ch15_03 - P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY...

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1240 CHAPTER 15 . . Second-Order Differential Equations 15-20 Multiplying this out, we have ( D 2 + cD + k ) F = ( + k ) f 0 sin ω t . Show that this gives the correct answer: that is, F ( t ) satisfies the equation F ±± + cF ± + kF = c ( f 0 sin ω t ) ± + k ( f 0 sin ω t ). 2. Spring devices are used in a variety of mechanisms, includ- ing the railroad car coupler shown in the photo. The coupler allows the railroad cars a certain amount of slack but applies a restoring force if the cars get too close or too far apart. If y measures the displacement of the coupler back and forth, then y ±± = F ( y ), where F ( y )is the force produced by the coupler. A simple model is F ( y ) = y d if y ≤− d 0i f d y d . y + d if y d This models a restoring force with a dead zone in the middle. Suppose the initial conditions are y (0) = 0 and y ± (0) = 1. That is, the coupler is centered at y = 0 and has a positive velocity. At y = 0, the coupler is in the dead zone with no forces. Solve y ±± = 0 with the initial conditions and show that y ( t ) = t for 0 t d .At this point, the coupler leaves the dead zone and we now have y ±± =− y + d . Explain why initial conditions for this part of the solution are y ( d ) = d and y ± ( d ) = 1. Solve this problem and determine the time at which the coupler reenters the dead zone. Continue in this fashion to construct the solution piece by piece. Describe in words the pattern that emerges. Then, explain in which sense this model ignores damping. Revise the function F ( y )to include damping. 15.3 APPLICATIONS OF SECOND-ORDER EQUATIONS In sections 15.1 and 15.2, we developed models of spring-mass systems with and without external forces. Surprisingly, the charge in a simple electrical circuit can be modeled with the same equation as for the motion of a spring-mass system. An RLC -circuit consists of resistors, capacitors, inductors and a voltage source. The net resistance R (measured in ohms), the capacitance C (in farads) and the inductance L (in henrys) are all positive. For now, we will assume that there is no impressed voltage. If Q ( t ) (coulombs) is the total charge on the capacitor at time t and I ( t )is the current, then I = Q ± ( t ). The basic laws of electricity tell us that the voltage drop across the resistors is IR , the voltage drop across the capacitor is Q C and the voltage drop across the inductor is LI ± ( t ). ~ Capacitance C Resistance R Voltage E ( t ) Inductance L These voltage drops must sum to the impressed voltage. If there is none, then LI ± ( t ) + RI ( t ) + 1 C Q ( t ) = 0 or LQ ±± ( t ) + RQ ± ( t ) + 1 C Q ( t ) = 0 . (3.1) Observe that this is the same as equation (1.2), except for the names of the constants. Example 3.1 works the same as the examples from section 15.1.
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15-21 SECTION 15.3 . . Applications of Second-Order Equations 1241 EXAMPLE 3.1 Finding the Charge in an Electrical Circuit A series circuit has an inductor of 0.2 henry, a resistor of 300 ohms and a capacitor of 10 5 farad. The initial charge on the capacitor is 10 6 coulomb and there is no initial current. Find the charge on the capacitor and the current at any time t .
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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_03 - P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY...

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