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Harmonic oscillator

  1. Dec 26, 2016 #1
    1. The problem statement, all variables and given/known data
    a mass is placed on a loose spring and connected to the ceiling. the spring is connected to the floor in t=0 the wire is cut
    a. find the equation of the motion
    b. solve the equation under the initial conditions due to the question
    שאלה לפורום.JPG
    2. Relevant equations
    ## \sum F=ma
    ##
    ## x(t)=Asin(\omega t + \phi ) ##
    3. The attempt at a solution
    a. due to the 2 law of newton: ## \sum F=ma_x
    ##
    ## mg-kx=ma ##
    ##
    mg-kx=m\ddot{x}
    \\
    \ddot{x}=g-\frac{k}{m}x ##

    b. first I'll find the point equilibrium
    ##
    c-kx=ma
    \\
    c-kx=m\cdot 0
    \\
    x_0=\frac{c}{k}

    ##

    then I'll define ## y=x-x_0 ##

    How do I go from here?
    thanks
     
  2. jcsd
  3. Dec 26, 2016 #2

    TSny

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    OK

    What does c stand for?

    Rewrite the differential equation in terms of y instead of x.
     
  4. Dec 26, 2016 #3
    As TSny hints, you tripped finding x0. Think again what condition will be true at equilibrium and define x0 again. There won't be an unknown c.
     
  5. Dec 28, 2016 #4
    sorry I meant ## x_0 = \frac{mg}{k}##
    ok and then ## \ddot{y}=g-\frac{k}{m}y ## ?
     
    Last edited: Dec 28, 2016
  6. Dec 28, 2016 #5

    BvU

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    ##
    \ddot{y}=g-\frac{k}{m}y## can't be right. Compare with ##
    \ddot{x}=g-\frac{k}{m}x##
     
  7. Dec 28, 2016 #6
    I might have a barrier but I do not understand how to build a differential equation for y,,,,
    I understand that ## \dot{x}=\frac{dx}{dt}=\frac{dy}{dt}=\dot{y} \\ a=\ddot{x}=\frac{d^2x}{dt^2}=\frac{d^2y}{dt^2}=\ddot{y} ##
     
  8. Dec 28, 2016 #7
    ??? I must be thick this morning, but if positive y is down that looks ok to me.
     
  9. Dec 28, 2016 #8
    Oh, good grief. It took me a minute!!
     
  10. Dec 28, 2016 #9
    Much later after it stops oscillating and y is y0 what is the acceleration? g?
     
  11. Dec 28, 2016 #10

    BvU

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    Simple: the second derivatives on the left are equal alright. But you need to substitute your expression for y in terms of x. Not just y = x, but: ...:rolleyes:
     
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