Simple Harmonic Motion: Block dropped onto a spring

[gravity_dress]

Homework Statement

A 2-kg block is dropped from a height of 0.45 m above an uncompressed spring. The spring has an elastic constant of 200 N/m and negligible mass. The block strikes the end of the spring and sticks to it.
(a) Determine the speed of the block at impact
(b) Determine the period of SHM that ensues
(c) Determine the distance that the spring is compressed at the instant the speed of the block is maximum
(d) Determine the maximum compression in the spring
(e) Determine the amplitude of the SHM

Homework Equations

v^2 = 2ax
T = 2pi[rt(m/k)]
Vmax = (angular frequency omega)(amplitude x naught)
mgh=(1/2)(k)(x^2)
x(t)=(amplitude)[cos(omega t)]
v(t)=(-omega)(amplitude)sin(omega t)
a(t)=(-omega^2)(amplitude)cos(omega t)

The Attempt at a Solution

Using kinematics' v^2=2ax, obtained 2.97 m/s for part (a)
Using T = 2pi[rt(m/k)], obtained 0.628 sec for the period in part (b) --> Am I right that omega = rt(m/k) in this case? Or should it be rt(k/m)? What determines which it is?
Using conservation of energy, obtained x=0.41 m or 41 cm for part (d)

...But I'm clueless on (c) and (e). I guessed that, to solve (c), I should find what t was when sin(omega t)=1; But that gives me something preposterous near 15 seconds. How should I go about finding amplitude? How should I use the equations for x(t) and v(t)? Am I wrong to ignore the displacement factor usually included at the end of these equations?

 

[gravity_dress]

Never mind, got it!

For anyone wondering, or with a similar problem: for part (c), set the forces equal: use ma=kx. That x is equilibrium.
For part (e), find the difference between Xmax--what you found using conservation of energy in d (for me it was 0.41 m)--and the x at equilibrium found in (c). In my case, the amplitude is 0.312 m.

Always use rt K/M in simple mass-spring systems.