Oscillating, hanging spring-block system

[Color_of_Cyan]

Homework Statement

A hanging spring stretches by 30.0 cm when an object of mass 500 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 16.5 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later?

Homework Equations

Fs = -kx
Fs = mg

x(t) = Acos(ωt + ∅)

w = (k/m)^1/2
w² = k/m

T = 2∏/w

E = (1/2)kA²
E = (1/2)mv² + (1/2)kx²

v = =- [ (k/m)(A² - x²) ]^1/2

The Attempt at a Solution

m = 0.5 kg

All I can do right off the bat is solve for k

Fs = (9.8 m/s2)(0.5kg) = 4.9N

4.9N = -kx

4.9N = (-k)(-0.3m)

k = 16.33 N/m

Now solve for ω ?
w = [ (16.33 N/m) / 0.5 kg ]^1/2

ω = 5.7 s^-1

Not sure where to go from here regarding energy and I'm afraid that if I plug into x(t) = Acos(ωt + ∅) it would be wrong?

 

[RTW69]

you know ω and t. Do you know how to find the amplitude-A?

 

[Color_of_Cyan]

is A 16.5 cm ?

so the problem is over if i use x(t) = Acos(ωt + ∅)

 

[Doc Al]

is A 16.5 cm ?

so the problem is over if i use x(t) = Acos(ωt + ∅)

Yes. (What will ∅ equal?)

 

[Color_of_Cyan]

i would have to guess and say that ∅ = 0

 

[Doc Al]

i would have to guess and say that ∅ = 0

Correct. (But you shouldn't have to guess!)