# Figure 8 36 shows the potential energy curve for a

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Unformatted text preview: he zero of potential energy at X = 0, then ':-lax2 + ~bX4. (b) A graph of U(x) for X ~ 0, .:,..... . 3 =5 N/m, b = 2 N/m and x is in meters, is <(Note that the potential energy is symmetric, '==:)J(x), but that only positive displacements idered in this problem.) The conservation. of an be written in terms of the total energy, (dx/dt)2 + U(x), so that dx/dt = ..,.U(x)J/m. The maximum speed occurs when anlinimumj i.e., dU/dx = 0, and d2Ufdx2 > o. the derivative, one finds 0 = -ax + bx3, which tionsx = 0 and x = :f:VO]b = :f:.j512 m = The second derivative d2U jdx2 = -a + 3bx2 IVEdorx = 0, which is a local maximum, but is .~f(}r = -i:..,JO:Jb, which are minima with )U(:f:VOJb) = -a2/4b = -(25/8) J = -3.13 J. i~teal physical motion (K ~O) for total energy iii; The turning points (where dx/dt = 0) can from the equation U(x) = Ej there are four positive) for energies with Umin < E < solutions (one positive) for E > O. ~~!~;JIJ;~5 ~ ..L+ ..J. ; x(m) =~":::": :r:J~.{~):.'C.t ... 5 : 6 t.. L.tl . j -3 FIGURE .. . 8-36 Problem 44. x Solution The f...
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## This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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