A 4 kg particle is moving along the x-axis under the action of the force F = -(π216)xN. At t = 2 sec, the particle passes through the origin and at t = l0 sec its speed is 42 m/s. The amplitude of the motion is:

a = -(π264)x   ⇒ω = π264 = π8⇒T = 2πω = 16 sec.There is a time difference of T2 between t = 2 sec to t = l0 sec. Hence particle is again passing through the mean position of SHM where its speed is maximum.i.e., Vmax= Aω = 42⇒A = 42π8 = 322πm.