A block whose mass is 2 kg is fastened to a spring whose spring constant is 100 Nm-1. It is pulled to a distance of 0.1 m from over a frictionless surface and is released at t = 0. Calculate the kinetic energy of the block when it is 0.05 m away from its mean position.

The block executes SHM, so its angular frequency,                    ωω=km=100 Nm-12 kg=7.07rads-1Its displacement at any time t is           x(t)=a cos ωt=0.1 cos (7.07t)When the particle is 0.05m away from the mean position.             0.05 = 0.1 cos (7.07t)or                          cos (7.07 t) = 0.5or         sin (7.07t)=32=0.866Velocity of the block at x = 0.05 m is                    v=Aω sinωt                       =0.1×7.07×0.866=0.61 ms-1Hence,    KE=12mv2=12×2(0.61)2=0.37 J