At what displacement from the equilibrium is the total energy of a simple harmonic oscillator one - sixths KE and five - sixth PE? Answer should be in terms of the amplitude A.

1 answer

  1. Let A be the full amplitude displacement and k be the sping constant.

    The maximum total energy is
    Et = (1/2) k A^2
    The potential energy for other displacements is
    Ep = (1/2) k x^2

    Ep/Et = 5/6 when (x/A)^2 = 5/6
    x/A = 0.91287

Answer this Question

Still need help?

You can ask a new question or browse more physics questions.