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Undamped Harmonic Motion (ODE problem)

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A 24-lb weight, attached to the end of a spring, stretches it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.


    2. Relevant equations
    [tex]\frac{d^{2}{x}}{dt^2}+\frac{k}{m}x=0[/tex]
    [tex]F=ma[/tex]


    3. The attempt at a solution

    First I found the spring constant k using [itex]F=ma=ks[/itex] at the point of equilibrium where the weight isn't moving. s is the displacement the weight causes on the spring, which is 4 inches, or 1/3 feet. So that equation turns out to be the weight equals the mass times the acceleration of gravity, which equals the spring constant times the displacement the weight caused, which is [itex]24=k(\frac{1}{3})[/itex].

    I found mass, m, by using [itex]F=ma[/itex] which is [itex]24lb=m(32\frac{ft}{s^2})[/itex] which gives me [itex]\frac{3}{4} slug[/itex]

    Putting that into my relevant equation at the top, it gives me [itex]\frac{d^{2}x}{dt^2}+96x=0[/itex] And using that one method (I forgot the name), I get [itex]m^2+96=0[/itex] which gives me [itex]m=\pm\sqrt{96}i[/itex]

    I use that for [itex]C_{1}cos\sqrt{96}t+c_{2}sin\sqrt{96}t[/itex]

    Now I have to find initial conditions to solve for [itex]C_{1}[/itex] and [itex]C_{2}[/itex]

    My initial position is [itex]\frac{1}{4}ft[/itex] above the equilibrium point, so does that mean I set that equation equal to [itex]-\frac{1}{4}[/itex] and then solve for when t = 0?

    And to get the other one, do I need to find the equation for the velocity by deriving the position equation? I'm not sure what I would set that equation equal to.

    Thanks.
     
  2. jcsd
  3. Mar 20, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    Yes, if x(t) is the position then you know x(0)=(-1/4) (if you want displacements above equilibrium to be negative). Since it's release "from rest" you also know velocity at t=0 is 0. So set x'(0)=0.
     
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