U 7 80x 17x2 11 where u is in joules and x m meters

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: energy, satisfies Equation 8-7 with total energy E = K + U = ~in(dx/dt? + U{x). The turning points are the solutions of this equation whendx/dt = OJ that is, E = U(x) = 3.5 = 7.0 8.0x + 1.7x2 (where energy is in joules and displacement in meters). The quadratic formula gives x = (1.7)-1(4.0-:t:. v'lO.l)m= 0.488 m and 4.22 m. Solution The potential energy is symmetric about x = Ox (U( -x) = U(x, so we need only plot it for positive x. The zeros of U are at x= -:t:.y'b7a = :f:l.83 m. The maxima and minimum can be found from the zeros of the derivative, dUjdx = (2x/e2)(b+ ac2 - ax2)e-x2/e2, which areXmin = 0 and Xmax = :1:Je2 + b/a = :1:3.51m. The corresponding energies are Umin= -b == -5.0 J, and Umax = ac2e-(I+blae2) = 3.43 J. We finally note that U(8 m) = 91 e-7.1l J = 7.43xlO-2 J. One can see that the particles with energies less than Umaxare confined to the well, with turning points on either side of :1:1.83ill, respectively. The speed ofa particle is v = J2(E - U)/m. Since U -t 0 for x -t 00, the speed of the particle with E = 4 J approaches J2(4 J)/(1 kg) == 2.83 mls for la...
View Full Document

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.

Ask a homework question - tutors are online