Speed of object with springs and equilibrium?

A 0.390-kg mass is attached to a spring with a force constant of 25.9 N/m and released from rest a distance of 3.20 cm from the equilibrium position of the spring. Calculate the speed of the mass when it is halfway to the equilibrium position.Speed of object with springs and equilibrium?
Total energy TE = 1/2 kdX^2; where k = 25.9 N/m and dX = .032 m.





From the conservation of energy TE = 1/2 kdX^2 = pe + ke = 1/2 k (dX/2)^2 + 1/2 mv^2 = TE; where m = .39 kg and v = ? the velocity you are looking for.





Therefore 1/2 k dX^2 (1 - 1/4) = 1/2 mv^2 and v^2 = 3/4 k dX^2/m and v = sqrt(3/4 k dX^2/m) = sqrt( 3/4 25.9*(.032)^2/.39) = ? You can do the math.





The physics is this. The system started with only the PE of the spring when at dX = .032. But as the spring/mass contracted, some of the PE converted into ke and some left over pe because the spring was still a bit strecthed (dX/2).