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