Help with particle in a potential

[Schmetterling]

Hello,

I would like you to check this problem.

Homework Statement

A particle of mass m is moving in a potential V(x) = $\frac{cx}{x^2+a^2}$, where a, c > 0.

1. Plot V(x)
2. Find the position of stable equilibrium and the period (small oscillations)
3. Find the values of velocity for which the particle: (a) oscillates, (b) escapes to $\infty$, (c) escapes to $-\infty$.

Homework Equations

V(x) = $\frac{cx}{x^2+a^2}$, a, c > 0.

The Attempt at a Solution

1. Well, I started by doing x = 0 to find the value of V in the origin.
Next, x = a --> V(x) = $\frac{c}{2a}$, x = na (n>1) --> V(x) = $\frac{nc}{a(n^2+1)}$. With c = 1 I can plot V(x) for x = a/2, a, 2a, 3a.
But if I do x = -na, V(x) turns negative... I think V(x) should be positive along the entire x-axis, shouldn't it? Otherwise, how can the particle oscillate?

2. I suppose the position of stable equilibrium is x = 0, isn't it?
Period = ?

3. Velocity. In order to the particle doesn't go to infinity (i.e. it stays in the region -a$\leq$ x $\leq$ a), its kinetic energy K needs to be $\leq$ V, right? Then, with K = $\frac{mv^2}{2}$ = V(x=a), (the maximum value of) v = $\sqrt{ \frac{2V}{m} }= \frac{1}{\sqrt {ma}}$ . For v > $\frac{1} {\sqrt{ma}}$ the particle escapes to infinity. But, what happens when x is negative?

Thank you.

 

[JesseC]

1) the way i'd think about sketching that potential is:

i) its an odd function, thus V(-x) = - V(x). Or in words, the function for positive x will look look identical to that for negative x when reflected in x axis, then in V(x) axis.
ii) At large x, the function can be approximated to c/x, thus is must tend to zero at both positive and negative infinity.
iii) When x is small (less than 1), the x^2 term can be neglected, thus the function could be approximated to V(x) = (c/a)x about zero.

Have you actually sketched this function? It would help you with the rest of the answer if you did because it is clearly not positive along the whole x-axis.

2) A particle is in stable equilibrium when it experiences no *net* force. If you know F = - dV/dx then it should follow how to find the value of x for which it is in equilibrium. (Particles in stable equilibrium reside at the bottom of potential wells) By the way, it is not at x=0.

To find the period of small oscillations you might want to think of a way to simplify the potential about the point of equilibrium. Heard of taylor or binomial expansion or something similar? Perhaps consider what happens when you displace the mass a small distance from equilibrium.

Do those first then think about velocity.

 

[mathfeel]

1. Plot V(x) ... I think V(x) should be positive along the entire x-axis, shouldn't it? Otherwise, how can the particle oscillate?

Not it is not. $V(x) < 0$ when $x<0$

2. I suppose the position of stable equilibrium is x = 0, isn't it?

Not it is NOT. Equilibrium is when the derivative of the potential is zero. Stable equilibrium is when in addition the second derivative is positive. If you plot V(x) as part 1 asked, it is where V(x) has a minimum and looks locally like a bowl. I can tell without plotting that V(x) has a minimum at some negative x, perhaps x = a.

Since $\frac{c x}{x^2 + a^2}$ is an odd function, $x=0$ is NOT a stable equilibrium. Actually in this case it is not even an deflection point.

3. Velocity.

Again, make a plot of V(x). In order to escape to $-\infty$, the kinet energy has to over come the depth of the bowl. In order to escape to $\infty$, it has too over come the hump on the positive x side.

As for frequency of small oscillation. Suppose that you found that V(x) has a minimum at $x=x_0$. Write $x = x_0 + \epsilon$ and do a series expansion of V(x) around $x=x_0$. Keep up to the second order term, whose coefficient gives you an effective "spring constant" for this potential for small oscillation.

 

[Schmetterling]

Thank you very much, JesseC and mathfeel.

Have you actually sketched this function? It would help you with the rest of the answer if you did because it is clearly not positive along the whole x-axis.

Yes, I did it. But I had the idea of "V(x) positive", so I didn't see immediately what about -x...

iii) When x is small (less than 1), the x^2 term can be neglected, thus the function could be approximated to V(x) = (c/a)x about zero.

Is it a or a^2?

To find the period of small oscillations you might want to think of a way to simplify the potential about the point of equilibrium. Heard of taylor or binomial expansion or something similar? Perhaps consider what happens when you displace the mass a small distance from equilibrium.

Do those first then think about velocity.

OK.

Since $\frac{c x}{x^2 + a^2}$ is an odd function, $x=0$ is NOT a stable equilibrium. Actually in this case it is not even an deflection point.

