a .5 kg mass is attached to a spring of spring constant 20 along a horizaontal, frictionless surface. the object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. at what location are the kinetic energy and the potential energy the same?

Question

anonymous · Answer

At center, you have no potential energy because the string is not stretched.
Therefore, the total energy of the system at the equilibrium position is
U(total) = 1/2 m v^2 = 1/2 * 0.5 * 1.5^2 = 0.5625
We want to know a point where half of that is in the spring energy and half of that is in the kinetic energy. This wil be where the spring itself has 0.5625/2 J of energy = 0.28125 Joules.
1/2 k x^2 = 0.2815
1/2 * 20 x^2 = 0.2815
Solve for x
x = +-  0.168 m. 
This means that when the mass is 0.168 m in front of or behind the spring's eq position, it will have the total energy of the system divided equally bbetween kinetic energy and spring energy