# A position of stable equilibrium, and the period of small oscillations

#### Davidllerenav

Problem Statement
A particle of mass $m$ moves under the action of a potential
$V(x)=\frac{cx}{x^2+a^2}$
where $a$ and $c$ are positive constants. Find the positions of stable equilibrium and the period of the small oscillations around those points.
Relevant Equations
The equations of the harmonic oscillator:
$\ddot{x}=\omega^2x=0$
Period: $\tau\simeq \frac{2\pi}{\omega}$
I tried by taking the derivative of the potential to find the critic points and the I took the second derivative to find which of those points are minimum points. I found that the point is $x=- a$. I don't understand how to calculate the period, since I haven't seen anything about the harmonic oscillator.

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#### haruspex

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What force acts at x? So what is the acceleration there?

#### Davidllerenav

What force acts at x? So what is the acceleration there?
The potential force?

#### haruspex

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The potential force?
There is a force resulting from the potential, yes. How is it obtained from V(x) algebraically?

#### Davidllerenav

There is a force resulting from the potential, yes. How is it obtained from V(x) algebraically?

#### haruspex

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Yes. So what is that as a function of x?

#### Davidllerenav

Yes. So what is that as a function of x?
Well, it is the vector that contains the partial derivatives as coordinate functions, but I don't know its physical meaning.

#### haruspex

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Well, it is the vector that contains the partial derivatives as coordinate functions, but I don't know its physical meaning.
There seems to be only one dimension here, so just the derivative... which is?

#### Davidllerenav

There seems to be only one dimension here, so just the derivative... which is?
The slope of the tangent line at that point of the curve.

#### haruspex

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The slope of the tangent line at that point of the curve.
Yes, yes... but write the actual function of x.

#### Davidllerenav

Yes. So what is that as a function of x?
$v'(x)=\frac{a^2c-x^2c}{(x^2+a^2)^2}$

#### haruspex

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$v'(x)=\frac{a^2c-x^2c}{(x^2+a^2)^2}$
Right, so what is the acceleration at x?

#### haruspex

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I don't know, sorry.
From posts #5 and #11 you know the force, and you are given the mass.

#### Davidllerenav

So, the force is $v'(x)=\frac{a^2c-x^2c}{(x^2+a^2)^2}=m A$, where $A$ is the acceleration. so $A= \frac{m(a^2c-x^2c)}{(x^2+a^2)^2}$ right?

#### haruspex

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So, the force is $v'(x)=\frac{a^2c-x^2c}{(x^2+a^2)^2}=m A$, where $A$ is the acceleration. so $A= \frac{m(a^2c-x^2c)}{(x^2+a^2)^2}$ right?
Retry that last step.

#### Davidllerenav

Retry that last step.
Sorry, I noticed my mistake. $A= \frac{a^2c-x^2c}{m(x^2+a^2)^2}$

#### haruspex

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Sorry, I noticed my mistake. $A= \frac{a^2c-x^2c}{m(x^2+a^2)^2}$
Ok, so write that as a differential equation. Then consider that you are interested in small perturbations from x=-a.

#### Davidllerenav

Ok, so write that as a differential equation. Then consider that you are interested in small perturbations from x=-a.
I haven't seen differential equations yet, so I have no idea how to do that.

#### haruspex

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I haven't seen differential equations yet, so I have no idea how to do that.
The relevant equation you quoted ($\ddot x=...$) is a differential equation.

#### Davidllerenav

The relevant equation you quoted ($\ddot x=...$) is a differential equation.
Yes, I know, but I have't take that class yet. My teacher said that it wasn't necessary to solve the differential equation to solve this problem.

#### haruspex

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Yes, I know, but I have't take that class yet. My teacher said that it wasn't necessary to solve the differential equation to solve this problem.
Yes, you do not have to solve the differential equation because the relevant equations you posted show how to write the period down directly from the differential equation without solving it. But you still need to get the acceleration equation you have in post #11 into the form $\ddot x+\omega^2x=0$. This will involve making an approximation for x close to the equilibrium value you found.
This will be a lot easier if you replace x by equilibrium value plus a variable displacement.

#### Davidllerenav

Yes, you do not have to solve the differential equation because the relevant equations you posted show how to write the period down directly from the differential equation without solving it. But you still need to get the acceleration equation you have in post #11 into the form $\ddot x+\omega^2x=0$. This will involve making an approximation for x close to the equilibrium value you found.
This will be a lot easier if you replace x by equilibrium value plus a variable displacement.
How can I find the approximation? Using a taylor polynomial?

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#### Davidllerenav

Ok, and of what degree?

"A position of stable equilibrium, and the period of small oscillations"

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