A particle executes S.H.M with amplitude of 10cm and period of 10s. Find the velocity and acceleration of the particle at a distance 5cm from the equilibrium position

Answer 1

y = 0.10 sin (2 pi * t/10) = 0.10 sin( pi t/5)
v = 0.10 (pi/5) cos (pi t/5)
a = -0.10 (pi/5)^2 sin(pi t/5) = -(pi/5)^2 y
now
find t for y = 0.05
use that t for your v and a