Asked by ami
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.
Answers
Answered by
drwls
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
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
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