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nagle_differential_equations_ISM_Part36

# nagle_differential_equations_ISM_Part36 - Exercises 4.10...

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Exercises 4.10 Thus, a general solution is given by y ( t ) = e - 1 . 25 t ( c 1 cos βt + c 2 sin βt ) + A cos t + B sin t e - 1 . 25 t ( c 1 cos βt + c 2 sin βt ) + 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. 0 . 05 = y (0) = c 1 + A 0 = y (0) = - 1 . 25 c 1 + βc 2 + B c 1 = 0 . 05 - A 0 . 04689 , c 2 = 1 . 25 c 1 - B β 0 . 00848 . Hence, y ( t ) e - 1 . 25 t (0 . 04689 cos βt + 0 . 00848 sin βt ) + 0 . 00311 cos t + 0 . 00016 sin t . To find the resonance frequency, we use the formula (15) in the text. 1 2 π γ r = 1 2 π k m - b 2 2 m 2 = 1 2 π 5 g - 25 8 1 . 0786 ( sec - 1 ) . 14. In the equation, governing this motion, my + by + ky = F ext , we have m = 8, b = 3, k = 40, and F ext ( t ) = 2 sin(2 t + π/ 4). Thus, the equation becomes 8 y + 3 y + 40 y = 2 sin(2 t + π/ 4) = 2 (sin 2 t + cos 2 t ) . Clearly, this is an underdamped motion, and the steady-state solution has the form y p ( t ) = A sin 2 t + B cos 2 t y p ( t ) = 2 A cos 2 t - 2 B sin 2 t y p ( t ) = - 4 Asin 2 t - 4 B cos 2 t . Substituting these formulas into the equation and collecting similar terms yields (8 A - 6 B ) sin 2 t + (6 A + 8 B ) cos 2 t = 2 (sin 2 t + cos 2 t ) 8 A

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nagle_differential_equations_ISM_Part36 - Exercises 4.10...

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