Solving Energy of a Spring: Find Velocity at 2/3 E

[ML1]

Homework Statement

An object of mass m attached to a spring of force constant oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the object's velocity when its potential energy is 2/3 E ?

Homework Equations

E=KE+PE_s

The Attempt at a Solution

1/2kA^2 = 2/3(1/2kA^2) + 1/3(1/2mv^2)

solving for v...

1/6mv^2 = 1/2kA^2 - 1/3kA^2

1/6mv^2 = 1/6 kA^2

mv^2 = kA^2

v= sqrt(k/m) A

But it's wrong...any help? Thanks!

 

[Delphi51]

velocity when its potential energy is 2/3 E

when PE = 2/3*E then KE = 1/3*E
At maximum displacement, v = 0, KE = 0, so E = PE =1/2*k*A²

1/2kA^2 = 2/3(1/2kA^2) + 1/3(1/2mv^2)

This says E = 2/3*(maxPE) + 1/3*(maxKE) and is only true when v is the maximum velocity. Not the velocity you are looking for in the question.