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

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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...
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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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