How to find the stability of equibrium points

[cloud360]

Homework Statement

I want to know if i got answer to C correct
[PLAIN]http://img822.imageshack.us/img822/7804/equilbrium.gif

Homework Equations

The Attempt at a Solution

[PLAIN]http://img703.imageshack.us/img703/508/equilbriumsol.gif

 

[cloud360]

I am thinking it is more simple then i think.

Do we just sub values into V’, whatever gives V’=0 is stable. So x=1 and x=2 are stable?

 

[Mark44]

You are mostly on the right track, but you have a couple of errors.

The critical points are at x = 1 and x = 2 (correct).
V''(x) = (4 - 3x)/x³, which is correct and is what you have. However, you have (4 - 3x)/x³ --> 4 - 3x. The two expressions aren't equal, and V''(x) \neq 4 - 3x. I don't understand the significance of the arrow, -->. You should remove it and the 4 - 3x expression. That's one of the errors I referred to.

The second error is that you calculated V''(x) using 4 - 3x, not (4 - 3x)/x³, which is equal to 4/x³ - 3/x². You have V''(1) = 1 (correct) and V''(2) = -2 (incorrect). V''(2) = 4/8 - 3/4 = 1/2 - 3/4 = -1/4. Your calculation for V''(1) would have been incorrect if x had been any value other than 1.

 

[cloud360]

You are mostly on the right track, but you have a couple of errors.

The critical points are at x = 1 and x = 2 (correct).
V''(x) = (4 - 3x)/x³, which is correct and is what you have. However, you have (4 - 3x)/x³ --> 4 - 3x. The two expressions aren't equal, and V''(x) \neq 4 - 3x. I don't understand the significance of the arrow, -->. You should remove it and the 4 - 3x expression. That's one of the errors I referred to.

The second error is that you calculated V''(x) using 4 - 3x, not (4 - 3x)/x³, which is equal to 4/x³ - 3/x². You have V''(1) = 1 (correct) and V''(2) = -2 (incorrect). V''(2) = 4/8 - 3/4 = 1/2 - 3/4 = -1/4. Your calculation for V''(1) would have been incorrect if x had been any value other than 1.

Hi, thanks again for your reply.

So was i right to assume that V''(x)<0 ,means that point x is unstable, and

V''(x)L>0, means point x is stable?

or am i wrong here. i am not sure if a stable point is just when V'(x)=0? or whether it is what i said above?

 

[Mark44]

Hi, thanks again for your reply.

So was i right to assume that V''(x)<0 ,means that point x is unstable, and

V''(x)L>0, means point x is stable?

Truthfully, I don't know, but how you have it seems reasonable to me. If V''(x) < 0, the graph of V(x) is concave down, and the particle is at a local maximum point for potential energy. The particle could go left or right to get to places with lower potential energy.

On the other hand, if V''(x) > 0, the graph of V(x) is concave up, the the particle is at a local minimum point for potential energy. Whether the particle goes left or right, it "wants" to get back to the point of minimum potential energy.

I could be wrong, but that's how it seems to me.

Your textbook should have an explanation of how to determine whether a critical point of V(x) is stable or unstable.

or am i wrong here. i am not sure if a stable point is just when V'(x)=0? or whether it is what i said above?

 

[cloud360]

Truthfully, I don't know, but how you have it seems reasonable to me. If V''(x) < 0, the graph of V(x) is concave down, and the particle is at a local maximum point for potential energy. The particle could go left or right to get to places with lower potential energy.

On the other hand, if V''(x) > 0, the graph of V(x) is concave up, the the particle is at a local minimum point for potential energy. Whether the particle goes left or right, it "wants" to get back to the point of minimum potential energy.

I could be wrong, but that's how it seems to me.

Your textbook should have an explanation of how to determine whether a critical point of V(x) is stable or unstable.

my lecturer had this to say (see attachment)

would you say i have interpreted his notes properly?

 

[Mark44]

Yes.

You should also change the "circular imaginary function" part. It's the roots of the characteristic equation that are imaginary, which give rise to the circular functions sine and cosine. The functions themselves are not imaginary.

 

[cloud360]

No. Aside from what your lecturer is talking about characteristic equations (I don't know what they are for this problem), he's saying that V''(x_e) < 0 ==> x_e is stable, and that V''(x_e) > 0 ==> x_e is unstable. That's the opposite of what you said and what I thought.

You should also change the "circular imaginary function" part. It's the roots of the characteristic equation that are imaginary, which give rise to the circular functions sine and cosine. The functions themselves are not imaginary.

I am pretty sure he is saying if V''(x)<0 its unstable. that's what i said, can you please check again, kindly. as now i am worried :(

 

[cloud360]

I have attached his entire notes, please see pg11, it has the stuff about equilibrium points. as i am not convinced that my statements are exact opposite of my lecturers

 

[Mark44]

I am pretty sure he is saying if V''(x)<0 its unstable. that's what i said, can you please check again, kindly. as now i am worried :(

Sorry, what you had was correct. I guess my brain stopped working momentarily. If V''(x) > 0, you get stablity, and if V''(x) < 0, you get instability, which is what I surmised earlier (post #5).

Also, now I see where the characteristic equation is coming from.