Two particles execute simple harmonic motion of the same amplitude and time period along the same straight line. The first particle starts from the mean position moving in the positive direction, while the second particle starts from the positive extreme position. If they cross each other for the first time at t = 2 s , the time period of their oscillations is

Step-by-step solution

Let the amplitude of both particles be A and their angular frequency be . The displacement of the first particle (starting from the mean position) is: x₁ = A ( t) The displacement of the second particle (starting from the positive extreme position) is: x₂ = A ( t) The particles cross each other when their displacements are equal: x₁ = x₂ A ( t) = A ( t) ( t) = 1 For the first meeting, the phase angle is: t = 4 We know that = 2 T and it is given that t = 2 s . Substituting these values: ( 2 T ) (2) = 4 4 T = 4 Solvi