A 4.9 kg particle is in simple harmonicmotion in one dimension and moves accordi

Question

A 4.9 kg particle is in simple harmonicmotion in one dimension and moves according to the followingequation, with t in seconds. x = (3.7m)cos[(/2 rad/s)t -/2 rad] (a) At what value of x is thepotential energy of the particle equal to half the totalenergy?
1 m

(b) How long does it take the particle to move to this positionx from the equilibrium position?
2 s x = (3.7m)cos[(/2 rad/s)t -/2 rad] (a) At what value of x is thepotential energy of the particle equal to half the totalenergy?
1 m

(b) How long does it take the particle to move to this positionx from the equilibrium position?
2 s

Explanation / Answer

m = 4.9 kg x = (3.7m)cos[(/2rad/s)t - /2rad] amplitude A = 3.7 m
a) when the potential energy kx2/2 equals to halfthe total energy kA2/2, kx2/2 =(1/2)*(kA2/2) x = A/2 = ±2.616 m

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