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April 1, 2015

April 1, 2015

Posted by **Joe** on Wednesday, April 18, 2012 at 2:50pm.

How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

- Physics -
**Elena**, Wednesday, April 18, 2012 at 5:56pmx=A•sinωt

v =dx/dt = A•ω•cosωt

PE = KE

kx^2/2 = mv^2/2

k = m•ω^2

m•ω^2• (A•sinωt)^2/2 = m•( A•ω•cosωt)^2/2

(sinωt)^2 = (cosωt)^2

(tan ωt)^2 = 1

ωt = π/4

x=A•sinωt = A•sin(π/4) =0.707•A

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