Solve for Displacement in Harmonic Oscillator

[flower76]

Can someone check my work please I'm pretty sure I don't have the right answer but I can't figure out what I have wrong.

The question is:
A simple harmonic oscillator has total energy E=1/2kA^2
where A is the amplitude of oscillation.
For what value of the displacement does the kinetic energy equal the potential energy?

So I figure that if KE is equal to PE, then PE=1/2E

Therefore:

1/2kx^2 =1/2(1/2kA^2)
kx^2 = 1/2kA^2
x = 1/4A

Any ideas?

 

[Doc Al]

So I figure that if KE is equal to PE, then PE=1/2E

Good.

Therefore:

1/2kx^2 =1/2(1/2kA^2)

Good.

kx^2 = 1/2kA^2

Good.

x = 1/4A

Not good.

 

[flower76]

I think I see my error.

Is the answer x = 0.71A ?

 

[Doc Al]

Yep. $x = (1/\sqrt{2}) A$