Two particles execute simple harmonic motion (SHM) along close parallel lines. SHM of the both the particles have same frequency and same amplitude. When they pass each other moving in opposite direction each time, their displacement is half their amplitude. Then, their phase difference is

Step-by-step solution

The two SHM have same amplitude and same frequency. aligned Let x₁ & = a t & and x₂= a ( t+ ) aligned (where, is the phase difference.) When they cross each other, x₁=x₂= a 2 At x₁= a 2 , we get, a 2 =a t t= 1 2 = 6 t= 6 At x₂= a 2 , we get, a 2 =a ( t+ ) aligned & & ( t+ ) & = 1 2 = 6 or 5 6 & & t+ & = 6 or 5 6 aligned (i) When t+ = 6 , then, = 6 - t= 6 - 6 =0 which is not possible. aligned & (ii) When t+ = 5 6 , & then = 5 6 - 6 = 4 6 = 2 3 & = 2 3 & aligned