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Phys 325 Spring 2011 Lecture 3

# Phys 325 Spring 2011 Lecture 3 - Physics 325 Lecture 3...

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Physics 325 Lecture 3 Equation of motion for F=F(x) case, Energy (cont.) Recall that the Taylor expansion of the potential about an equilibrium point x = 0 is given by (Equation 4.11)   2 3 2 3 2 3 0 0 1 1 2 3! x x d U d U U x x x dx dx (3.1) We write leading term in the expansion as   2 1 2 U x kx with 2 2 k d U dx evaluated at x = 0. The stability of the motion depends on the sign of the second derivative, that is the sign of k . If k >0 , the equilibrium is stable and if k <0 the equilibrium is unstable. Let’s look at the stable case, k >0 : We assume that we have known initial conditions x 0 and v 0 that determine E 0 2 2 0 0 0 1 1 2 2 E mv kx We plug this in to the quadratic equation for the velocity (Equation 4.8) to find   2 2 0 0 2 1 1 2 2 v x E kx E kx m m With the following substitutions 0 2 k m A E k we can write   2 2 v x A x And with the integral form for the time evolution (Equation 4.9): 0 2 2 dx t t A x

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The integral can be done with the clever substitution x =A sin : 0 2 2 2 cos 1 sin dx A d d A x A   So we have 0 0 t t . Or defining a phase angle 0 0 t we write 1 sin x t A Therefore   sin x t A t (3.2) This is a simple harmonic oscillator with amplitude A and phase that are determined from the initial conditions. For example, the amplitude A is determined by the total energy E 0 the oscillator moves between the turning points. We continue with our exploration of different types of forces. e) F F v : In one dimension:     dv F v mx m dt m dv dt F v (3.3) The procedure is the same as we have done before. Assuming we can integrate the left hand side, we do so and then invert t ( v ) to find v ( t
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