A 1.54 x 10-2-kg block is resting on a horizontal friction-less surface and is attached to a horizontal spring whose spring constant is 129 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 8.84 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Answer 1

v=Aw
A=v/w
w=sqrt of k/m
w=91.52
A=8.84/91.52
A=0.0966m

Answer 2

initial KE=1/2*m v^2
ending PE=1/2 k A^2
set them equal

1/2 k A^2=1/2 m v^2
A^2=mass/k v^2
= 1.54E-2/129 * 8.84^2 == 0.00932900961

A=sqrt (0.00932900961)
A== 0.0966 meters