Ok... This was because thinking about it as a positive potential...

Equilibrium is when the derivative of the potential is zero. Stable equilibrium is when in addition the second derivative is positive. If you plot V(x) as part 1 asked, it is where V(x) has a minimum and looks locally like a bowl. I can tell without plotting that V(x) has a minimum at some negative x, perhaps x = a.

Yes! I sketched V(x) again and it looks nearly like a "bowl" in x = -a!

First attempt (wrong...):
[PLAIN]http://img8.imageshack.us/img8/1402/vx1n.jpg

Second attempt (better?):
[PLAIN]http://img23.imageshack.us/img23/6619/vx2r.jpg

Now, I will go on trying with the rest of this problem.

 

[Schmetterling]

Write $x = x_0 + \epsilon$ and do a series expansion of V(x) around $x=x_0$.

Sorry, what is $\epsilon$?Oh...

Perhaps consider what happens when you displace the mass a small distance from equilibrium.

Is it $\epsilon$?

 

[JesseC]

Is it a or a^2?

Yes technically it is a^2, but what is more important than the constants is the general shape of the function. Your second attempt at sketching it is good! :) Let us know if you have any more problems.

 

[JesseC]

Sorry, what is $\epsilon$?


Oh... Is it...

Yes :D your hunch is correct!

 

[Schmetterling]

Yes technically it is a^2, but what is more important than the constants is the general shape of the function.

You are right! :)

Your second attempt at sketching it is good! :) Let us know if you have any more problems.

Yes, thank you again!

Yes :D your hunch is correct!

Great! :D

 

[mathfeel]

Yes! I sketched V(x) again and it looks nearly like a "bowl" in x = -a!

You sketch looks like there is a "kink" at x = -a. This function should be smooth in all derivatives.

Nearly looking like a "bowl" is the whole point here! Any stable equilibrium looks locally like a bowl. And you can expand the potential in series around any point $x_0$:$$V(x) = V(x_0) + V^{\prime}(x_0) (x-x_0) + \frac{1}{2} V^{\prime\prime}(x_0)(x-x_0)^2 + \cdots$$
In particular, you can expand around the point of equilibrium so that some of these terms vanishes. Then $V(x)$ is nearly given by a parabola if you ignore all higher order terms. This is justified if $\epsilon \doteq x - x_0$ is small.

Now, what other potential is given by a parabola? And what kinds of motion does that have? How do you compute the frequency in that case?

 

[Schmetterling]

You sketch looks like there is a "kink" at x = -a. This function should be smooth in all derivatives.

Oops...

And you can expand the potential in series around any point $x_0$ ... In particular, you can expand around the point of equilibrium so that some of these terms vanishes. Then $V(x)$ is nearly given by a parabola if you ignore all higher order terms. This is justified if $\epsilon \doteq x - x_0$ is small.

I did it. I got:
$V(x) = -\frac{c}{2a}+\frac{1}{2} \frac{c}{2a^3}\epsilon^2$

Now, what other potential is given by a parabola? And what kinds of motion does that have? How do you compute the frequency in that case?

Ok... Harmonic oscillator?

$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2ma^3}{c}}$

Now I understand why it's necessary to expand the potential in series

Next step: velocity!

 

[JesseC]

I did it. I got:
$V(x) = -\frac{c}{2a}+\frac{1}{2} \frac{c}{2a^3}\epsilon^2$

Ok... Harmonic oscillator?

$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2ma^3}{c}}$

That looks right to me so long as you've got the algebra for V($\epsilon$) right. Have a stab at the velocity part.

 

[Schmetterling]

That looks right to me so long as you've got the algebra for V($\epsilon$) right.

I hope the algebra and the derivatives are right...

Next step: velocity!

I have to find the values of v for 3 cases: the particle escapes $\infty$, the particle escapes to $-\infty$, the particle oscillates... I feel a little confused...

1. $\infty$

In order to escape to $\infty$, it has too over come the hump on the positive x side.

The hump on the positive x side is on x = a, so kinetic energy K > V(x=a), ok?

$K = \frac{1}{2}mv^2 > \frac{c}{2a},$
$mv^2 > \frac{c}{a},$
$v > \sqrt{\frac{c}{am}}.$

2. $-\infty$

In order to escape to $-\infty$, the kinetic energy has to over come the depth of the bowl.

The bowl is in x = -a, ---->K > V(x=-a), but I think it only isn't enough if the particle wants to leave the potential well... What else does it need to escape? K > 0, maybe?

Sorry, I have to go right now... To be continued...

 

[JesseC]

Suppose the particle is sitting at the bottom of the "bowl", where would you define the "edges"? Are the edges the same height in either direction?

Don't forget you can define the zero point of potential to be wherever you please, what is important is the change in potential :)

 

[Schmetterling]

Ok...