Max extension of spring/ magnitude of its acceleration

[WPCareyDevil]

Homework Statement

An ideal spring has a spring constant k = 30 N/m. The spring is suspended vertically. A 1.1 kg body is attached to the unstretched spring and released. It then performs oscillations.


(a) What is the magnitude of the acceleration of the body when the extension of the spring is a maximum in m/s^2

(b) What is the maximum extension of the spring in m?

Homework Equations

kx=mg
=> x=mg/k

a(max)=(A)(w^2)
w=2pi/T
T=2pi*(sqrt(m/k))

The Attempt at a Solution

k=30
m=1.1
a=?
x=?

I have to solve part B (amplitude) before I can solve part A. So, simple algebra sets

x=mg/k
x=(1.1)(9.8) / 30
=.35933 m

But this is wrong according to web assign. Could you tell me what I am doing wrong? Perhaps units or something?

Thank you!

 

[LowlyPion]

You might want to account for the conservation of energy:

PE = PE

m*g*x = 1/2*k*x²

 

[ak1948]

At the minimum extension (starting point) all energy is gravitational; at the maximum extension, all energy is in the spring. This is how you solve b.

Only at the equilibrium point midway between is kx = mg.