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Energy conservation and periodic motion

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-3-2_20-34-40.png
    Four weightless rods of length ##l## each are connected by hinged joints and form a rhomb (Fig. 48). A hinge A is fixed, and a load is suspended to a hinge C. Hinges D and B are connected by a weightless spring of length ##1.5l## in the undeformed state. In equilibrium, the rods form angles ##\alpha _0 = 30° ## with the vertical. Determine the period T of small oscillations of the load.

    2. Relevant equations
    ##U = mgh ##
    ##U = \frac{1}{2} kx^2 ##

    3. The attempt at a solution
    I used energy conservation law in this case. But this gave me crazy results. Surely, I have made a mistake to apply this law. Would you please help me to find out the mistake ?
    upload_2015-3-2_20-43-2.png
    So, ## h = 2l cos \alpha## .... (i) ;
    ##y = 2l sin \alpha## ;
    Expansion (or compression) of the spring, ## x = y - 1.5 l = l ( 2sin \alpha - 1.5)## .....(ii)
    Let the spring constant be ##k## and the mass of the load be ##m##;
    Applying energy conservation law :
    ## \frac{1}{2} kx^2 - mgh = constant ## [when h increase, gravitational potential decreases]
    ## kx \frac {dx}{dt} - mg \frac {dh}{dt} = 0##
    ##kx - mg \frac {dh}{dx} = 0## ... (iii)
    But, from eq. (i) & (ii), ## \frac{dh}{dx} = - tan \alpha ##
    So, (iii) >>
    ## kx = -mg tan \alpha ## ;
    ## k = - \frac{mg tan \alpha}{x} = - \frac{mg tan \alpha}{ l ( 2sin \alpha - 1.5)} ##
    So, ##k## becomes variable.
     
  2. jcsd
  3. Mar 2, 2015 #2

    PeroK

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    I don't follow your equations. What is x?

    I would take x = h - h0 (where h0 is the equilibrium point).
     
  4. Mar 2, 2015 #3
    x is the expansion of the spring.
     
  5. Mar 2, 2015 #4

    PeroK

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    Okay. I missed that. Where's your Kinetic Energy?
     
  6. Mar 2, 2015 #5
    Oh! Sorry! How could I make such a silly mistake!!:))
    Thanks for your help.
     
  7. Mar 2, 2015 #6

    DEvens

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    Oscillation means that neither x nor h are constant. So you don't want the equation you wrote.

    Harmonic oscillation comes from a restorative force proportional to the distance away from equilibrium. So you want to try to figure out the force pushing the system back to equilibrium. You have the equilibrium position. Suppose you push the mass up by a short distance, call it d. What will the force be that is required to hold the mass there? It should be some constant times d. And the direction should be such that the system is trying to push the mass back to its equilibrium. That is, the system will provide a force of the following form.

    F = - Keff d

    Here Keff is the effective spring constant of the system.

    Once you have that, then you can do the usual things for harmonic motion to get the period.
     
  8. Mar 2, 2015 #7

    PeroK

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    In this case the restorative force is not proportional to the displacement.
     
  9. Mar 2, 2015 #8

    DEvens

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    For small displacements it should be. Otherwise it will not produce simple harmonic oscillation. Presumably what will be needed are the small angle approximations to the sin and cos functions.
     
  10. Mar 2, 2015 #9

    PeroK

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    It looks a lot easier to use energy conservation, as the OP intended.
     
  11. Mar 3, 2015 #10
    I solved this problem using this technique. Thanks for your help.
     
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