A 0.25 kg harmonic oscillator has a total oscillation energy of 9.9 J. If the oscillation amplitude is 20.0 cm?

KE= (1/2) m v^2 PE = (1/2) k x^2 k = 30.3 N/m v(max) = 3.42 m/s m=2.55kg Since there are no net forces acting on the system and there is no friction, the mechanical energy is conserved. At the equilibrium point, there is no PE because the spring is not compressed at all (x=0), so its mechanical energy is entirely kinetic. E(total) = KE + PE = KE + 0 = (1/2) (2.55kg) (3.42 m/s)^2 = 14.91291 J When the spring is extended to its maximum point, the amplitude (x=A), the energy is entirely potential.

KE (@ equilibrium) = PE (@ amplitude) 14.91291 J = (1/2) k x^2 14.91291 J = (1/2) * 30.3 N/m * x^2 x ~= +-0.992144 ~ +- 1.0 meters Because mechanical energy is conserved in this setup, the total mechanical energy never changes, so the answer to part B) is still 14.91291 J, some of which is in kinetic and some in potential, but the sum is still the same.

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