SHM - Horizontal Spring Question

[GX255]

1. A 41g object is attached to a horizontal spring with a spring constant of 15N/m and released from rest with an amplitude of 22.4cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.

Homework Equations

W=Square Root of (K/m)
1m=100cm
V^2+(wx)^2=(wA)^2
w= omega
A= Amplitude

The Attempt at a Solution

K= 15N/m
m= 41kg
A=22.4cm=0.224m
W=sq root of K/m = 0.605

V^2=(wA)^2
V = 0.0677

 

[Noein]

Since equilibrium halfway x should = 0 ?

Halfway to the equilibrium position would be half the amplitude.

 

[fanie1031]

I have the same problem. I set it up two different ways.
v^2=k/mA^2-K/mx^2, then square root
A-full amplitude
X-half the amplitude
or
v=[(k/m)(A^2*X^2)]^(-1/2)

but none worked.

I'm pretty sure I'm setting up the amplitude stuff wrong.