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Maths behind non-linear dynamics, driven damped oscillator more specifically.

  1. Jan 29, 2013 #1
    I am investigating the mathematics behind driven damped oscillators, I will then simulate it in matlab and observe the unpredictable long term behavior of the system.

    In order to create non-linearity in a oscillating spring I can no longer use hookes law but a form of it by introducing a power to the x term... for example F=-kx^3. Am I right?

    In order to find the force i will differentiate the potential energy equation with respect to position(x) because the F = Work done/ Distance moved. Or I will find the potential energy by integrating the force with respect to position.

    V(potential energy) = 1/4 - x^2/2 + x^4/4
    dV/dx = F = x + x^3............. this is the double well potential no? How do I derive the formula for potential energy and/or the force.

    Now, because the force = x+x^3 and F=-kx^3 can I set these equal to each other and cancel some terms? So, x+x^3 = -kx^3..... no that would leave = x = -k.... F=-kx, the linear form of hookes law.

    Can someone point me in the right direction, either a good website outlining to mathematical concepts involved with this or can we have a discussion on the matter? I dont want to write too much right now.

    Thanks alot!
     
  2. jcsd
  3. Jan 29, 2013 #2

    Mute

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    Yes, you could introduce such a nonlinearity to the force. However, let me ask you, do you want to be using a purely nonlinear spring force in this problem, or the linear Hooke's law with an additional non-linear term. i.e., do you want ##F = -k_3x^3## or ##F = -k_1x + k_3x^3##? I ask this because you appear to be mixing them together later in your post.

    This potential energy would correspond to a force of ##x - x^3##. Remember that force is ##F = -\frac{dV}{dx}##. You don't really need the constant term, either - it just sets the zero level of your potential.

    You need to be more careful with your minus signs. The relative sign between your x and x^3 terms keeps changing throughout your post, and the different signs will give you different behaviors.

    Let me suggest that you write your potential energy function as

    $$V(x) = \frac{1}{2}kx^2 + \frac{1}{4}bx^4,$$
    where we assume k is positive and b can be either positive or negative. Depending on the sign and/or size of b you can get two possible kinds of potentials: a single well potential or a double well potential. Pick a value for k and play around with b to see when you get which kind of potential.

    I'm not sure what you are trying to do here. You have two different forces, one which is Hooke's law plus a nonlinear term and you are trying to equate it to a purely cubic nonlinear force law. You don't need to do that - you have either one of those forces or the other. You don't need to try to equate them.

    For an example of an oscillator with a cubic nonlinear term (as well as damping), look up the Duffing equation. See if you can understand some of the concepts there and see how they relate to what you are trying to do. And, of course, continue to post on the forum to ask for help if you get stuck.
     
  4. Jan 29, 2013 #3
    OK great! I was getting mixed up as to what force is what.

    I think it would be better to add a non linear term to hookes law good idea!

    I'm going to use the duffing type oscillator yeah. I've been trying to find how to derive the general duffing equation and the potential energy equation. But no luck yet. I need to derive the maths for my specific system is all.

    For the double well potential... why does it generally have a depth of 1/4?

    How do I find the minimum velocity of the system to get over the hump in the double well?

    Also, is the 2nd derivative of the potential energy you gave me 1/3*b*x^3? That is what I use to determine the stationary points?
     
    Last edited: Jan 29, 2013
  5. Jan 29, 2013 #4

    Mute

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    These equations all generally come from Newton's second law: ##\sum_i F_i = ma##. As usual, ##a = \ddot{x}## - so that gives you the second derivative term in your differential equation - so you just need to decide what forces are acting on your system and add them together (being careful with the signs). For example, one of those forces for the simple harmonic oscillator is the Hooke's law force, ##-kx##. This gives you a linear spring. If you want to study a nonlinear spring, one of the simplest nonlinearities you could add to the force would be a term like bx^3, as I mentioned in my previous post. (One could add a term like x^2 to the force, but that would make for a weird spring, as compression vs. stretching by the same amounts would yield different forces). Another one of the forces acting on your system might be a drag force.

    I'm not sure what you mean here. If you have a double well potential of the form kx^2/2+bx^4/4, the locations of the minima and their depths will be controlled by the ratio ##|k/b|##, and the depth will not necessarily be 1/4.

    Consider how much energy does your oscillator initially have (it will depend on the starting position). How might you use that information to figure out how fast the oscillator needs to move to make it over the hump?

    No, you're missing a term and you make analgebra mistake. The potential I suggested is

    $$V(x) = \frac{1}{2}kx^2 + \frac{1}{4}bx^4.$$

    You're missing the second derivative of the first term (it's not zero), and when you differentiate the second term twice you should be left with an x^2, not x^3. The coefficient (you wrote 1/3) is also incorrect, so be careful there when calculating the derivatives again.
     
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