A 4.1 kg particle is in simpleharmonic motion in one dimension and moves accordi

Question

A 4.1 kg particle is in simpleharmonic motion in one dimension and moves according to thefollowing equation, with t in seconds. x = (2.8m)cos[(/4rad/s)t - /4rad] (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 = (2.8m)cos[(/4rad/s)t - /4rad] (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

Given mass m = 4.1kg and equation x = (2.8m)cos[(/4rad/s)t - /4rad]        ---------( 1) compare this with x = A cos ( wt - ) we get amplitude A = 2.8 m angular frequency w = / 4 rad / s (a). P.E = ( 1/ 2) T.E ( 1/ 2) m w ^ 2 x ^ 2 = ( 1/ 2) ( 1/ 2) m w ^ 2 A ^2                        x ^ 2 = ( 1/ 2) A ^ 2 from thsi x = ± A / 2                 = ± 2.8 m / 2                = 1.979 m (b). substitue this x value in eq ( 1 ) weget 1.979 = 2.8 cos (0.785 t - 0.785 ) cos ( 0.785 t - 0.785 ) = 0.707        0.785 t -0.785 =0.785                            t = 2 s

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