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Finding Displacements for forced oscillations

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle of mass [tex] m [/tex] is at rest at the end of a spring (force constant [tex] = k[/tex]) hanging from a fixed support. At [tex] t = 0 [/tex], a constant downward force [tex] F [/tex] is applied to the mass and acts for a time [tex] t_0 [/tex]. Show that, after the force is removed, the displacement of the mass from its equilibrium position ([tex] x = x_0 [/tex], where [tex] x [/tex] is down) is
    [tex] x - x_0 = \frac{F}{k}[cos \omega_0(t-t_0) - cos \omega_0 t] [/tex]​
    where [tex] \omega_0^2 = \frac{k}{m} [/tex].


    2. Relevant equations
    Newton's 2nd law [tex] F = ma [/tex] and knowing how to solve second order linear ODEs.



    3. The attempt at a solution
    Okay, so I have no idea if what i'm doing is right, but i've been working on this (trying different things) for about 6 hours now.. It's painful. So, any assistance would be appreciated.

    First, I split this thing into two parts.. one after the force is removed and one while the force is applied.

    After force is removed: ([tex] t \geq t_0 [/tex] and [tex] t = 0 [/tex])
    [tex] mg - kx = m \ddot{x} [/tex]
    [tex] \ddot{x} + \omega_0^2 x = g [/tex]
    We know [tex] x(t) = Acos(\omega_0 t - \phi ) + \frac{g}{\omega_0^2} [/tex].
    We also know that at [tex] t = 0, x = x_0, [/tex] so:
    [tex] x_0 = Acos\phi + \frac{g}{\omega_0^2} [/tex]
    Now, the displacement from the equilibrium is:
    [tex] x - x_0 = Acos(\omega_0 t - \phi ) - Acos\phi [/tex]
    Simplifying, [tex] x - x_0 = A[cos(\omega_0 t - \phi ) - cos\phi] [/tex]

    While the force is applied: ([tex] 0 < t \leq t_0 [/tex])
    [tex] F + mg - kx = m \ddot{x} [/tex]
    [tex] \ddot{x} + \omega_0^2 x = \frac{F + mg}{m} [/tex]
    So, [tex] x_2 (t) = Acos(\omega_0 t - \phi) + \frac{F+mg}{m\omega_0^2} [/tex]



    Now, at [tex] t = t_0 [/tex], we know [tex] x = x_2 [/tex], so,
    [tex] Acos(\omega_0 t_0 - \phi ) + \frac{g}{w_0^2} = Acos(\omega_0 t_0 - \phi) + \frac{F+mg}{m\omega_0^2} [/tex]
    But this gives me [tex] F = 0 [/tex] which is of course true, because [tex] t = t_0 [/tex] is the time when the force is removed. I just do not know where to go and if this is helpful at all. Eventually, i want to have [tex] A = \mbox{something} [/tex] so that i can plug that into my equation for displacement from the equilibrium [tex] x - x_0 [/tex] above.

    What am I doing wrong? What should I do? Where should I go?
     
  2. jcsd
  3. Oct 7, 2009 #2

    gabbagabbahey

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    If the force is applied at t=0, for a time [itex]t_0[/itex], doesn't that mean

    [tex]F+mg-kx=m\ddot{x}, \;\;\;\;\; 0\leq t\leq t_0[/tex]

    [tex]mg-kx=m\ddot{x}, \;\;\;\;\; t> t_0[/tex]


    so that your equilibrium position is actually [itex]x_0=A\cos(\phi) + \frac{F+mg}{m\omega_0^2}[/itex]?
     
  4. Oct 7, 2009 #3
    okay, so, using your advice.. I have, as you said..

    [tex] x_0 = A\cos(\phi) + \frac{F+mg}{m\omega_0^2} [/tex]

    Therefore, our displacement from equilibrium is:

    [tex] x - x_0 = A[cos(\omega_0 t - \phi) - cos\phi] - \frac{F}{m\omega_0^2} [/tex]

    Still though, when we have time [tex] t = t_0 [/tex], [tex] x = x_2 [/tex]. This still cancels the [tex] Acos(\omega_0 t_0 - \phi) [/tex] from both sides and leaves [tex] F = 0 [/tex]. Again, I don't know where to go from here to clean up the expression for [tex] x - x_0 [/tex].
     
  5. Oct 9, 2009 #4

    gabbagabbahey

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    I think you are confusing yourself by using the same variables to represent the amplitude and phase of the oscillation during both intervals...

    During the interval [itex]0\leq t\leq t_0[/itex] you have:

    [tex]x (t) = A_1\cos(\omega_0 t - \phi_1) + \frac{F+mg}{m\omega_0^2} [/tex]

    and during the interval [itex]t\geq t_0[/itex], you have:

    [tex]x (t) = A_2\cos(\omega_0 t - \phi_2) + \frac{g}{\omega_0^2} [/tex]

    Use the fact that [itex]x(0)=x_0[/itex] and [itex]x(t)[/itex] and [itex]x('t)[/itex] must be continuous at [itex]t_0[/itex]....
     
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