Animation 1 | Animation 2

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We begin by considering the classical simple harmonic motion of a mass on a spring. We have chosen the mass of the ball on the spring to be 0.5 kg and the spring constant to be 2 N/m (position is given in meters and time is given in seconds).  Given Hooke's law, we can write the force as F_x = −kx in one dimension. This also allows us to write

F_x = ma_x = m d²x/dt² = −kx,

which using ω = (k/m)^1/2, can be written as

d²x/dt² = −ω ²x . (12.1)

This equation has the general solution x(t) = Acos(ωt) + Bsin(ωt), where the coefficients A and B are determined by the initial conditions (x₀ and v₀).

Animation 1 shows the graphs of kinetic and potential energy versus time. They have the form of cos² (the potential energy) and the form of  sin² (the kinetic energy).  We know from simple harmonic motion that if the object is initially displaced from equilibrium with no initial velocity that the solution to the above equation is

x = x₀cos(ωt) and v = dx/dt = −ωx₀sin(ωt). (12.2)

Given the form of the kinetic energy (T) and the potential energy (V), we have that

T(t) = (kx₀²/2) sin²(ωt) and V(t) = (kx₀²/2) cos²(ωt).

Animation 2 shows the graphs of kinetic and potential energy vs. position. The potential energy can be found from

V = −∫ F dx = (1/2)kx² = (1/2)mω²x².

Since the total energy is the sum of the kinetic and the potential energies, we have that

E = p²(t)/2m + mω²x²(t)/2.  (12.3)

In the animation, the energy starts out all potential, at the equilibrium position the energy is all kinetic, and at maximum compression the energy is all potential again. The classical particle on a spring is never allowed beyond the point where all of its energy is potential (otherwise its kinetic energy would be negative), as it is classically forbidden.

Original script authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.