 2T = 8 5 The phase difference between the two particles is: = _A - _B = 4 - 8 5 = 12 5 Answer: 12 5