nagle_differential_equations_ISM_Part36

nagle_differential_equations_ISM_Part36 - Exercises 4.10...

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Unformatted text preview: Exercises 4.10 Thus, a general solution is given by y ( t ) = e- 1 . 25 t ( c 1 cos t + c 2 sin t ) + A cos t + B sin t e- 1 . 25 t ( c 1 cos t + c 2 sin t ) + 0 . 00311 cos t + 0 . 00016 sin t. We now find constants c 1 and c 2 such that y ( t ) satisfies the initial conditions. . 05 = y (0) = c 1 + A 0 = y (0) =- 1 . 25 c 1 + c 2 + B c 1 = 0 . 05- A . 04689 , c 2 = 1 . 25 c 1- B . 00848 . Hence, y ( t ) e- 1 . 25 t (0 . 04689 cos t + 0 . 00848 sin t ) + 0 . 00311 cos t + 0 . 00016 sin t. To find the resonance frequency, we use the formula (15) in the text. 1 2 r = 1 2 r k m- b 2 2 m 2 = 1 2 r 5 g- 25 8 1 . 0786 ( sec- 1 ) . 14. In the equation, governing this motion, my 00 + by + ky = F ext , we have m = 8, b = 3, k = 40, and F ext ( t ) = 2 sin(2 t + / 4). Thus, the equation becomes 8 y 00 + 3 y + 40 y = 2 sin(2 t + / 4) = 2 (sin 2 t + cos 2 t ) ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part36 - Exercises 4.10...

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