# Oscillating, hanging spring-block system

1. Jan 17, 2012

### Color_of_Cyan

1. The problem statement, all variables and given/known data

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?

2. Relevant 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
3. 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 in to x(t) = Acos(ωt + ∅) it would be wrong?

2. Jan 18, 2012

### RTW69

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

3. Jan 18, 2012

### Color_of_Cyan

is A 16.5 cm ?

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

4. Jan 18, 2012

### Staff: Mentor

Yes. (What will ∅ equal?)

5. Jan 18, 2012

### Color_of_Cyan

i would have to guess and say that ∅ = 0

6. Jan 18, 2012

### Staff: Mentor

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