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

[Davidllerenav]

Homework 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.

 

[haruspex]

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]

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?

It is the gradient, right?

 

[haruspex]

It is the gradient, right?

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]

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]

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]

$v'(x)=\frac{a^2c-x^2c}{(x^2+a^2)^2}$

Right, so what is the acceleration at x?

 

[Davidllerenav]

Right, so what is the acceleration at x?

I don't know, sorry.

 

[haruspex]

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]

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]

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]

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]

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?

 

[haruspex]

How can I find the approximation? Using a taylor polynomial?

Yes.

 

[Davidllerenav]

Yes.

Ok, and of what degree?

 

[haruspex]

Ok, and of what degree?

Just the first term.

 

[Davidllerenav]

Just the first term.

I think that the polynomial is $F=mA=v'(a)+v''(a)(x-a)$, it would be $ma=0+\frac{c}{2a^3}(x-a)$ right?

 

[haruspex]

I think that the polynomial is $F=mA=v'(a)+v''(a)(x-a)$, it would be $ma=0+\frac{c}{2a^3}(x-a)$ right?

I assume the a on the left should be A, the acceleration, not to be confused with the given a, a coordinate.
Almost right, but you have a sign error.

 

[Davidllerenav]

I assume the a on the left should be A, the acceleration, not to be confused with the given a, a coordinate.
Almost right, but you have a sign error.

Well, I thought that that $a$ was supposed to be the value I found, $x=-a$.

 

[haruspex]

Well, I thought that that $a$ was supposed to be the value I found, $x=-a$.

In post #27 you correctly wrote mA= In one equation, A being the acceleration, but in the next equation you changed it to ma=.

 

[Davidllerenav]

In post #27 you correctly wrote mA= In one equation, A being the acceleration, but in the next equation you changed it to ma=.

Oh, sorry, I meant $mA=0+\frac{c}{2a^3}(x-a)$. Since I'm plugging in $x=-a$, would it be $mA=0-\frac{c}{2a^3}(-a-a)$?

 

[haruspex]

Oh, sorry, I meant $mA=0+\frac{c}{2a^3}(x-a)$. Since I'm plugging in $x=-a$, would it be $mA=0-\frac{c}{2a^3}(-a-a)$?

You don't want to plug in x=-a.
As I posted, it would be rather clearer if you were to rewrite the equation I am terms of displacement from -a, e.g. x=-a+y. Then you would be looking for a differential equation of the form $\ddot y+\beta y=0$ for some constant beta.
Try that approach and maybe it will fix your sign error.

 

[Davidllerenav]

You don't want to plug in x=-a.
As I posted, it would be rather clearer if you were to rewrite the equation I am terms of displacement from -a, e.g. x=-a+y. Then you would be looking for a differential equation of the form $\ddot y+\beta y=0$ for some constant beta.
Try that approach and maybe it will fix your sign error.

Sorry for taking so long answer. I don't undertand how should I should I do that, should I just replace $x$ by $-a+y$ on the taylor polynomial?

 

[haruspex]

Sorry for taking so long answer. I don't undertand how should I should I do that, should I just replace $x$ by $-a+y$ on the taylor polynomial?

I would use that substitution in your equation in post #17. Then simplify a bit and discard terms beyond the first nonzero term in each sum.

 

[Davidllerenav]

I would use that substitution in your equation in post #17. Then simplify a bit and discard terms beyond the first nonzero term in each sum.

Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?

 

[haruspex]

Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?

Yes.

 

[Davidllerenav]

Yes.

I got $A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$. I need to expand this with Taylor, right?

 

[haruspex]

I got $A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$. I need to expand this with Taylor, right?

Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.

 

[Davidllerenav]

Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

 

[haruspex]

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

Close, but you forgot something in the denominator.

 

[Davidllerenav]

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

That it is squared, right? $\frac{2ayc}{4a^4}$?

 

[haruspex]

That it is squared, right? $\frac{2ayc}{4a^4}$?

Yes.
Now write out the differential equation entirely in terms of y. I.e. using $\ddot y$ instead of A. Careful with signs.

 

[Davidllerenav]

Yes.
Now write out the differential equation entirely in terms of y. I.e. using $\ddot y$ instead of A. Careful with signs.

Like this $\frac{2ayc}{4a^4}+\omega^2y=0$?

 

[haruspex]

Like this $\frac{2ayc}{4a^4}+\omega^2y=0$?

No. Don't try to bring omega into it yet.
A is the same as $\ddot x$. Using x=-a+y, write A in terms of $\ddot y$. Then equate it to the expression you found for A in post #41.

 

[Davidllerenav]

No. Don't try to bring omega into it yet.
A is the same as $\ddot x$. Using x=-a+y, write A in terms of $\ddot y$. Then equate it to the expression you found for A in post #41.

So $\ddot x= \frac {d^2}{dx^2}(-a+y)$, right?

 

[haruspex]

So $\ddot x= \frac {d^2}{dx^2}(-a+y)$, right?

No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?

 

[Davidllerenav]

No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?

So $\ddot x=\ddot y$.

 

[haruspex]

So $\ddot x=\ddot y$.

Right. Now write the differential equation that results from combining that with your expression for A in post #41.

 

[Davidllerenav]

Right. Now write the differential equation that results from combining that with your expression for A in post #41.

I can solve the differential equation for $\ddot x$ such that $\ddot x=-\omega^2 x$ and using the expression for A, I get $\frac{2ayc}{m4a^4}=-\omega^2 x$, right?

 

[haruspex]

I can solve the differential equation for $\ddot x$ such that
and using the expression for A, I get $\frac{2ayc}{m4a^4}=-\omega^2 x$, right?

You need to have only y or x, not both. Use the equation you had in post #47.
And you can simplify the expression on the left a little.
Can you now deduce an expression for ω?

Oh, and it is not true that $\ddot x=-\omega^2 x$. There were other terms when the equation was all in x. This is why we switched to y.