Program Explanation Given information: The stiffness matrix for the given mass and spring system is written as, K = [ − ( k 1 + k 2 ) k 2 k 2 − ( k 2 + k 3 ) ] The value of each mass and spring constant is written below. m 1 = 1 , m 2 = 2 k 1 = 2 , k 2 = 4 , k 3 = 4 The walls are not exist for the given spring mass system. Explanation: The matrix differential equation for the given system is written as, M x ″ = K x Here the mass matrix M is written as, M = [ m 1 0 0 m 2 ] M = [ 1 0 0 2 ] Similarly the stiffness matrix K is written as, K = [ − ( k 1 + k 2 ) k 2 k 2 − ( k 2 + k 3 ) ] K = [ − ( 2 + 4 ) 4 4 − ( 4 + 4 ) ] K = [ − 6 4 4 − 8 ] Therefore, matrix differential equation is written as, [ 1 0 0 2 ] x ″ = [ − 6 4 4 − 8 ] x Reduce the above differential equation in simplest form as, x ″ = [ 1 0 0 2 ] − 1 [ − 6 4 4 − 8 ] x x ″ = [ 1 0 0 0.5 ] [ − 6 4 4 − 8 ] x x ″ = [ − 6 4 2 − 4 ] x Compare the above equation with x ″ = A x , and obtain the matrix A as, A = [ − 6 4 2 − 4 ] So, obtain the characteristic equation of matrix A to obtain the eigenvalues as, | − 6 − λ 4 2 − 4 − λ | = 0 ( − 6 − λ ) ( − 4 − λ ) − 8 = 0 λ 2 + 10 λ + 24 − 8 = 0 λ 2 + 10 λ + 16 = 0 Further, solve the above equation as, ( λ + 2 ) ( λ + 8 ) = 0 λ 1 = − 2 or λ 2 = − 8 The relation between natural frequency and eigenvalue in the mass spring system is defined as ω = − λ . Therefore, natural frequency corresponding to λ 1 = − 2 is calculated as, ω 1 = − λ 1 ω 1 = 2 Also, natural frequency corresponding to λ 2 = − 8 is calculated as, ω 2 = − λ 2 ω 2 = 8 ω 2 = 2 2 Now, obtain the eigenvector corresponding to λ 1 = − 2 as follows: [ − 6 − ( − 2 ) 4 2 − 4 − ( − 2 ) ] [ a b ] = [ 0 0 ] [ − 4 4 2 − 2 ] [ a b ] = [ 0 0 ] [ − 4 a + 4 b 2 a − 2 b ] = [ 0 0 ] From the comparison of corresponding elements from both sides and simplification, it is obtained that a = b . For simplest vector, consider 1 for a and therefore b = 1 , and the eigenvector corresponding to λ 1 = − 2 is written as, v 1 = [ 1 1 ] Similarly, obtain the eigenvector corresponding to λ 2 = − 8 as follows: [ − 6 − ( − 8 ) 4 2 − 4 − ( − 8 ) ] [ a b ] = [ 0 0 ] [ 2 4 2 4 ] [ a b ] = [ 0 0 ] [ 2 a + 4 b 2 a + 4 b ] = [ 0 0 ] From the comparison of corresponding elements from both sides and simplification, it is obtained that a = − 2 b

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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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 6P

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Program Description: Purpose of the problem is to find the two natural frequencies of the system and discuss the natural modes of oscillation.

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Chapter 5.4, Problem 5P

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Chapter 5 Solutions

EBK DIFFERENTIAL EQUATIONS

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 find the two natural frequencies of the system and discuss the natural modes of oscillation.

Solution Summary: The author explains that the problem is to find the two natural frequencies of the system and discuss the natural modes of oscillation.

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Program Description: Purpose of the problem is to find the two natural frequencies of the system and discuss the natural modes of oscillation.

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Chapter 5 Solutions