# Homework Help: Differential Equation - Springs

1. Mar 8, 2012

### τheory

1. The problem statement, all variables and given/known data
A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.

Find the displacement function x(t)

2. Relevant equations
Fg = mg
Fs = -kx
Fnet = ma

3. The attempt at a solution
Fnet = Fg + Fs
ma = mg + (-kx)
mx'' = mg + (-kx)
mx'' + kx = mg

Converting Units:
m = 1 kg = 2.2 lbs
g = -9.8 m/s^2 = -32.1 ft/s^2
k = 100 lbs/ft

Solving Second Order Non-Homogeneous DE:
mx'' + kx = mg
2.2x'' + 100x = -70.6

Particular Solution:
Xp = A
X'p = 0
X''p = 0

2.2*0 + 100*A = -70.6
A = -7.06

2.2x'' + 100x = 0
Roots = 0 ± 6.742i

x = -7.06 + C1*cos(6.742t) + C2*sin(6.742t)

Using Conditions:
x = -1 inch or -0.083 feet when t = 0
x' = 0 ft/s when t = 0

-0.083 = -7.06 + C1*cos(0) + C2*sin(0)
-0.083 = -7.06 + C1
C1 = 6.976

x' = -6.742*C1*sin(6.742t) + 6.742*C2*cos(6.742t)
0 = -6.742*C1*sin(0) + 6.742*C2*cos(0)
0 = 6.742*C2
C2 = 0

Final Equation:
x = -7.06 + 6.976*cos(6.742t)

Is this right?

2. Mar 8, 2012

### τheory

Turns out, I incorrectly assumed the situation pertaining the the spring; the spring is not being hung from the ceiling as I initially presumed, it is set sideways. Thus, gravity is not a force to be considered in this problem. Another problem was that I utilized 1 lb as the mass. However, mass is in terms of grams and I neglected to convert weight to mass. In the end, after correcting these mistakes, I was able to submit the answer on Webwork and got it right.:grumpy:

Last edited: Mar 8, 2012