A frictionless spring with a 10-kg mass can be held stretched 1 meters beyond its natural length by a force of 40 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2.5 m/sec, find the position of the mass after t seconds.
How do I convert this to a second order equation and then solve it?
Spring constant would be 40nm = k * 1 meter -> k = 40
ATTEMPTS BASED ON COMMENTS/ANSWERS
10x'' + 40x = 0, when x(0) = 1 and x'(0) = 2.5 10r^2 + 40 = 0 10r^2 = -40 r^2 = -4 r = +- 2i y = Acos(2x) + Bsin(2x) = homogenous solution y' = -2Asin(2x) + 2Bcos(2x) 1 = Acos(2*0) + Bsin(2*0) = A 2.5 = -2Asin(2*0) + 2bcos(2*0) = 2B y = cos(2x) + 5/4sin(2x)