How to find the period of small oscillations given the potential?

[Davidllerenav]

Homework Statement

A particle of mass $m$ is located in a unidimensional potential field where the potential energy of the particle depends con the coordinate $x$ as $U(x)=U_0(1-\cos ax)$; $U_0$ and $a$ are constants. Find the period of small oscilations that the particle performs about the equilibrium position.

Relevant Equations

$x''+\omega^{2}x+0$
$T=2\pi /\omega$

I first found the equilibrium points taking the derivative of the potential. $U'(x)=U_0 a\sin(ax)$, and the equilibrum is when the derivative is 0, so $U_0 a\sin(ax)=0$ so $x=0$ or $x=\pi/a$. Taking the second derivative $U''(x)=U_0a^2 \cos(ax)$ I find that $x=0$ is a minimum point, since the second derivative is greater than 0, and $x=\pi/a$ is a maximum point. But if I replace any of those points on the first derivtive, I get 0. I don't know what to do.

 

[timetraveller123]

wait i am not sure what you are asking you got the points' derivate to be zero so obviously if you sub back the points into the derivate you are going to get zero or am i missing what you are asking?thanks

 

[PeroK]

Problem Statement: A particle of mass $m$ is located in a unidimensional potential field where the potential energy of the particle depends con the coordinate $x$ as $U(x)=U_0(1-\cos ax)$; $U_0$ and $a$ are constants. Find the period of small oscilations that the particle performs about the equilibrium position.
Relevant Equations: $x''+\omega^{2}x+0$
$T=2\pi /\omega$

I first found the equilibrium points taking the derivative of the potential. $U'(x)=U_0 a\sin(ax)$, and the equilibrum is when the derivative is 0, so $U_0 a\sin(ax)=0$ so $x=0$ or $x=\pi/a$. Taking the second derivative $U''(x)=U_0a^2 \cos(ax)$ I find that $x=0$ is a minimum point, since the second derivative is greater than 0, and $x=\pi/a$ is a maximum point. But if I replace any of those points on the first derivtive, I get 0. I don't know what to do.

What determines the motion for "small" oscillations about a local minimum in potential?

 

[Davidllerenav]

wait i am not sure what you are asking you got the points' derivate to be zero so obviously if you sub back the points into the derivate you are going to get zero or am i missing what you are asking?thanks

Yes, sorry you're wright. What I'm asking is what should I do next, after I get the equilibrium points.

 

[Davidllerenav]

What determines the motion for "small" oscillations about a local minimum in potential?

A small displacement?

 

[PeroK]

A small displacement?

It's the behaviour of the potential on a small interval about the minimum.

 

[Davidllerenav]

It's the behaviour of the potential on a small interval about the minimum.

Ok, so what should I do once I have the equilibrium points?

 

[PeroK]

Ok, so what should I do once I have the equilibrium points?

Have you heard of the Taylor series expansion of a function?

 

[Davidllerenav]

Have you heard of the Taylor series expansion of a function?

Yes, they are like this: $f(x)\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$.

 

[PeroK]

Yes, they are like this: $f(x)\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$.

Can you see how to use that here? Using what you know about $U$.

 

[Davidllerenav]

Can you see how to use that here? Using what you know about $U$.

Should I expand the potential using the Taylor expansion centered on the equilibrium point?

 

[PeroK]

Should I expand the potential using the Taylor expansion centered on the equilibrium point?

You could try that. But in your OP you had a relevant equation:

Relevant Equations: $x''+\omega^{2}x = 0$

How does $x''$ depend on the potential?

 

[Davidllerenav]

How does $x''$ depend on the potential?

I don't know. We know that $x''$ is the acceleration, so maybe I can use the fact that the force is related to the potential by $F=-grad \ U$, since $F=ma=mx''$.

 

[PeroK]

I don't know. We know that $x''$ is the acceleration, so maybe I can use the fact that the force is related to the potential by $F=-grad \ U$, since $F=ma=mx''$.

Yes, that's right. But, let's use $\ddot{x}$ for the second time derivative. So, you have:

$m \ddot{x} = -U'(x)$

Now, use the Taylor series.

