This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 325, Fall 2010 Prof. Susan Lamb Homework Assignment #2 Solutions 1a) F ( x ) = dU dx = αe − βx 2 parenleftBig 2 βx 2 1 parenrightBig 1b) An example is given in the figure below, where α and β are numerically equal to 1. x U(x)0.40.30.20.1 0.1 0.2 0.3 0.44321 1 2 3 4 The possible motion a particle can execute depends on its energy. If the energy is larger than zero, then the particle is unbound. If x > 0, then the particle comes in from the right, it slows down as it loses kinetic energy and gains potential energy. If the total energy is less than the maximum value of U ( x ), then the particle reaches a turning point and travels back to the right. If E > U max , then the particle continues traveling left, picking up speed as it travels through the low U region just to the left of the origin, and then slowing back down to its original speed as U → 0. If the particle comes in from the left with E > 0, the situation is similar, except it travels faster through the region of negative U before arriving at its turning point. If E > U max , it continues to the right. If E < 0, the particle is bound in the potential well with x < 0. If E > U min then the particle oscillates back and forth between two turning points. If E = U min then the particle remains stationary at the bottom. 1c) The equilibrium points are the places where F = 0. These would be x = + ∞ and x =∞ , if you were actually able to put a particle there (you cannot!), andif you were actually able to put a particle there (you cannot!...
View
Full
Document
This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Physics, mechanics, Work

Click to edit the document details