A particle oscillates in straight line simple harmonically with period 8 second and amplitude 4 2 ~m . Particle starts from mean position. The ratio of the distance travelled by it in 1^ st second of its motion to that in 2^ nd second is ( 45^ =1 / 2 , 2 =1 )

Step-by-step solution

Given: Period T = 8 s, amplitude A = 4 2 m, and the particle starts from the mean position. Angular frequency is = 2 T = 4 rad/s. The equation of motion is x(t) = A ( t) = 4 2 ( 4 t ) . At t = 1 s, displacement is x(1) = 4 2 ( 4 ) = 4 m. Since motion is unidirectional from x(0) = 0 , distance in the first second is d₁ = 4 m. At t = 2 s, displacement is x(2) = 4 2 ( 2 ) = 4 2 m. Distance in the second second is d₂ = |x(2) - x(1)| = 4( 2 - 1) m. The ratio is d₁ d₂ = 1 2 - 1 , or 1 : ( 2 - 1) . The correct choice is D