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Harmonic Oscillator - Mass With Initial Velocity

  1. Feb 22, 2012 #1
    For a harmonic oscillator with mass M, spring of stiffness k and displacement the force equation is:

    -kx = Md2x/dt2

    How do you handle the situation and work out a solution for x(t) when the mass has an initial velocity. E.g. a mass dropped onto the spring?
     
  2. jcsd
  3. Feb 22, 2012 #2
    It's a pretty simple second order differential equation with sins and coss as it's solutions, you then just solve for x'(0) = initial velocity
     
  4. Feb 22, 2012 #3

    Philip Wood

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    The first thing you do is to solve the DE in the general case, that is without specific initial conditions. You find
    x = A cos [itex]\omega[/itex]t + B sin [itex]\omega[/itex]t

    in which [itex]\omega[/itex] = [itex]\frac{force per unit x}{mass}[/itex]

    and A and B are arbitrary constants. It is these whose values accommodate the initial conditions. So, suppose you knew that at t = 0, x = 0 and [itex]\frac{dx}{dt}[/itex] = v[itex]_{0}[/itex].

    Substituting x = 0, t = 0 into the general solution gives

    x = B sin [itex]\omega[/itex]t. [A = 0.]

    So [itex]\frac{dx}{dt}[/itex] = B [itex]\omega[/itex] cos [itex]\omega[/itex]t

    Now imposing [itex]\frac{dx}{dt}[/itex] = v[itex]_{0}[/itex] when t = 0

    v[itex]_{0}[/itex] = B [itex]\omega[/itex] so B =[itex]\frac{v}{\omega}[/itex]

    Putting in this value for B, we finally have

    x = [itex]\frac{v}{\omega}[/itex] sin [itex]\omega[/itex]t.
     
  5. Feb 22, 2012 #4
    Thanks for the answer Philip!

    I was reading a question about Bungee ropes last night on Webanswers and this equation could probably go some way towards modeling the behavior of a jumper when they have reached the end of the rope which then starts to stretch.
     
  6. Feb 22, 2012 #5

    Philip Wood

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    Yes, it would be an excellent model. You've got to be careful over gravity, which displaces the equilibrium position below the point at which the bungee starts to stretch, but I doubt if this will cause too many problems for you.

    And – not something to worry about yet – the bungee won't obey Hooke's law perfectly.
     
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