s251ex2(fa03)

s251ex2(fa03) - MATH 251 Fall 2003 Exam 2 November 10, 2003...

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MATH 251 Fall 2003 Exam 2 November 10, 2003 Solution Key 1. (5 points) A spring-mass system, subject to an external force of 10 cos(2 t ) Newtons, is equipped with a spring with a Hooke’s constant 12 Newtons per meter. For what mass will the resonance occur? (a) 2 kg (b) 3 kg Correct (c) 6 kg (d) 10 kg Solution: The frequency of free oscillation ω is subject to the following relation: ω 2 = spring constant mass For resonance the frequency of the external force should be equal to the frequency of free oscillation. 2. (5 points) Of what form will the particular solution to the following differential equa- tion be? Do not solve the equation. y 00 - 4 y 0 + 4 y = e 2 t + t 2 e 3 t - sin(2 πt ) (a) Ae 2 t + Bt 2 e 3 t + Cte 3 t + De 3 t + E sin(2 πt ) + F cos(2 πt ) (b) Ate 2 t + Bt 2 e 3 t + Cte 3 t + De 3 t + E sin(2 πt ) + F cos(2 πt ) (c) At 2 e 2 t + Bt 2 e 3 t + Cte 3 t + De 3 t + E sin(2 πt ) + F cos(2 πt ) Correct (d) Ae 2 t + Bt 2 e 3 t + E sin(2 πt ) + F cos(2 πt ) Solution: The characteristic equation, associated with the homogeneous part of the equation, has root 2 of multiplicity 2 . Therefore the solution-to-be has the e 2 t term multiplied by t 2 . For the t 2 e 3 t part of the right–hand side we need to throw a full polynomial of degree 2 multiplied by the exponent function, not just Bt 2 . 3. (5 points) The second order linear equation y 00 - 4 y 0 + 4 y = 0 is equivalent to (a) ( x 0 1 = x 2 x 0 2 = - 4 x 1 + 4 x 2 Correct (b) ( x 0 1 = x 1 x 0 2 = - 4 x 1 + 4 x 2 (c) ( x 0 1 = x 2 x 0 2 = 4 x 1 - 4 x 2
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(d) ( x 0 1 = x 1 x 0 2 = 4 x 1 - 4 x 2 Solution: We think of x 1 as y and x 2 as y 0 . 4. (5 points) Which of the following functions corresponds to this graph: - t 6 y ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ 0 1 2 - 1 (a) tu 1 ( t ) - 1 (b) tu 1 ( t - 1) - 1 (c) ( t - 1) u 1 ( t ) - 1 Correct (d) ( t - 1) u 1 ( t - 1) - 1 Solution: We are ’turning’ at t = 1 , that’s why we must have u 1 ( t ) term. Before the turnpoint, we are at the level - 1 , that’s why we have to add constant - 1 . After the turnpoint, our function looks like t - 2 , therefore we write t - 1 in front of the Heaviside function (remember, we already have - 1 before the turn). 5. (14 points) Solve the following initial value problem: y 00 - 5 y 0 - 14 y = - 14 t 2 - 10 t - 26 , y (0) = 0 , y 0 (0) = 13 Solution: The associated homogeneous differential equation y 00 - 5 y 0 - 14 y = 0 has characteristic polynomial 2 points λ 2 - 5 λ - 14 = ( λ - 7)( λ + 2) with roots
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s251ex2(fa03) - MATH 251 Fall 2003 Exam 2 November 10, 2003...

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