A particle is executing simple harmonic motion. The displacement of the particle from the mean position in 2 second is equal to 1 2 times its amplitude. What is the period of oscillation of the particle ?

Step-by-step solution

The displacement of a particle in simple harmonic motion is given by x(t) = A ( t + ) . Assuming the particle starts from the mean position at t = 0 , we have = 0 , so x(t) = A ( t) . Given that at t = 2 s, x(2) = A 2 , we substitute these values into the equation: A 2 = A (2 ) (2 ) = 1 2 This implies 2 = 4 or 2 = 3 4 . Taking the simplest case for the first occurrence, 2 = 4 . Since = 2 T , we have 2 ( 2 T ) = 4 or 2 ( 2 T ) = 3 4 . For 2 = 4 , 4 T = 4 T = 16 s. For 2 = 3 4 , 4 T = 3 4 T = 16 3 s. Comparing with t