Consider the diagram
From the given diagram, it is clear that
(a) A `t t=(3T)/(4)` the displacement of the paritcle is zera. Hence, the particle execution SHM will be at mean position i.e., x=0. So, acceleration is zero and force also zero.
(b) A `t t=(4T)/(4)` displacement is maximum i.e., extreme position, so acceleration is maximum.
(c) A `t t=(T)/(4)` corresponds to mean postion, so velocitly will be maximum at this position.
(d) A `t t=(2T)/(4)=(T)/(2)` corresponds to extreme position, so `KE=0"and"PE=",maximum"`.