Potential wells and oscillations
Since V(x) = -  F(x) dx , we can invert this to have: F(x) = - dV(x)/dx .
If the plot of V(x) versus x has a dip (a potential well), then at the bottom of that dip the
slope will be zero, so dV(x)/dx = 0 at that point.
But all places that have dV/dx = 0 are not necessarily at the bottom of dips (or wells).
V(x) could have a slope of zero at the top of a peak, or at an inflection point (where the
slope of the slope changes sign).
If the slope of the slope is positive in the local area of the place where dV/dx is zero, then
we have a dip. If the slope of the slope is negative, then we have a peak. If the slope of
the slope is zero, then we have an inflection point. We will draw examples of each of
these three cases in class. See also the excel spreadsheet on Slope of the Slope.
At the bottom of a dip (or well), the curve will look at least approximately like that of a
parabola centered on the point where dV/dx is zero. Mathematically, we can show this
by using the power series expansion. If we call the location where dV/dx = 0 as "a",
then:
V(x) = V(a) + [dV/dx]x=a (x-a) + [d2V/dx2]x=a (x-a)2/2 + ...
The first term, V(a), is just a constant and is unimportant since potential energy is defined
with an arbitrary constant in it (the standard position). The second term is zero since
[dV/dx] evaluated at x=a is zero (it is at the bottom of the well). Hence, the potential
energy looks approximately like that of a harmonic oscillator: V(x) = (1/2)kx2 where k
= [d2V/dx2]x=a . If the object has some energy slightly above V(a), then it will oscillate
with an approximate frequency:  = (k/m)1/2 . If we get too far from x=a, then the higher
order terms will become more important and the oscillation will become more distorted
from a purely sine oscillation and the frequency will vary more from the simple  =
(k/m)1/2 .
Homework Problem: Problem #8 Consider a particle of mass m that has the
potential energy function: PE(x) = ax2 – bx3 , where a and b are positive
constants.
a) Graph PE(x) versus x. (Your choice for values for a and b as long as both a
and b are positive. Warning: be careful with your choice of values for a and b
and with your range of x when plotting this graph – if your range of x is too big
you may miss the “well”. Your values of a and b may affect the width and depth
of the “well” also.)
b) Is there a potential energy “well” near the origin? If so, find the critical velocity
for the particle when it is at the position x=0 for the particle to be able to escape
from this well.
c) Find the force, F(x) that is associated with this potential energy.
Homework Problem: Problem #9 Consider a potential energy of the form:
PE(x) = -a/x6 + b/x12 where a and b are positive constants. (This is one model
for the potential energy for the force between two atoms in a diatomic molecule.)
Assume one molecule is very heavy and doesn’t move much, and the other
molecule is very light in comparison with mass, m, and does move in one
dimension (x) based on the force due to the above potential energy function.
a) Graph PE(x) versus x. (Your choice for values for a and b as long as both a
and b are positive. Warning: be careful with your choice of values for a and b
and with your range of x when plotting this graph – if your range of x is too big
you may miss the “well”. Your values of a and b may affect the width and depth
of the “well” also.)
b) Find the equilibrium distance between the two molecules (the equilibrium
position of the lighter atom).
c) Find the frequency of small oscillations of the light atom about it’s equilibrium
position in the molecule.
HINT:  (dy /  [1-y²]) = arcsin(y) ; or HINT: let sin(A) = x  [k/2 E]