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Differential equation: Spring/Mass system of driven motion with damping

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data
    A 32 pound weight stretches a spring 2 feet. The mass is then released from an initial position of 1 foot below the equilibrium position. The surrounding medium offers a damping force of 8 times the instantaneous velocity. Find the equation of motion if the mass is driven by an external force of 2cos(5t).


    2. Relevant equations
    F=kx
    m=W/g

    m[itex]\frac{d^{2}x}{dt^{2}}[/itex]+[itex]\beta[/itex][itex]\frac{dx}{dt}[/itex]+kx=f(t)


    3. The attempt at a solution

    I found that k=16[itex]\frac{lb}{ft}[/itex] and m=1 slug. This gets me the following equation:

    [itex]\frac{d^{2}x}{dt^{2}}[/itex]+[itex]\beta[/itex][itex]\frac{dx}{dt}[/itex]+16x=2cos(5t)

    I'm at a loss for how to determine [itex]\beta[/itex], which is the damping force of 8 times the instantaneous velocity. I don't know how to determine instantaneous velocity. I know that once I have [itex]\beta[/itex], I can just use a LaPlace transform to find x(t). But [itex]\beta[/itex] is my stumbling block right now.

    As I was writing this, it occured to me that [itex]\frac{dx}{dt}[/itex]=instantaneous velocity and that would make [itex]\beta[/itex]=8. That in turn makes the problem very easy to solve. Am I correct in this thinking?

    We kind of rushed through this application in class the other day. Thanks in advance for any help you're able to provide.
     
  2. jcsd
  3. Aug 7, 2011 #2

    rock.freak667

    User Avatar
    Homework Helper

    Yeah I would agree that β=8 here. Just make sure your initial conditions are correct with the proper signs.
     
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