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# quiz6 - Quiz 6 Math 240 Calculus III Name Note In order to...

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Quiz 6 Name: Math 240 - Calculus III March 17, 2009 Note: In order to receive full credit, you must show work that justifies your answer. Suppose the position at time t of an object in motion is given by x ( t ), satisfying x - ω 2 x = - g sin( ωt ) with initial conditions x (0) = 0 and x (0) = v 0 . Note that ω and g are fixed constants. (a) (7 points) Find the equation of motion x ( t ). Solution : The complementary solution is x c = c 1 e ωt + c 2 e - ωt . The particular solution has the form x p = A sin( ωt ) + B cos( ωt ). Substituting x p into the differential equation, we have: x p - ω 2 x p = - 2 ω 2 ( A sin( ωt ) + B cos( ωt )) = - g sin( ωt ) . Thus, B = 0 and - 2 2 = - g , so A = g 2 ω 2 . The general solution is then x g ( t ) = c 1 e ωt + c 2 e - ωt + g 2 ω 2 sin( ωt ). The initial condition x (0) = 0 implies 0 = c 1 + c 2 , or c 1 = - c 2 . The initial condition x (0) = v 0 implies v 0 = ωc 1 - ωc 2 + g 2 ω = - 2 ωc 2 + g 2 ω . So c 2 = g - 2 ωv 0 4 ω 2 = - c 1 , and the equation of motion is x ( t ) = 2 ωv 0 - g 4 ω 2 e ωt + g - 2 ωv 0 4 ω 2 e - ωt + g 2 ω 2 sin( ωt ) .

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quiz6 - Quiz 6 Math 240 Calculus III Name Note In order to...

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