Classical Mechanics - Potential Energy Function

AI Thread Summary
The potential energy function V(x) = cx/(x²+a²) has been analyzed to find equilibrium points, revealing x = a and x = -a as solutions. To determine stability, the second derivative test is necessary, with one equilibrium point being stable and the other unstable. The potential energy at these points is evaluated as V(a) = c/2 and V(-a) = -c/2, indicating differing stability. The period of small oscillations can be calculated using the derived equilibrium points and the appropriate formulas for harmonic motion. Understanding these concepts is crucial for solving related problems in classical mechanics.
teme92
Messages
185
Reaction score
2

Homework Statement


The potential energy function of a particle of mass m is V(x) = cx/(x2+a2), where c and a are positive constants.

Qualitatively sketch V as a function of x. Find two equilibrium points: identify which is a position of stable equilibrium, and find the period of small oscillations about it.

Homework Equations


PE=0.5mv2=-0.5kx2

The Attempt at a Solution


I think I'm supposed to differentiate and let it equal to zero which gave me:

(c(x2+a2) - 2cx2) / (x2+a2)2 = 0
cx2 + ca2 - 2cx2 = 0
x = a

Putting that into the function gave me: V(x) = cx2/2x2 = c/2

I don't know if any of this is necessary but it doesn't answer the question. Also when it says sketch do I just sub value of x in? Any help would be greatly appreciated.
 
Physics news on Phys.org
It answers part of the question, i.e., one of the equilibrium points. In order to find the other you must find the second solution to your equation (it exists as it is a second order equation in x).

To know which is stable you must look at the second derivative to deduce which is a min and which is a max.
 
Hey Orodruin thanks for the reply. How do I find the second solution?
 
You start from the equation you obtained and solve it for the general case. You have x^2 = a^2. This has two possible solutions.
 
Ah yes sorry, so the other value for x is -a. But this gives the same solution as it gets squared?
 
I just realized I should be subbing a in for x so I get V(x) = ±c/2a. So from this how do I tell where the stable equilbrium is? Also how do I find the period?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top