(a) For the block from Newton's second law in projection form Fy = mwy
N - mg = my (1)
But from y = a(1 - cosωt)
we get y = ω2acosωt (2)
From eqn (1) and (2)
From Newtons' third law the force by which the body m exert on the block is directed vertically downward and equls
(b) When the body m starts, falling behind the plank or losing contact, N = 0, (because the normal reaction is the contact force). Thus from eqn. (3)
(c) We observe that the motion takes place about the mean position y = a. At the initial instant y = 0. As shown in (b) the normal reaction vanishes at a height (g/ω2) above the position of equilibrium and the body flies off as a free body. the speed of the body at a distance (g/ω2) from the equilibrium position is ω√(a2 - (g/ω2)2), so that the condition of the problem gives
Hence solving the resulting quadratic equation and taking the positive roof,