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Elastic PE, gravity and SHM

  1. Jun 2, 2008 #1
    My first post here! I signed up a few minutes to correct someone... then deleted my post when I realised I was on about something entirely different. Ahem.

    Anyway, have a look at this diagram:

    http://img2.freeimagehosting.net/uploads/c8911ecf7e.jpg

    Particle P is suspended on an elastic string fixed at O, natural length = 'l', equilibium point is ( l + d ) below O, and 'x' is the distance the particle has been pulled down to.

    Correct me if this is wrong. If you release P its motion will be simple harmonic, at least until it reaches the equilibrium point. But what happens to its motion after that? Once " OP < l ", its motion is like a normal particle under gravity but how does it move when " l < OP < l + d "?

    Thanks in advance!
     
    Last edited by a moderator: Jun 2, 2008
  2. jcsd
  3. Jun 2, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi pc0019! Welcome to PF! :smile:

    The equilibrium position is where the string is stretched, but the weight of the particle (particle?) exactly balances the tension in the string.

    The equations are exactly the same on both sides of the equilibrium position (for OP > l, of course).

    So it oscillates harmonically up and down past the equilibrium position. :smile:
     
  4. Jun 2, 2008 #3

    Doc Al

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    Interesting. Being an elastic string and not a rigid spring, I'd think that positive tension must be maintained in order to get SHM. Assuming the string is semi-Hookean (restoring force proportional to positive displacement), you'd get SHM as long as x < d. (It will oscillate about the equilibrium point, as tiny-tim says.)
     
  5. Jun 2, 2008 #4
    Thanks, tiny-tim!

    OK, if I understand both of you correctly, the particle (A level maths student :"]) will act simple-harmonically when " l < OP < l + d + x ", ie motion through 'd' mirrors the motion through 'x' until (x - d) and if d = x, the motion of the particle would be exactly the same as a spring on a frictionless horizontal surface?

    I think I must have got something horribly wrong because that doesn't add up (metaphorically speaking)? I was thinking that motion through 'd' would be similair to, but not precisely SHM.

    Thanks for the link btw Doc Al :smile:
     
    Last edited: Jun 2, 2008
  6. Jun 2, 2008 #5

    tiny-tim

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    (oh, it's a maths student on the end of the string … that makes more sense! :smile:)

    I'm confused … what doesn't add up? :confused:

    Have you written out the equation of motion? :smile:
     
  7. Jun 2, 2008 #6
    Well it will be soon, what with exams and all... :wink:

    What I don't understand, is how my statement "if d = x, the motion of the particle would be exactly the same as a spring on a frictionless horizontal surface" could be true.

    I don't know what you mean by 'equation of motion'. This is conjecture, although the question that got me thinking asked for the time between "OP = l + d + x ", to the point where the string becomes 'slack' - when "OP = l" I think...
     
  8. Jun 2, 2008 #7

    tiny-tim

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    Sorry … still can't see anything surprising about that. :confused:
    I mean an equation relating acceleration to forces and distances.

    Or an equation beginning x'' = …

    How will you solve this without an equation?
    oh, you didn't mention that x > d.

    But you still need an equation. :smile:
     
  9. Jun 2, 2008 #8
    Sorry, the question actually asks for the speed, not the time.

    x is not necceserally greater than d, although in the question it is. 'x' isn't displacement. I am using T = mg = [λ (x + d)]/l

    So, for x > d, this is happening:

    http ://img59.imageshack.us/img59/3192/shm2qa4.jpg

    Where, when the displacment is 0, P is at the equilibrium position, and the blue lines are when the string is slack and P moves freely under gravity. And where x = d there is full SHM, and where x < d, chop off the graph somewhere above the equilibrium position and stick it together. To use the proper technical terms. Sorry if I am repeating myself but I want to make sure I am understood correctly.


    In the question, the modulus of elasticity = 4g ~ 39.2, l = 0.8, d = 0.05, x-max = 0.1, T = (pi/7), m = 0.25.

    Using initial energy = final energy, EPE(0.55125) = GPE(0.3675) + KE, therefore v = 0.7root3.

    As such I don't need to model the motion, but I thought it would be useful to know.
     
  10. Jun 2, 2008 #9

    tiny-tim

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    If you're only asked for the speed, then just use the energy equation (which I think you have done).
     
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