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Mechanics - Elastic springs and strings

  1. Mar 9, 2006 #1
    A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length [itex]\frac{5l}{3}[/itex]. Show that if P is projected vertically downwards from A with speed [itex]\sqrt(\frac{3gl}{2})[/itex], P will come to instantaneous rest after moving a distance [itex]\frac{10l}{3}[/itex].

    I thought about working all this out by finding the energies before and after the projection.

    Before:-
    [itex]KE =\frac{3gl}{4}[/itex]
    [itex]GPE = 0[/itex]
    [itex]EPE = 0[/itex]

    After:-
    [itex]KE = 0[/itex]
    [itex]GPE = -(y - 5/3L)g[/itex] where y is the full length of stretched string.
    [itex]EPE = \frac{gl(Y- 5/3L)^2}{2L(Y- 5/3L)}[/itex]

    i cant seem to get the anwer using these, are they correct?
     
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 10, 2006 #2
    This seems like more of a physics related question. I'm sure if you post it in the physics section someone would help you out a little faster. I haven't taken diffeq in awhile, but this seems like a pretty straightforward diffeq question. I could definitely be wrong though
     
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