 

[Davidllerenav]

Now, use the Taylor series.

With the potential, centered at a=0?

 

[PeroK]

With the potential, centered at a=0?

Of course. That's the point of interest for small oscillations.

 

[Davidllerenav]

Of course. That's the point of interest for small oscillations.

Ok, of what degree?

 

[PeroK]

Ok, of what degree?

I'll leave that to you to decide.

 

[Davidllerenav]

I'll leave that to you to decide.

Ok, I did it of degree 3, and I go this: $U'(x)\approx U_0a\sin(0)+U_0a^2\cos(0)(x)-\frac{U_0a^3\sin(0)}{2!}(x)^2-\frac{U_0a^4\cos(0)}{3!}(x)^3$, so $U'(x)\approx U_0a^2x-\frac{U_0a^4x^3}{6}$. What should I do know?

 

[PeroK]

Ok, I did it of degree 3, and I go this: $U'(x)\approx U_0a\sin(0)+U_0a^2\cos(0)(x)-\frac{U_0a^3\sin(0)}{2!}(x)^2-\frac{U_0a^4\cos(0)}{3!}(x)^3$, so $U'(x)\approx U_0a^2x-\frac{U_0a^4x^3}{6}$. What should I do know?

If $x$ is small, then the linear term dominates and you get the equation for a simple harmonic potential by dropping the term term in $x^3$.

 

[Davidllerenav]

If $x$ is small, then the linear term dominates and you get the equation for a simple harmonic potential by dropping the term term in $x^3$.

Would you explain that a bit more please?

 

[PeroK]

Would you explain that a bit more please?

This is the standard approach when expanding as a Taylor series for "small" $x$. Note that in this case the quantity $ax$ is dimensionless, so when $ax$ is small we have $ax \gg a^2x^2 \gg a^3x^3 \dots$.

This leaves what's called a linear approximation for the function. Examples include $\sin \theta = \theta$, which is often used for a simple pendulum for small oscillations.

 

[Davidllerenav]

This is the standard approach when expanding as a Taylor series for "small" $x$. Note that in this case the quantity $ax$ is dimensionless, so when $ax$ is small we have $ax \ll a^2x^2 \ll a^3x^3 \dots$.

This leaves what's called a linear approximation for the function. Examples include $\sin \theta = \theta$, which is often used for a simple pendulum for small oscillations.

Wouldn't $a^2x^2$ be less than $ax$ when x is small? This only works for small oscillations, right?

 

[PeroK]

Wouldn't $a^2x^2$ be less than $ax$ when x is small? This only works for small oscillations, right?

Yes. I've corrected the typo. The analysis in this particular problem is the same in general for any potential:

Local minimum; first derivative is 0; second derivative is positive; linear approximation of $U'$ for small oscillations; SHM for small oscillations.

It's useful to remember this whole idea as it is used all over physics.

 

[Davidllerenav]

Yes. I've corrected the typo. The analysis in this particular problem is the same in general for any potential:

Local minimum; first derivative is 0; second derivative is positive; linear approximation of $U'$ for small oscillations; SHM for small oscillations.

It's useful to remember this whole idea as it is used all over physics.

I see. So x is small because of small oscilations right? Since x is the coordinate, it would be small.

 

[PeroK]

I see. So x is small because of small oscilations right? Since x is the coordinate, it would be small.

Effectively, yes.

 

[Davidllerenav]

Effectively, yes.

Ok, so whenever I need to expand a potential for small oscillations, I just need the term with a linear $x$?
After getting that $U'(x)\approx U_0a^2x$, I replace that on the equation and get that $F=-U_0a^2x$ what do I do next? Do I need the harmonic oscillator equation?

 

[PeroK]

Ok, so whenever I need to expand a potential for small oscillations, I just need the term with a linear $x$?
After getting that $U'(x)\approx U_0a^2x$, I replace that on the equation and get that $F=-U_0a^2x$ what do I do next? Do I need the harmonic oscillator equation?

That is the equation of SHM. $F = -kx$, for some constant $k$.

 

[Davidllerenav]

That is the equation of SHM. $F = -kx$, for some constant $k$.

From that I have $k=U_0a^2$, how do I use the constant to find the period?

