A particle executes a simple harmonic motion with a periodic time 8 second. At time t = 0 , it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is ( 45^ = 45^ = 1 2 , 90^ = 0^ = 1 )

Step-by-step solution

The time period of the simple harmonic motion is T = 8 s. The angular frequency is = 2 T = 2 8 = 4 rad/s. Since the particle starts from the mean position at t = 0 , its displacement equation is given by: x(t) = A ( t) = A ( 4 t ) The distance traveled by the particle in the 1st second ( t = 0 to t = 1 s) is: D₁ = x(1) - x(0) = A ( 4 ) - 0 = A 2 The distance traveled by the particle in the 2nd second ( t = 1 to t = 2 s) is: D₂ = x(2) - x(1) = A ( 2 ) - A ( 4 ) = A - A 2 The ratio of the distance traveled in the 2nd