Asked by Abdela
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?
Answers
Answered by
Damon
y = .012 sin (2pi t/.4)
v = dy/dt = (.012)(2 pi/.4)cos (2 pi t/.4)
max v = .012 * 2 * pi / .4
v = dy/dt = (.012)(2 pi/.4)cos (2 pi t/.4)
max v = .012 * 2 * pi / .4
Answered by
Mohammed sultan
4₹cm/s
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