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Mass-spring-dampened system

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A mass-spring-dampener system is applied a force [itex]mg[/itex] and is immediatly removed, setting the system in motion. The system is constantly applied force [itex]Mg[/itex] and is static at [itex]y=y_0[/itex].
    Find a formula for both [itex]A[/itex] and [itex]\phi[/itex]


    2. Relevant equations

    [itex]\ddot{y}+2\delta\dot{y}+w_0^2y=0[/itex]
    [itex]\frac{2\pi}{w_0}=T_0[/itex]
    [itex]\delta = \frac{3}{5}w_0[/itex]
    [itex]F_f=-b\dot{y}[/itex]
    [itex]Mg=ky_0[/itex]

    3. The attempt at a solution

    from this i find [itex]k[/itex] and [itex]b[/itex]. No problem, not part of my question.

    when the force is applied, the system 'moves' in y direction and is set in motion, given function:

    [itex]y(t)=Ae^{-\delta t}cos\left(w_d t+\phi\right)[/itex]
    [itex]w_d=\sqrt{w_0^2+\delta^2}[/itex]

    I'm to find [itex]A[/itex] and [itex]\phi[/itex]

    my try:

    I understand [itex]\dot{y}(0)=0[/itex] gives:
    [itex]\dot{y}=-\delta Ae^{-\delta t}cos(w_d t+\phi)-w_d A e^{-\delta t}sin(w_d t + \phi)[/itex]
    gives:
    [itex]\phi=\arctan{\frac{-\delta}{w_d}}[/itex]

    however i do not find a substitute for [itex]y(0)[/itex]. The solution says [itex]y(0)=\frac{m}{M}y_0[/itex], but i don't see the logic in that at all

    Sorry if its a bit caotic, this is only part of the assignment. ask and i will provide!
     
  2. jcsd
  3. Nov 15, 2012 #2
    nvm, found the solution
     
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