exam1_sol - 2 km , the roots are complex (with negative...

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Exam 1 solutions (a) d 2 x dt 2 c m dx dt k m x = h t m (b) This is a linear second-order ordinary differential equation with constant coefficients. (c) We assume a solution of the form x = Ce rt , giving for r the quadratic equation: r 2 + (c/m)r + (k/m) = 0 Using the quadratic formula: r = − c ± c 2 4 k m  2 m . If these roots are distinct, the general solution is x = C 1 e r 1 t C 2 e r 2 t . If c = 2 km , there is only one root and the general solution is x = C 1 e ct / 2m C 2 te ct / 2m . (d) For distinct roots, we have C 1 + C 2 = b and r 1 C 1 + r 2 C 2 = 0 . So C 1 = -(r 2 /r 1 )C 2 = b - C 2 → (1 - r 2 /r 1 )C 2 = b → C 2 = b / (1 - r 2 /r 1 ) , C 1 = b/(1 - r 1 /r 2 ) . If there is only one root, the condition on x(0) gives C 1 = b and the condition on x'(0) gives C 2 = bc/2m . (e) If c 2 km , there are two negative real roots, so the solution is the combination of two exponential decay curves. If c = 2 km , the solution also decays but has a component that is proportional to t times an exponential decay. If c
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Unformatted text preview: 2 km , the roots are complex (with negative real part), and the solution has oscillations (sine/cosine waves) superimposed on an exponential decay. (f) h(t) and its derivatives has a one-member family of basis functions, { e t } . So we look for a solution of the form Ce t . Substituting this in the differential equation, we find that x t = e t m c k is a solution. ( (m - c + k) = 0 if and only if e t is a homogenous solution.) (g) Laplace transform of each term in the equation yields m s 2 x s m s x m x' c s x s x k x s = a . Putting in the given initial conditions, m s 2 x s c s x s k x s = a or x s = a m s 2 c s k ....
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This note was uploaded on 08/30/2011 for the course CGN 2200 taught by Professor Glagola during the Spring '11 term at University of Florida.

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