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Unformatted text preview: Chapter 4 Section 4.1 1. The differential equation is linear. 2. The differential equation is nonlinear. 3. The differential equation is linear. 4. The differential equation is linear. 5. The differential equation is nonlinear. 6. We will use the equation mg kL = 0. If the mass is 8 lb, then mg = 8. If the mass stretches the spring 6 inches, then L = 6. Therefore, 8 6 k = 0 which means k = 4 / 3 lb/inch. 7. Again, we use the equation mg kL = 0. Here the mass is 10 kg. The force due to gravity is g = 9 . 8 m/s 2 . Therefore, mg = 98 Newtons. The mass stretches the spring . 7 meters. Therefore, k = 98 /. 7 = 140 N/m. 8. The spring constant is k = 2 / (1 / 2) = 4 lb/ft. The mass m = 2 / 32 = 1 / 16 lb s 2 /ft. The equation of motion is 1 16 y 00 + 4 y = 0 or y 00 + 64 y = 0 with initial conditions y (0) = 1 / 4 ft, y (0) = 0 ft/sec. 9. The spring constant is k = . 98 /. 05 = 19 . 6 N/m. The mass m = . 1 kg. The equation of motion is . 1 y 00 + 19 . 6 y = 0 or y 00 + 196 y = 0 with initial conditions y (0) = 0 m, y (0) = . 1 m/sec. 10. The spring constant is k = 3 / (1 / 4) = 12 lb/ft. The mass m = 3 / 32 lb s 2 /ft. The equation of motion is 3 32 y 00 + 12 y = 0 or y 00 + 128 y = 0 with initial conditions y (0) = 1 / 12 ft, y (0) = 2 ft/sec. 11. The inductance L = 1 henry. The resistance R = 0. The capacitance C = 0 . 25 10 6 farads. Therefore, the equation for charge q is q 00 + (4 10 6 ) q = 0 with initial conditions q (0) = 10 6 coulombs, q (0) = 0 coulombs/sec. 1 12. The spring constant is k = . 196 /. 05 = 3 . 92 N/m. The mass m = . 02 kg. The damping constant is = 400 dynesec/cm = . 4 Nsec/cm. Therefore, the equation of motion is . 02 y 00 + . 4 y + 3 . 92 y = 0 or y 00 + 20 y + 196 y = 0 with initial conditions y (0) = . 02 m, y (0) = 0 m/sec. 13. The spring constant is k = 16 / (1 / 4) = 64 lb/ft. The mass m = 1 / 2 lb s 2 /ft. The damping coefficient is = 2 lbsec/ft. Therefore, the equation of motion is 1 2 y 00 + 2 y + 64 y = 0 or y 00 + 4 y + 128 y = 0 with initial conditions y (0) = 0 ft, y (0) = 1 / 4 ft/sec. 14. The spring constant is k = 3 /. 1 = 30 N/m. The mass m = 2 kg. The damping coefficient is = 3 / 5 Nsec/m. Therefore, the equation of motion is 2 y 00 + 3 5 y + 30 y = 0 or y 00 + . 3 y + 15 y = 0 with initial conditions y (0) = . 05 m, y (0) = . 1 m/sec. 15. The inductance L = 0 . 2 henry. The resistance R = 3 10 2 ohms. The capacitance C = 10 5 farads. Therefore, the equation for charge q is . 2 q 00 + 300 q + 10 5 q = 0 or q 00 + 1500 q + 500 , 000 q = 0 with initial conditions q (0) = 10 6 coulombs, q (0) = 0 coulombs/sec. 16. We know the net force acting on the mass satisfies F net = mx 00 ( t ) where x 00 is the acceleration of the mass. Here, we assume the spring is pulling the mass to the left. We will consider that to be the negative direction. The force due to the spring is given by F s = k ( L + x ) ....
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This note was uploaded on 09/27/2010 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.
 Spring '08
 comech
 Equations

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