Figure 8 36 shows the potential energy curve for a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online