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Dynamic system - initial displacement

  1. Nov 20, 2013 #1
    Please see attached for question.

    My problem is the initial displacement.

    i am told the first step is this

    kx + c(dx/dt) = unit impulse

    laplace transform

    kX + c(sX-x0)=1

    from that i can see the initial displacement is included in the damper but why isnt it included in the spring term e.g i would expect it to be k(X-x0)

    can anybody explain this?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2013 #2

    rude man

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    First, you have to make an assumption as to whether the system is over - or - underdamped. Let's assume underdamping since that's more fun ...

    Second, you should understand what is meant by "an impulse δ". Actually, it's δ(t), signifying that an impulse is applied at t = 0. More importantly, the dimension of δ(t) is T-1 so your F = ma equation will be dimensionally incorrect if you don't modify the impulse.

    Your actual impulse is bδ(t) where b = 1 newton-second if you're using the SI system. I suggest retaining the "b" in your work so you can check dimensional consistency term-by-term.

    OK, so now you write your F = ma equation, transform to the frequency (Laplace) domain, and solve for X(s) and then x(t).

    I hope you have a Laplace table that includes the underdamped second-order case. If not you'll have to use partial-fraction expansion with complex roots. Yuk!

    OR, you can just assume overdamped in which case the poles are both real and you can use partial fraction expansion easily.
     
  4. Nov 23, 2013 #3
    Thank you for your reply Rudeman, but i am still unsure specifically about including the initial displacement in the equation

    would the spring term include the initial displacement e.g k(x-x0)

    or would it just be kx

    with an explanation of why if possible,

    Thanks
     
  5. Nov 23, 2013 #4

    rude man

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    Ʃ
    Let x = 0 be the position of the bar before the impulse is applied. So your initial conditions are x(0) = x'(0) = 0.

    EDIT: the fact that the bar is massless makes this a first-order system, making things a whole lot easier. So your equation is ƩF = ma = 0. This makes the math very easy. Sorry that didn't dawn on me earlier. So also ignore my bit about having to assume underdamping or overdamping. There are no oscillations possible with m = 0.
     
    Last edited: Nov 23, 2013
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