Physics 325 Lecture 5

# Physics 325 Lecture 5 - Physics 325 Lecture 5 So now...

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Physics 325 Lecture 5 So now, Equation 4.10 becomes () 23 00 11 ! xx dU Ux x x dx dx == =+ L + (5.1) We write leading term in the expansion as 2 1 2 k x with 2 kdU d x = 2 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 22 0 E mv kx We plug this in to Equation 4.8 to find 21 1 2 2 vx E k x E k x mm ⎛⎞ =−= − ⎜⎟ ⎝⎠ With the following substitutions 0 2 km A E k ω we can write A x = And with Equation 4.9 0 dx tt A x −= The integral can be done with the clever substitution x=A sin θ : 0 2 cos 1s i n dx A d d Ax A θθ = −− ∫∫ So we have 0 0 −=− . Or defining a phase angle 0 t 0 φ = + we write 1 sin x t A ωφ +== Therefore ( ) ( ) sin xt A t = + (5.2)

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This is a simple harmonic oscillator with amplitude A and phase φ 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) In one dimension: () FF v = rr r dv Fv m x m dt m dv dt = =→ = ∫∫ && (5.3) 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) and integrate this a second time to find x(t) . There are many forces of this time, air resistance for example.
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## 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.

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Physics 325 Lecture 5 - Physics 325 Lecture 5 So now...

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