Final answer:
At 1/3 and 5/6 of the period T after passing through the equilibrium point, an object's potential energy in simple harmonic motion is half of its kinetic energy.
Step-by-step explanation:
An object of mass m attached to a spring with spring constant k exhibits simple harmonic motion (SHM). The question asks at what fraction of a period after passing through the equilibrium point the object's potential energy (PE) will be half its kinetic energy (KE). In simple harmonic motion, the energy of the object oscillates between kinetic and potential energy, with the sum remaining constant. At the equilibrium point, all the energy is kinetic, given by KE = 1/2 mv². In SHM, the potential energy is defined by PE = 1/2 kx².
When the potential energy is half the kinetic energy, we have PE = 1/2 KE. Substituting the expressions for kinetic and potential energy and simplifying, we find that x (the displacement from the equilibrium) is equal to ±(1/2)m²/k. Using the relationship between the displacement and the phase of the oscillation, we can determine the fraction of the period at which this condition occurs.
Without going into the detailed calculations, we know from the theory of simple harmonic motion that this occurs at 1/3 and 5/6 of the period T, from passing through the equilibrium point, reflecting the symmetry of the position's cosine function over one period.