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Damped harmonic motion

  1. Jun 9, 2009 #1
    http://img13.imageshack.us/img13/9091/53337497.th.jpg [Broken]

    Can someone please help me with the problem above? I am unable to start it. Clearly, using the constant acceleration "suvat" equations, 0.5ft^2 is the distance obtainined, however I am unable to proceed.

    Thanks in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 9, 2009 #2

    HallsofIvy

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    So you are able to do part (a), then. At any moment, t, the force on the particle is man2 times the extension of the spring. That is y and, using (a), [itex]y=(1/2)ft^2-x[/itex].

    [tex]m\frac{d^2x}{dt^2}= man^2((1/2)ft^2- x)[/tex]

    Are you sure about "modulus of elasticity man2"? That "a" doesn't seem to fit.
     
    Last edited by a moderator: May 4, 2017
  4. Jun 9, 2009 #3
    For the tension, you forgot to divide by the natural length = a, and hence the a seems to fit! Thanks alot for the help though, it makes sense now! How would I do the last part of finding the maximum tension in the spring?

    Thanks
     
    Last edited: Jun 9, 2009
  5. Jun 12, 2009 #4
    Could anyone please help with the last part of the question please?

    Thanks
     
  6. Jun 12, 2009 #5

    Cyosis

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    The tension is maximum when the string's extension is at a maximum. You have an expression for the displacement. How do you find the maximum of this function?
     
  7. Jun 13, 2009 #6
    Would I say y = 0.5ft^2 - (particular solution of the equation), and then find when dy/dt = 0, to find the time at which y is a max, and then subv this time back in, to find ymax, and then use the tension equation?

    Thanks
     
  8. Jun 13, 2009 #7

    Cyosis

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    Sounds like a good idea.
     
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