A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are v1 and v2, respectively. Its time period is

Let A be the amplitude of oscillation, then                   v12=ω2A2-x12                                             …(i)                   v22=ω2A2-x22                                            …(ii)On subtracting Eq. (ii) from Eq. (i), we get          v12-v22=ω2x22-x12⇒    ω=v12-v22x22-x12⇒2πT=v12-v22x22-x12⇒T=2πx22-x12v12-v22