A 28.0 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 4.70 s. What is the position of the mass 3.854 s after the mass is released? T=2pi*sqrt m/k KE1+PE1=KE2+PE2 from this you get mv^2=KA^2 x=Asin *omega*t omega=2pi*frequency frequency =1/T I used the first equation get k=3.92 then I use that k in the second equation, but I dont know how to get v to then get A. Any help? Would I then take that A and plug it into the third equation to get my X?