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If a particle moves under the influence of a resistive force proportianal to velocity and a potential $U$,

$$F(x,\dot x)=-b\dot x-\frac {\partial U}{\partial x}$$ Where b>0 and $U(x)=(x^2-a^2)^2$

My thoughts were to make $F=0$, which would result in: $$\dot x= \frac 1b(4x^3-4a^2x)$$ This means that the points of equilibrium are a function of time?


So I set up as suggested, using $F=m\ddot x$ to get $$0=m\ddot x+b\dot x+4x(x^2-a^2)$$ I am trying to solve for $x(t)$. I know how to solve the equation that has only x, but I do not know how to solve for the x^3 term. I am curious how to combine these to get x(t)

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Related or similar question – ja72 Sep 19 '13 at 22:33
Maybe using Laplace transforms, but it has been a really long time since I did something like this. – ja72 Sep 19 '13 at 22:38
@ja72 different question. This is looking for points of equilibrium when the force has a velocity dependance. The other post is asking for equations of motion given a potential – yankeefan11 Sep 19 '13 at 23:13
@yankeefan11: This problem has a simple trick to solve it in a single step without manipulating differential equations. I added it as an answer, although by now you probably already know it, seeing as this was asked several months ago. – DumpsterDoofus Mar 18 '14 at 1:02

The stable point of equilibrium is at $x=0,x=\pm a$. This becomes obvious when you realize that for these kinds of problems with linear friction, you can actually ignore the friction term when computing the equilibrium state of the system.

Why? If the system is at equilibrium, then it is both at rest ($\dot{x}=0$) and has no net force acting on it ($F=0$). Combining these two statements yields $$0=F=-\frac{\partial U}{\partial x}=4 x \left(x^2-a^2\right)$$ which implies $$x\in\{-a,0,a\}.$$

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You can determine equilibrium points by one of two approaches:

  1. Use Newton second law, which means replace $F$ by $m\ddot x$. The equation becomes a 2nd order ODE. Solve it for $x$.

  2. Using your formulation by setting $F = 0$ you obtained a 1st order ODE. Solve it for $x$ as function of time. At the time when the velocity becomes zero, the force is already zero and the potential gradient is zero as well. So the particle doesn't experience any force and has zero momentum. The value of $x$ at that time is the equilibrium position.

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this gets really ugly, is it possible to do this easier? – yankeefan11 Sep 19 '13 at 3:16
I can't think of another way, you can do option 2 by going to, you will get a nice function x(t). You can use "step-by-step" option if you made a trial account. – Gotaquestion Sep 19 '13 at 8:32
Or go to option 1 and solve it using MATLAB or Mathematica. If the equation doesn't have a closed form solution solve it numerically. There are many commands in MATLAB for doing so – Gotaquestion Sep 19 '13 at 8:48… Wolframalpha wont spit out a solution for me – yankeefan11 Sep 19 '13 at 16:44

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