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Homework Help: Non Ideal spring

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data
    a block of mass m rests on a horizontal frictionless surface. The block is attached to a horizontal spring.

    The spring is not ideal. The force exerted on the block by the spring is F= -kx-Bx^3 where B is a positive constant.

    Calculate the first-order (largest non-zero) correction to the hooke's law period when the amplitude of the motion of the block is small, but finite

    2. Relevant equations

    If it was ideal the period of the oscillation would be T= 2*pi * sqrt of (m/k)
    but it is not idea
    3. The attempt at a solution

    I know the period has to be shorter because the restraining force is greater now with the added -Bx^3 term but the questions is how much shorter is the period?

    I just tried solving for k=-F/x - Bx^2 and then plugging it back into the period formula but that doesn't seem right

    Ok I know that -kx-Bx^3= m*x" (2nd deriv of x with respect to time)

    so I thought maybe I could solve it like an ode

  2. jcsd
  3. Oct 14, 2008 #2
    You have a force equation as a function of spring stretch. Remember force relates to acceleration, so you really have acceleration as a function of spring stretch. To relate to period, you really want to find the speed (as a function of spring stretch). You can do that a few ways: via acceleration or via the "potential" surface and conservation of energy (the method I tend to prefer and suggest). It's typically at that point that you decide to call the speed x-dot and separate separate variables to find angular frequency, which can be related to period.

    I suggest following the text case with the non-altered force equation to work through this process... then do again with the added force (you might be able to think of it as a unique driving force).
  4. Oct 14, 2008 #3
    This is what I tried

    I know its periodic so I simply assumed that solution is x=Acos(wt) then I took the first and second derivatives for x' and x"

    then I simply plugged those back into the equation keeping in mind that I have cos^3 term for x^3

    using a trig identity I solved for it and got w= sqrt of [(k +.75BA^2)/mass]

    please tell me if I am way off the mark
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