A 0.27 kg mass is suspended on a spring that stretches a distance of 4.9 cm. The mass is then pulled down an additional distance of 12.5 cm and released. What's the displacement from the equilibrium position with the mass attached (in cm) after 0.42 s? Take up to be positive and use g = 9.81 m/s2.

**2. Relevant equations**

F=kx

ω(angular velocity)= (k/x)^0.5

**3. The attempt at a solution**

First I solved for k:

k=mg/x

k=(0.27*9.81)/4.9

k=0.54

Then I solved for the angular velocity:

ω=(0.54/0.27)^0.5

ω=squareroot 2

Then I tried to come up with an equation for displacement using the format:

x(t) = Asin(ωt)

i know that ω=root2

t=0.42

but how do i solve for A?? or what is A? should it be 4.9 or 12.5 or the sum of the two btw please correct me if im wrong in my calculation. thanks!