 

[PeroK]

From that I have $k=U_0a^2$, how do I use the constant to find the period?

You posted the relevant equation in your OP.

 

[Davidllerenav]

You posted the relevant equation in your OP.

$x''+\omega^{2}x=0$?

 

[PeroK]

$x''+\omega^{2}x=0$?

Yes, and the other one.

 

[Davidllerenav]

Yes, and the other one.

$T=2\pi/\omega$? One question, can't I just write $F$ as $mx''$ such that $mx''=-U_0a^2x$? Thus $x''=-\frac {U_0a^2}{m}x$ and since $x''+\omega^{2}x=0\Rightarrow x''=-\omega^2x$ then $\omega^2=\frac {U_0a^2}{m}$?

 

[PeroK]

$T=2\pi/\omega$? One question, can't I just write $F$ as $mx''$ such that $mx''=-U_0a^2x$? Thus $x''=-\frac {U_0a^2}{m}x$ and since $x''+\omega^{2}x=0\Rightarrow x''=-\omega^2x$ then $\omega^2=\frac {U_0a^2}{m}$?

Yes, that's the idea.

 

[Davidllerenav]

Yes, that's the idea.

And how do I get to that with the constant $k$?

 

[PeroK]

And how do I get to that with the constant $k$?

You have:

$T=2\pi/\omega$

$\omega^2=\frac {U_0a^2}{m}$?

You don't need anything else.

 

[Davidllerenav]

You have:
You don't need anything else.

Ok, using that I get $\omega=\sqrt {\frac {U_0a^2}{m}}$ thus $T=2\pi\sqrt {\frac {m}{U_0a^2}}\Rightarrow T=\frac {2\pi}{a}\sqrt {\frac {m}{U_0}}$, is that right?

 

[Davidllerenav]

@PeroK I have one question. I can also use the formula $\omega^2=\frac{k}{m}$ right? Also, how can I solve a problem like this if the equilibrium point is not 0?

 

[PeroK]

@PeroK I have one question. I can also use the formula $\omega^2=\frac{k}{m}$ right? Also, how can I solve a problem like this if the equilibrium point is not 0?

Yes.

The Taylor series can be expanded about any point. As you posted in post #9 for a point $a$.

 

[Davidllerenav]

Yes.

The Taylor series can be expanded about any point. As you posted in post #9 for a point $a$.

The Taylor series can be expanded about any point. As you posted in post #9 for a point $a$.

But for example if the equilirbium point is $y$ then would end up with something like $U_0a^2(x-y)$, what do I do with that y?

 

[timetraveller123]

at the equilibrium point there would no linear terms in x so
but if your quadratic term is non-zero(positive) then you can do what you did before the coefficient is the "spring constant" if you shift your coordinate system to the y point then then it is similar to what you did before

edit:
just realized you were talking about force so there would be linear force and hence linear term in x so once again you can shift your coordinate system origin to y this makes it similar to what you solved before

 

[Davidllerenav]

at the equilibrium point there would no linear terms in x so
but if your quadratic term is non-zero(positive) then you can do what you did before the coefficient is the "spring constant" if you shift your coordinate system to the y point then then it is similar to what you did before

edit:
just realized you were talking about force so there would be linear force and hence linear term in x so once again you can shift your coordinate system origin to y this makes it similar to what you solved before

I understand, thanks. But what i meant was if the equilibrium point was not 0, but anotber constante, say $c$, such that after using Taylor series I end up with $U_0a^2(x-c)$, I would have $U_0a^2x-U_0a^2c$, right? What do I do? Do I just replace that on the equation?

 

[timetraveller123]

no what i am saying is even if the equilibrium point is not at zero you can make it zero by shifting your coordinate system to that point or you don't even have to shift . The new oscillation is just same as the one about the equilibrium about 0 but just that now it is at c. you still use the coefficient in front of the linear force term . in this case the oscillation frequency would be same because of the sine nature of the potential so in this case there is another stable equilibrium point at $\frac{2 \pi}{a}$ the ball oscillates about that point the same way as in the point at origin as the ball doesn't which equilibrium point it is in too all the stable points are the same