Final answer:
The maximum potential energy in a simple harmonic oscillator is equal to the total energy of the system which is conserved. At the maximum displacement, the entire energy is potential, with an equation of U_max = 1/2 kA², where k is the spring constant and A is the amplitude.
Step-by-step explanation:
The maximum value of potential energy in a simple harmonic oscillator is equal to the total energy of the system since the total energy is conserved and is the sum of kinetic and potential energy. At the point of maximum displacement (amplitude), the entire energy is potential, as kinetic energy is zero because the velocity of the mass is zero at this point. Using the equation for kinetic energy (KE = 1/2 mω²A²), where m is mass, ω is angular frequency, and A is amplitude, we can equate this to the maximum potential energy (U = 1/2 kx²) at maximum displacement x = A. In a mass-spring system with mass m and spring constant k, and no dissipative forces, the angular frequency ω is connected to the spring constant and mass by the relationship ω = √(k/m). Therefore, the maximum potential energy U_max can be given by U_max = 1/2 kA², which is the same as the total energy since k is related to ω and m by k = mω². When a mass is attached to the spring and set into simple harmonic motion, at the amplitude A, the potential energy is at its maximum because the mass has momentarily come to rest before reversing direction.