A particle executing simple harmonic motion of amplitude 5 cm and period 2 s. Find the speed of the particle at a point where its acceleration is half of its maximum value

Answer 1

x = 5sin(πt)
x" = -5π^2 sin(πt)

x" = 5/2 at two times. You can use the graph at

http://www.wolframalpha.com/input/?i=sin(t)+%3D+1%2F(2pi%5E2)

to see which value to use.