Two particles 'A' and 'B' execute SHMs of periods 'T' and ' . 3 T 2 .' . If they start from the mean position then the phase difference between them, when the particle 'A' completes two oscillations will be

Step-by-step solution

Given the periods T_A = T and T_B = 3T 2 , both particles begin at the mean position with initial phase zero. Particle A completes two oscillations after t = 2T_A = 2T seconds. The angular frequencies are _A = 2 T_A = 2 T and _B = 2 T_B = 4 3T . At t = 2T , their phases are: _A = _A t = 2 T 2T = 4 _B = _B t = 4 3T 2T = 8 3 The phase difference is: = |4 - 8 3 | = 4 3 Final answer: C