The motion of the baby can be described by the equation

x=-Acos(wt), where w=(k/m)^(1/2)

We know that x=0.20m, A=0.45m (the distance the system was initially displaced), and we
can calculated w to be [(98/0.3)/10]^(1/2)=5.7

Now we plug the numbers in:

0.20=(-0.45)cos(5.7t), which becomes
-(4/9)=cos(5.7t)

Now take the inverse cosine of both sides, and solve for t.

arccos(-4/9)=5.7t
t = [arccos(-4/9)]/5.7
t = 0.35s

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