# class6 - 1.6 Equilibrium stability and bounded motion...

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1.6 Equilibrium, stability and bounded motion Consider a potential energy that is only a function of the position x in one-dimension: U = U ( x ). The energy conservation equation is E = 1 2 mv 2 + U ( x ) implying (56) v ( t )= dx dt = ± ± 2 m [ E - U ( x )] (57) which implies periodic oscillations for bounded motion. We can integrate t 2 - t 1 = ± m 2 ² x 2 x 1 dx ³ E - U ( x ) (58) Looking at this equation, it is clear that only those values of x with E - U ( x ) positive are possible. example: A particle of mass 2 moves on the positive x -axis under the force F =4 /x 2 - 1. Initially, the particle is released from rest at the point x = 4. Find the turning points and the period of motion. solution: The potential is U ( x ) = 4 /x + x , so that energy conservation states: 1 2 (2) v 2 + 4 x + x = E (59) The initial condition v = 0 when x = 4 gives E = 5, so that v 2 =5 - 4 x - x (60) The turning points occur when v = 0 , or x =1 , 4. To ±nd the period, write v = dx/dt in the last equation and take the square roots: dx dt = ± ± ( x - 1)(4 - x ) x (61) The period is the time it takes to get from x

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## This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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class6 - 1.6 Equilibrium stability and bounded motion...

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