HW8 - 293 HW 8 Geoffrey Recktenwald 3.4-8 The clock loses...

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Unformatted text preview: 293 HW 8 Geoffrey Recktenwald 9-13-2005 3.4-8 The clock loses 10 minutes/day which means it takes approx. 24 hrs 10 minutes to go the normal n number of cycles. Thus, nP s = 1450 minutes where P s is the period period of the pendulum for the slow clock. We note that for P n (the period of the normal clock) nP n = 1440 minutes The period P is determined by P = 2 π ω where ! = p g=L . Thus, P 1 = 2 π p g=L 1 where L 1 = 30in. The ratio P 2 =P 1 is P 2 P 1 = 2 π p g/L 2 2 π p g/L 1 = p L 2 p L 1 Which means L 2 = P 2 P 1 2 L 1 = 1440min 1450min 2 30in = 29 : 59in 3.4-14 Parameters: m = 25, c = 10, and k = 226 Initial Conditions: x (0) = 20 and x ′ (0) = 41 Part A) We are solving the ODE my ′′ + cy ′ + ky = 0 assume a solution x ( t ) = e rt to get the characteristic equation mr 2 + cr + k = 0. Solve for r to find r = − 1 5 ± 3 i . Thus, x ( t ) = e − 1 5 t ( A cos 3 t + B sin 3 t ) Use the initial conditions to find x (0) = A = 20 x ′ (0) = − 1 5 ( A ) + (3 B ) = 41 Thus,...
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HW8 - 293 HW 8 Geoffrey Recktenwald 3.4-8 The clock loses...

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