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Spring and Hammer

  1. Oct 3, 2004 #1
    Problem: A mass m is at rest on the end of a spring of spring constant k. At t = 0 it is given an impulse J by a hammer. Write the formula for the subsequent motion in terms of m, k, J, and t.

    Would ma = -kx + J/t be an acceptable answer?
     
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  3. Oct 3, 2004 #2

    arildno

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    No, it is not!
    The force associated by impulse J should be modeled by Dirac's delta function; solve the problem with Laplace Transforms.
     
    Last edited: Oct 3, 2004
  4. Oct 3, 2004 #3
    Please note that the problem is from a general physics textbook. Assume as many simplifying assumptions as possible.
     
  5. Oct 3, 2004 #4

    arildno

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    In that case, solve it as follows:
    Just after the impulse J, the mass has an initial velocity [tex]v_{0}=\frac{J}{m}[/tex]
    In the subsequent problem, your diffferential equation is:
    [tex]-kx=m\ddot{x}[/tex]
    whereas initial conditions are:
    [tex]x(0)=0,\dot{x}(0)=\frac{J}{m}[/tex]
     
    Last edited: Oct 3, 2004
  6. Oct 3, 2004 #5
    Hmm...Why didn't I think of that? I guess that does it for that problem. Thanks.
     
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