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A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0 s, the block has a displacement of -0.90 m, a velocity of -0.80 m/s, and an acceleration of +2.9 m/s2.

What is the force constant of the motion?

Added Sat, 27 Jun '15

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dispalcemnt of the block executing SHM if given by x = A cos (wt+phi)

here x = -0.9 m

initial velocity v = -0.8 m/s

acclearation a = 2.9 m/s^2

apply relation between initial acclearationa and dispalcemnt as

a = -w^2x

-w^2 = (2.9/-0.9)

W^2 = 3.222

W = 1.795 rad/s

for spring mass system W^2 = K/m

K = 1.795*1.795 *4.8

K = 15.46 K/m

Added Sat, 27 Jun '15

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