A particle performing linear S.H.M. has period 8 second. At time t=0 , it is in the mean position. The ratio of the distances travelled by the particle in the 1^ st and 2^ nd second is ( 45^ =1 / 2 )

Step-by-step solution

The motion is described by x = A ( t) with period T = 8 , s , giving angular frequency = 2 T = 4 , rad/s . First second: The displacement at t = 1 , s is x₁ = A ( 4 ) = A 2 . Since motion starts from the mean position, the distance traveled is d₁ = A 2 . Second second: The displacement reaches x₂ = A ( 2 ) = A at t = 2 , s . The distance covered from t = 1 , s to t = 2 , s is d₂ = A - A 2 = A (1 - 1 2 ) . The ratio is d₁ : d₂ = A 2 : A (1 - 1 2 ) = 1 : ( 2 - 1) .