Question

A particle that undergoes simple harmonic motion has a period of 0.4 s and an amplitude of 12 mm. The maximam velocity of the particle is?

Answer 1

y = .012 sin (2pi t/.4)

v = dy/dt = (.012)(2 pi/.4)cos (2 pi t/.4)

max v = .012 * 2 * pi / .4

Answer 2

4₹cm/s