Let #m# be the mass of the block placed on the connected block moving with a period of #T=0.8s#. The block of mass also will have same period if it does not slip. The slipping is stopped due to the force of friction acting between the two surfaces of the blocks in contact. If the coefficient of limiting friction be #mu# then the limiting value of static friction will be #mumg#, where g is the acceleration due to gravity.The maximum force acting on the mass due to its amplitude #a# will be #momega^2a#,where #omega=(2pi)/T#.When the amplitude #a=a_"max"# the maximum force operating on the block will just be in equilibrium with the force of limiting friction to stop slipping of the block.
So #momega^2a_"max"=mumg#
#=>a_"max"=(mug)/omega^2=(mugT^2)/(4pi^2)=(mug(0.8)^2)/(4pi^2)=(0.16 mug)/pi^2~~0.16mu#
