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Homework Help: Equation of motion for system of springs- SHM

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    I don't want to write the whole question as it is very long and I just have one query.. Basically the question involves a particle attached to three different springs which have fixed end points. In the question the mass of the particle is m, and the other constants used throughout are k (stiffness of one of the springs), g, and l0 which is the natural length of one of the springs. All the lengths in the question are given in terms of l0.
    My question is this- my answer for the equation of motion doesn't contain the constant l0 in the part that determines the amplitude, only g, m and k. I've checked over and over and I can't find a mistake but to me this seems impossible. The answer I have includes l0 as the 'starting point' but the term involving cos does not depend on the length. Does this definetely mean I've made a mistake?


    2. Relevant equations
    Hooke's Law, method for solving linear differential eq.


    3. The attempt at a solution
    see above.
     
  2. jcsd
  3. Mar 29, 2010 #2

    rock.freak667

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    Do you have a diagram associated with it? That would help much in visualizing your problem.
     
  4. Mar 29, 2010 #3
    Here you go..
    To clarify my actual question, looking at the system below, (for the case where the mass starts from rest at x = 8/7l0) is it possible for the amplitude of the motion to not be dependant on l0.
     

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    Last edited: Mar 29, 2010
  5. Mar 29, 2010 #4

    rock.freak667

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    How did you derive your equation of motion, can you show your steps?
     
  6. Mar 29, 2010 #5
    I'll write it out tomorrow if necessary as I need to go to bed now. But I'd really rather a 'yes/no' answer to my question- its a theoretical point. If it does definetely mean there is a mistake (as I think it does) then I have to really find it myself cos this is coursework. I just don't want to waste lots of time looking for a mistake that isn't there...
    Thanks for replying though =].
     
  7. Mar 29, 2010 #6

    rock.freak667

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    Your spring constants aren't the same, and the masses attached to springs A and B aren't the same, so I am to say that you have a system with 3 degrees of freedom?
     
  8. Mar 30, 2010 #7
    The spring constants are different yes, but theres only one mass (m) in the system (ie so I've been looking simply at the forces acting on the 'particle' with mass m.)
    Afraid I've never used the term 'degrees of freedom' so don't know what that implies..
     
  9. Mar 31, 2010 #8

    rock.freak667

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    Well in my experience with questions with such springs, none of the equations of motion I've derived contains the the natural length of the spring.
     
  10. Mar 31, 2010 #9

    vela

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    Your solution is probably fine. The equilibrium length of the springs enters into determining where the equilibrium point is, but not into determining how the mass moves when displaced slightly from that point.
     
  11. Apr 4, 2010 #10
    Great (well I sent it in now anyway so I hope its right!)...
    Out of interest, is this actually the case in 'real life' or is it subject to some modelling constraints? I find it very strange that in the system above the springs could be 20m long or 1cm and the equation of motion says the amplitude of the oscillation would be identical.. I guess the spring has to be long enough for it never to become slack, is that the only difference?
     
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