33b practice final

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 9

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online