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The Differential Equation of Free Motion

  1. Oct 3, 2009 #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.


    2. Relevant equations
    mx"+kx=0


    3. The attempt at a solution

    I understand that I need to set up a differential equation but I'm not quite sure if this is right.

    K = 100 lb/ft m = 1 lb

    thus plugging into the equation

    x" + 100x = 0

    use characteristic equation,

    I get r= +/- 10i

    y[tex]^{g}[/tex](t)= cos (10t) + sin (10t)


    So far is everything is alright?

    When I tried to solve for period. I set the inside, 10t=2pi. At least that's how my professor does it but apparently it's wrong when I submit my answers online.

    I'm also not understanding how the initial conditions work?
     
  2. jcsd
  3. Oct 3, 2009 #2

    LCKurtz

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    You need constants A and B in front of your trig terms to get the general solution. You are given the initial position is one inch (watch your units), plus or minus depending on your coordinates, and apparently the initial velocity is zero. Use these two facts to determine your A and B.
     
  4. Oct 3, 2009 #3
    Thanks anyway but I solved this problem.

    I rethought about the mass and realized that 1 pound is equal to mg. Thus 1/32 is the actual mass.

    So the equation should have been

    1/32x" + 100x=0, x(0)=1/12, x'(0)=0

    So yeah, now everything makes sense.
     
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