What is the potential energy at 1/2A in simple harmonic motion?

[Eagles342]

A particle of mass is connected to a spring with a force constant K. The particle undergoes simple harmonic with an amp A. What is the potential energy of the partic when the position is (1/2A)?

Homework Equations

E=1/2kA^2
1/2kdelta^2=1/2mv^2+1/2kx^2

The Attempt at a Solution

U=1/2kx^2=1/2k(1/2A)^2=1/4(1/2KA^2)=1/4E= K=E-U= 3/4? Not really sure at all can somebody help i don't think i did it correctly

 

[Simon Bridge]

Welcome to PF;
You started strong - the formula for PE is correct.
You substituted x=A/2 fine.

You have been asked to find U - so stop when you have.

 

[dauto]

Besides, E-U = 3/4 makes no sense.

 

[BiGyElLoWhAt]

1/4(1/2KA^2) doesn't equal 1/4E, it equals 1/4 Umax.
gotta be careful.