Question

In an angular shm angular amplitude of oscillation is pi rad and time period

Answer 1

The angular velocity will be shm be like w.  

So then w=2π/T.  

If suppose we have taken t=0 the particle will be mean the position to an displacement at given time is

X=Asinwt

At some point t, X = A ( Particle is at mean extremity)

Thus sinwt=1 hence wt = π/2

t=T/4

Therefore the SHM of a time from a mean is A/2 which will be remains same for half amplitude.