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Change in energy in springs

  1. Mar 6, 2007 #1
    I am doing a project on tightly-coupled processive molecular motors. My 1D model consists of motors which literally step discrete step sizes of size U while executing a biased random walk along a filament which moves along its longitudinal axis with negligible drag. The motors are attached to a spring which is anchored at one end.

    If you can imagine the motor has legs and is attached to a spring (~) of stiffness k.


    I can't draw the springs but if you can imagine them attached to the motors like the one drawn above. Then a two motor system looks a bit like this....

    -x dir ____________/\_____________/\___________ +x dir

    One property of this system is that the filament moves at a constant velocity and so it must have zero force on it at all times.

    So if one motor steps along, the filament and the motors move in the -x direction by u/n ( where n is the number of motors),

    So for 2 motors if any of the motors steps forward, then the filament has moves back u/2, the motor which has stepped has increased the extension in its spring by u/2 and the motor which has stayed still has compressed its spring by u/2 hence the total force on the filament is zero.

    To generalise this to n motors, for any motor that takes a forward step,
    1) the filament moves back u/n
    2)the motor that has stepped has increased the extension in its spring by U(n-1)/n
    3)the motor that didnt step decreased the extension in its spring by u/n

    Now my question is......

    What is the change in the energy in the spring when a motor steps?

    The change in the energy of the motor that HASN'T stepped:
    If the initial extension in its spring is xj then the final extension is xj-u/n
    then the change in energy dE

    Efinal - Einitial = sumj k/2(xj-U/n) - sumj k/2(xj)^2 right? (sumj = sum of all xj)

    which simplifies down to sumj = k/2((U^2)/(n^2) - 2*xj*U/n)

    The change in the energy of the motor that HAS stepped:

    if initial extension is xi

    dE = Efinal - Einitial = sumi k/2(xi+U-U/n)^2 - sumi k/2*(xi)^2

    which I THINK simplifies down to

    dE = sumi k/2*(2*xi*U*(n-1)/n+(U^2)/(n^2)(1-2n+n^2)

    The trouble is I don't think this works!


    Do I need to sum the extensions of all the springs? or is it just the extensions of the springs of the motors that havent stepped which goes into sumi and those which did step goes into sumj?
  2. jcsd
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