ch15_04 - P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY...

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15-29 SECTION 15.4 . . Power Series Solutions of Differential Equations 1249 glucose/100 ml blood two hours after the injection and 78 mg glucose/100 ml blood three hours after the injection. 40. Show that the data in exercise 39 are inconsistent with the case 0 <ω<α . 41. Consider an RLC -circuit with capacitance C and charge Q ( t ) at time t . The energy in the circuit at time t is given by u ( t ) = [ Q ( t )] 2 2 C . Show that the charge in a general RLC -circuit has the form Q ( t ) = e ( R / L ) t / 2 | Q 0 cos ω t + c 2 sin ω t | , where Q 0 = Q (0) and ω = 1 2 L ± R 2 4 L / C . The relative energy loss from time t = 0t o time t = 2 π ω is given by U loss = u (2 π/ω ) u (0) u (0) and the inductance quality factor is defined by 2 π U . Using a Taylor polynomial approximation of e x , show that the inductance quality factor is approximately ω L R . EXPLORATORY EXERCISES 1. In quantum mechanics, the possible locations of a particle are described by its wave function ± ( x ). The wave function sat- isfies Schr¨odinger’s wave equation ¯ h 2 m ± ±± ( x ) + V ( x ) ± ( x ) = E ± ( x ) . Here, ¯ h is Planck’s constant, m is mass, V ( x )is the potential function for external forces and E is the particle’s energy. In the case of a bound particle with an infinite square well of width 2 a , the potential function is V ( x ) = 0 for a x a . We will show that the particle’s energy is quantized by solv- ing the boundary value problem consisting of the differential equation ¯ h 2 m ± ±± ( x ) + v ( x ) ± ( x ) = E ± ( x ) plus the boundary conditions ± ( a ) = 0 and ± ( a ) = 0. The theory of boundary value problems is different from that of the initial value prob- lems in this chapter, which typically have unique solutions. In fact, in this exercise we specifically want more than one solution. Start with the differential equation and show that for V ( x ) = 0; the general solution is ± ( x ) = c 1 cos kx + c 2 sin , where k = 2 mE / ¯ h . Then set up the equations ± ( a ) = 0 and ± ( a ) = 0. Both equations are true if c 1 = c 2 = 0, but in this case the solution would be ± ( x ) = 0. To find nontrivial solutions (that is, nonzero solutions), find all values of k such that cos ka = 0or sin = 0. Then, solve for the energy E in terms of a, m and ¯ h . These are the only allowable energy levels for the particle. Finally, determine what happens to the energy levels as a increases without bound. 2. Imagine a hole drilled through the center of the Earth. What would happen to a ball dropped in the hole? Galileo conjec- tured that the ball would undergo simple harmonic motion, which is the periodic motion of an undamped spring or pendu- lum. This solution requires no friction and a nonrotating Earth. The force due to gravity of two objects r units apart is Gm 1 m 2 r 2 , where G is the universal gravitation constant and m 1 and m 2 are the masses of the objects. Let R be the radius of the Earth and y the displacement from the center of the Earth.
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This note was uploaded on 10/04/2010 for the course CHE2C 929102 taught by Professor Carter during the Spring '10 term at UC Davis.

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ch15_04 - P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY...

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