F = -kx

-2(9.8) = -k(0.25)

k = 78.4 N/m

w = (k/m)1/2 = (78.4/2) = 6.26 rad/s
 

For a mass released from rest above the equilibrium position the equation is a cosine curve.

y = A cos wt = 0.2cos6.26t

-0.1 = 0.2cos6.26t

6.26t = 120º or -120º

For the second time it is at -10 cm, the angle is -120º or 240º (see the figure)

6.26t = 240º = 4.189 rad

t = 4.189/6.26 = 0.669 s
 

v = -Awsinwt = -0.2(6.26)sin6.26t

At t = 0.669 s

v = -0.2(6.26)sin(6.26)(0.669) = -0.2(6.26)sin4.189 rad = -0.2(6.26)sin240º

v = -1.08 m/s
 

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