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# ch04 - Chapter 4 Section 4.1 1 The dierential equation is...

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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) = 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 (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 (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) = 0 coulombs/sec. 1

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12. The spring constant is k = . 196 /. 05 = 3 . 92 N/m. The mass m = . 02 kg. The damping constant is γ = 400 dyne-sec/cm = . 4 N-sec/cm. Therefore, the equation of motion is . 02 y 00 + . 4 y 0 + 3 . 92 y = 0 or y 00 + 20 y 0 + 196 y = 0 with initial conditions y (0) = . 02 m, y 0 (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 γ = 2 lb-sec/ft. Therefore, the equation of motion is 1 2 y 00 + 2 y 0 + 64 y = 0 or y 00 + 4 y 0 + 128 y = 0 with initial conditions y (0) = 0 ft, y 0 (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 γ = 3 / 5 N-sec/m. Therefore, the equation of motion is 2 y 00 + 3 5 y 0 + 30 y = 0 or y 00 + . 3 y 0 + 15 y = 0 with initial conditions y (0) = . 05 m, y 0 (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 0 . 2 q 00 + 300 q 0 + 10 5 q = 0 or q 00 + 1500 q 0 + 500 , 000 q = 0 with initial conditions q (0) = 10 - 6 coulombs, q 0 (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 ) . The force due to air resistance, F d , is acting in the direction opposite to the direction of motion of the mass, and is proportional to the velocity of the mass. Therefore, F d = - γx 0 ( t ) 2
where γ 0. In particular, if the mass is moving to the left, then the velocity x 0 ( t ) < 0 in which case, F d 0, while if the mass is moving to the right, then the velocity x 0 ( t ) > 0 in which case, F d 0. Finally, we are assuming that an external force F ( t ) is acting on the mass. Putting together these factors, we have F net = F s + F d + F ( t ) = mx 00 ( t ) = - k ( L + x ) - γx 0 ( t ) + F ( t ) .

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