Actually the following question came in an exam i gave yesterday. I know the site does not allow home-work type questions, but as I am having a conceptual problem in solving the question, I am posting it anyway. The question is:

A particle is acted upon by a force given by $F= -ax^3 -bx^4$ where $a$ and $b$ are positive constants. At point $x=0$, the particle is $1)$ in stable equilibrium$/ 2.) $in unstable equilibrium$/ 3.)$in neutral equilibrium$/4.)$ not in equilibrium. Which option is correct.

My question how to figure out from the force-displacement equation, whether a particle is in equilibrium or not and in what type of equilibrium. I guess it has something to do with potential energy but cannot figure it out.

  • $\begingroup$ Whether you solve it or not, these questions are a suite to "homework" tag $\endgroup$ – Waffle's Crazy Peanut Nov 5 '12 at 2:22

You gave an expression for the force, whereas Prathyush's answer treats it as the potential. If that expression really is the force, then the answer is different.

You don't really need to do any calculation in this case to see what is going on. Clearly $F(0) = 0$, so this is indeed a point of equilibrium (the question should probably read something like "the particle is placed at rest at the point $x=0$...", to be unambiguous). To check stability, think about displacing the particle slightly in one direction or the other. If $x$ is very close to zero, then $|x^4| \ll |x^3|$, so we can safely ignore the second term, and consider $F \sim -a x^3$. So if $x$ is small and positive, the force pushes the particle in the negative direction, and vice-versa; $x=0$ is therefore stable.

  • $\begingroup$ I agree that the answer should treat $F$ as a force, given that is what it is stated to be in the question. $\endgroup$ – Daniel Blay Nov 5 '12 at 9:19
  • $\begingroup$ Yes I did not notice, you are quite correct. $\endgroup$ – Prathyush Nov 5 '12 at 15:37

Firstly Sketch the force and It will help you find out which points are stable equilibrium and which ones are not. The points at which the force is zero are points where the particle is at equilibrium.

$F=-a x^3 -b x^4$

At x = 0, the force is zero therefor the particle is in equilibrium.

To analyze stability, took at the sign of the force infinitesimally away from the points of equilibirum. for small displacements is $x^3$ is much larger than $x^4$.

The small x, the force has a sign that is opposite to that of its displacement. Implying that any due to any small deviation from equilibrium, the particle will experience a force tending to equilibrium. Which means it is a stable equilibrium.

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    $\begingroup$ $F$ was defined in the question as a Force, not a potential. $\endgroup$ – Daniel Blay Nov 5 '12 at 10:11
  • $\begingroup$ I cant seem to delete my post. $\endgroup$ – Prathyush Nov 5 '12 at 15:38
  • $\begingroup$ @Prathyush that is probably because it was accepted. I would suggest editing it to fix any errors. $\endgroup$ – David Z Nov 6 '12 at 18:31
  • $\begingroup$ Edit: corrected answer, force was mistakenly taken to be potential $\endgroup$ – Prathyush Nov 6 '12 at 21:16

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