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Homework Help: Springs and shm

  1. May 18, 2010 #1
    Need to show that the vertical displacement of a particle on a spring is

    [tex] \ddot{x} + 100x = 0 [/tex]

    [tex] \frac{1}{2}kx^2 + \frac{1}{2}m\dot{x}^2=C [/tex]

    [tex] k\dot{x}x + m\dot{x}\ddot{x}=0[/tex]

    Then since m=0.4, k=40.

    [tex] \ddot{x}+100x=0 [/tex]

    what has happened to gravitational potential energy? why isnt it included in the potential and kinetic energy?
  2. jcsd
  3. May 18, 2010 #2


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    Homework Helper

    Well you see if you include it, it will cancel out due to equilibrium conditions

    [tex]\frac{1}{2}k(x+\delta)^2 -mgx + \frac{1}{2}m\dot{x}^2 = C[/tex]

    At equilibrium kδ=mg or kδ-mg = 0
  4. May 18, 2010 #3

    ah i see how it vanishes now. but does
    [tex] \frac{1}{2}k\delta ^2 [/tex]
    mean anything? does it disappear?

    it does because its a constant right?
  5. May 18, 2010 #4


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    Homework Helper

    Remember that when you have a spring and then you suspend a mass from it, there will be an initial displacement. δ is this displacement such that when you displace the mass a distance 'x', the spring extends by 'δ+x'

    Also it disappears after you differentiate the energy equation.
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