Physics

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

asked by Anushka
  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.

    posted by Steve

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