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Differential Equation - Springs

  1. Mar 8, 2012 #1
    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. jcsd
  3. Mar 8, 2012 #2
    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
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