Understanding Potential Energy Graphs for Two-Particle Systems

[AsadaShino92]

Homework Statement

The potential energy V(R) of a two particle system exhibiting oscillatory behavior near a local minimum at the equilibrium separation Ro. V(R)= -(A/R)+(B/R^2) , where R is the interparticle separation.

A) Sketch V(R), what happens to V(R) as R→0
B) At what value of R is there a minima in the potential?
C) For very small oscillations about this equilibrium point, Ro, write the force on the particle F=-k(R-Ro), define k.

Homework Equations

V(R)= -(A/R)+(B/R^2)

The Attempt at a Solution

I must apologize in advance because I feel that I'm about to ask a stupid question. But how can I plot this function if I don't know what the values of A and B are?

 

[TSny]

It's a good question.

What if you plot $V$ in units of $A^2/B$ and $R$ in units of $B/A$?

[Edited to correct a mistake.]

 

[PeroK]

I would have just taken $A = B = 1$ for the purpose of a sketch.

 

[AsadaShino92]

It's a good question.

What if you plot $V$ in units of $A^2/B$ and $R$ in units of $B/A$?

[Edited to correct a mistake.]

Sorry, I'm not sure how I would plot the function in those terms such as A^2/B and B/A. Do you think I can fix the values like PeroK suggested? My only concern is that the function slightly changes based on what A and B are. But I see that the function runs off to infinity as R approaches 0.

 

[TSny]

Sorry, I'm not sure how I would plot the function in those terms such as A^2/B and B/A. Do you think I can fix the values like PeroK suggested? My only concern is that the function slightly changes based on what A and B are. But I see that the function runs off to infinity as R approaches 0.

Yes, you can go with PeroK's suggestion. The graph will be the same as I was suggesting.

Note that $A^2/B$ has the dimension of energy and $B/A$ has the dimension of length.

So, you can introduce dimensionless quantities $\tilde{V} = \frac{V}{A^2/B}$ and $\tilde{R} = \frac{R}{B/A}$.

If you write the equation in terms of these dimensionless quantities $\tilde{V}$ and $\tilde{R}$, you should find that $A$ and $B$ disappear.

Graphing $\tilde{V}$ versus $\tilde{R}$ will give the same graph as graphing $V$ versus $R$ with $A= B = 1$.

 

[AsadaShino92]

Yes, you can go with PeroK's suggestion. The graph will be the same as I was suggesting.

Note that $A^2/B$ has the dimension of energy and $B/A$ has the dimension of length.

So, you can introduce dimensionless quantities $\tilde{V} = \frac{V}{A^2/B}$ and $\tilde{R} = \frac{R}{B/A}$.

If you write the equation in terms of these dimensionless quantities $\tilde{V}$ and $\tilde{R}$, you should find that $A$ and $B$ disappear.

Graphing $\tilde{V}$ versus $\tilde{R}$ will give the same graph as graphing $V$ versus $R$ with $A= B = 1$.

Thanks for your explanation. I wasn't used to this method so I couldn't see the relationship before. For part B where it asks to find the minimum, taking A=B=1, I found an equilibrium point at R=1. Would it be correct for me to use the first derivative test?

 

[TSny]

For part B where it asks to find the minimum, taking A=B=1, I found an equilibrium point at R=1. Would it be correct for me to use the first derivative test?

For A = B = 1, I don't think the equilibrium is at R = 1. But, I'm pretty sure that you are meant to express the equilibrium value of R in terms of A and B.

Yes, use differentiation to find the minimum.

 

[AsadaShino92]

For A = B = 1, I don't think the equilibrium is at R = 1. But, I'm pretty sure that you are meant to express the equilibrium value of R in terms of A and B.

Yes, use differentiation to find the minimum.

So then the equilibrium value of R is at R=(B/A)? I found this by leaving A and B as variables and setting V(R)=0.

 

[TSny]

The equilibrium condition is not V(R) = 0. Equilibrium is where the force is zero.

 

[AsadaShino92]

The equilibrium condition is not V(R) = 0. Equilibrium is where the force is zero.

Since V(R) is given as the potential energy function in the problem, I can find my force function by using F(x)=-du/dx. Is that the right idea?

 

[TSny]

Since V(R) is given as the potential energy function in the problem, I can find my force function by using F(x)=-du/dx. Is that the right idea?

Yes. Good.