33b practice final

# 33b practice final - MATH 33B - Lecture 1 - Summer 2009...

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Unformatted text preview: MATH 33B - Lecture 1 - Summer 2009 Solutions to Practice Final - September 3, 2009 NAME: STUDENT ID #: This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 7 problems for a total of 100 points. 1 2 1. (15 points) A mass of 4 kg is attached to a spring with a spring constant of k= 52 kg/s 2 . The system is placed in a medium that provides a damping constant = 16 kg/s . The mass is displaced 0 . 1 m from its equilibrium position and imparts an instantaneous downward velocity of 1 . 9 m/s . Find the position of the mass as a function of time. Solution. m = 4 ,k = 16 , = 12 ,x (0) = 0 . 1 ,x (0) = 1 . 9 4 x 00 + 16 x + 52 x = 0 = x 00 + 4 x + 13 x = 0 2 + 4 + 13 = 0 = =- 2 3 i x ( t ) = e- 2 t ( C 1 cos(3 t ) + C 2 sin(3 t )) . 1 = x (0) = C 1 1 . 9 = x (0) =- 2 C 1 + 3 C 2 ; C 2 = 0 . 7 x ( t ) = e- 2 t (0 . 1cos(3 t ) + 0 . 7sin(3 t )) / 3 2. (15 points) Check that y 1 ( t ) = t is a particular solution to the equation t 2 y 00- t (2 + t ) y + (2 + t ) y = 0 . Then, find the general solution of this second-order differential equation. Solution. Check that y 1 is a solution: t 2 y 00 1 ( t )- t (2 + t ) y 1 ( t ) + (2 + t ) y ( t ) =- t (2 + t ) + (2 + t ) t = 0 Let y 2 ( t ) = v ( t ) y 1 ( t ) = v ( t ) t . Then y 2 = v t + v and y 00 2 = v 00 t +2 v . Plugging y 2 into the LHS of the equation , we obtain t 2 ( v 00 t + 2 v )- t (2 + t )( v t + v ) + (2 + t ) vt = t 3 v 00- t 3 v Since we expect v is chosen such that y 2 is a solution, v should satisfy...
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## 33b practice final - MATH 33B - Lecture 1 - Summer 2009...

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