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 $k$?