physics
posted by AK on .
A 27.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.30 s. What is the position of the mass 3.526 s after the mass is released?
I tried using the equation x(t)=x*cos(ωt+φ) however I keep getting the answer to be 0.3486 m and according to the homework that is incorrect. Any guidance would be appreciated.

Thanks for showing some evidence of your own work.
The mass will be released when the spring & mass reach the equilibrium position, (x=0), together. That takes 1/4 of the period, or 1.075 s. After release, the object travels at constant velocity. It is not clear whether by "release" they mean separation from the spring or the release of the object from the initial position.
For the velocity of the object when it is released (V) , used conservation of energy.
(1/2)kX^2 = (1/2)M V^2.
X is the initial spring compression.
You can figure out what k is using the period, P, as follows:
P = 2*pi*sqrt(m/k)