Program Explanation Given information: The matrix differential equation for the given system is shown below. x ″ = [ − c 1 c 1 0 c 2 − 2 c 2 c 2 0 c 3 c 3 ] x Here coefficient c i is defined as, c i = k i m i The value of mass m 1 and mass m 2 for the given system is written below. m 1 = 2000 slugs m 2 = 500 slugs m 3 = 2000 slugs The value of spring constant is written as, k 1 = 0 , k 4 = 0 k 2 = k k 3 = k k = 8000 lb / ft The initial conditions for the system are written as, x 1 ( 0 ) = 0 , x 2 ( 0 ) = 0 , x 3 ( 0 ) = 0 x ′ 1 ( 0 ) = v 0 , x ′ 2 ( 0 ) = 0 , x ′ 3 ( 0 ) = 0 Explanation: The matrix differential equation for the given system is written as, x ″ = [ − c 1 c 1 0 c 2 − 2 c 2 c 2 0 c 3 − c 3 ] x The matrix A for the system is written below. A = [ − c 1 c 1 0 c 2 − 2 c 2 c 2 0 c 3 − c 3 ] A = [ − k m 1 k m 1 0 k m 2 − 2 k m 2 k m 2 0 k m 3 − k m 3 ] Substitute the corresponding values from the given information and write as, A = [ − 4 4 0 16 − 32 16 0 4 − 4 ] Use calculator or matrix program to find the eigenvalues and corresponding eigenvectors for the matrix A and obtain the result as, λ 1 = 0 ⇒ v 1 = [ 1 1 1 ] T λ 2 = − 4 ⇒ v 2 = [ − 1 0 1 ] T λ 3 = − 36 ⇒ v 3 = [ 1 − 8 1 ] T Therefore, the solution vector is written as, x ( t ) = ( a 1 + b 1 t ) v 1 + ( a 2 cos − λ 2 t + b 2 sin − λ 2 t ) v 2 + ( a 3 cos − λ 3 t + b 2 sin − λ 3 t ) v 3 x ( t ) = ( a 1 + b 1 t ) [ 1 1 1 ] + ( a 2 cos 2 t + b 2 sin 2 t ) [ − 1 0 1 ] + ( a 3 cos 6 t + b 3 sin 6 t ) [ 1 − 8 1 ] Now, substitute the given initial condition and obtain the value of arbitrary constants as follows: a 1 = 0 , a 2 = 0 , a 3 = 0 b 1 = 4 v 0 9 , b 2 = − v 0 4 , b 3 = v 0 108 Substitute the arbitrary constants value in the equation and obtain the complete particular solution as, x ( t ) = ( 0 + 4 v 0 9 t ) [ 1 1 1 ] + ( 0 cos 2 t − v 0 4 sin 2 t ) [ − 1 0 1 ] + ( 0 cos 6 t + v 0 108 sin 6 t ) [ 1 − 8 1 ] x ( t ) = 4 v 0 9 t [ 1 1 1 ] − v 0 4 sin 2 t [ − 1 0 1 ] + v 0 108 sin 6 t [ 1 − 8 1 ] { x 1 ( t ) = 4 v 0 9 t + v 0 4 sin 2 t + v 0 108 sin 6 t x 2 ( t ) = 4 v 0 9 t − 2 v 0 27 sin 6 t x 3 ( t ) = 4 v 0 9 t − v 0 4 sin 2 t + v 0 108 sin 6 t The above equations of motion hold only till x 2 ( t ) − x 1 ( t ) < 0 and x 3 ( t ) − x 2 ( t ) < 0

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., Penney, David, Calvis

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 5.4, Problem 24P

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Program Description: Purpose of the problem is to show that railway cars separated after π/2 seconds and proceed with constant velocities, also the values of these constant velocities are x′1(t)=−19v0 and x′2(t)=x′3(t)=89v0 .

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Chapter 5.4, Problem 23P

Chapter 5.4, Problem 25P

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MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. 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Program Description: Purpose of the problem is to show that railway cars separated after π / 2 seconds and proceed with constant velocities, also the values of these constant velocities are x ′ 1 ( t ) = − 1 9 v 0 and x ′ 2 ( t ) = x ′ 3 ( t ) = 8 9 v 0 .

Solution Summary: The author explains that the problem is to show that railway cars separated after pi /2 seconds and proceed with constant velocities.

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Program Description: Purpose of the problem is to show that railway cars separated after π/2 seconds and proceed with constant velocities, also the values of these constant velocities are x′1(t)=−19v0 and x′2(t)=x′3(t)=89v0 .

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Chapter 5 Solutions