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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.) 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.) 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]

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neutrino oscillations, Simple Harmonic Motion, natural frequency, forced oscillations, solar neutrinos, electron neutrino, neutrino masses, restoring force, driving force, equilibrium position

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posted: | 5/22/2011 |

language: | English |

pages: | 2 |

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