Let U be the potential energy (with the zero at zero displacement)and K be the kinetic energy of a simple harmonic oscillator.U_avg and K_avg are the average values over a cycle.Then: A)K_avg U_avg B)K_avg

Question

Let U be the potential energy (with the zero at zero displacement)and K be the kinetic energy of a simple harmonic oscillator.U_avg and K_avg are the average values over a cycle.Then: A)K_avg > U_avg B)K_avg < U_avg C)K_avg = U_avg D)K = 0 when U = 0 E)K + U = 0

Quizplus · Accepted Answer

In a simple harmonic oscillator, the potential energy and kinetic energy are continuously interchanging. At the equilibrium position, the kinetic energy is maximum and potential energy is minimum. At the maximum displacement from the equilibrium position, the kinetic energy is minimum and potential energy is maximum. Therefore, over a cycle, the average value of kinetic energy is equal to the average value of potential energy, i.e., K_avg = U_avg. Thus, option C is correct. Option A and B can be rejected easily as they compare kinetic and potential energies without considering their mutual relationship. Option D is incorrect because the kinetic energy is maximum at the equilibrium position and potential energy is minimum. Option E is incorrect because the sum of kinetic and potential energies is constant and equal to the total energy of the oscillator.

Let U Be the Potential Energy (With the Zero at Zero