Program Explanation Given information: The value of mass and spring constants are written below. m 1 = 1 , m 2 = 1 2 k 1 = 75 , k 2 = 25 , k 3 = 0 The walls are not exist for the given spring mass system. 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 given external forces F 1 ( t ) and F 2 ( t ) acting on the masses m 1 and m 2 respectively as, F 1 ( t ) = 0 F 2 ( t ) = 100 cos 10 t The initial conditions for the system is written as, x ( 0 ) = 0 x ′ ( 0 ) = 0 Explanation: The matrix differential equation for the given system is written as, M x ″ = K x + F Here the mass matrix M is written as, M = [ m 1 0 0 m 2 ] M = [ 2 0 0 1 2 ] Similarly the stiffness matrix K is written as, K = [ − ( k 1 + k 2 ) k 2 k 2 − ( k 2 + k 3 ) ] K = [ − ( 75 + 25 ) 25 25 − ( 25 + 0 ) ] K = [ − 100 25 25 − 25 ] Similarly the external force vector F is written as, F = [ F 1 ( t ) F 2 ( t ) ] F = [ 0 100 cos 10 t ] F = [ 0 100 ] cos 10 t Therefore, matrix differential equation for the given system is written as, [ 2 0 0 1 2 ] x ″ = [ − 100 25 25 − 25 ] x + [ 0 100 ] cos 10 t Reduce the above differential equation in simplest form as, x ″ = [ 2 0 0 1 2 ] − 1 [ − 100 25 25 − 25 ] x + [ 2 0 0 1 2 ] − 1 [ 0 100 ] cos 10 t x ″ = [ 0.5 0 0 2 ] [ − 100 25 25 − 25 ] x + [ 0.5 0 0 2 ] [ 0 100 ] cos 10 t x ″ = [ − 50 12.5 50 − 50 ] x + [ 0 200 ] cos 10 t Consider the periodic particular solution as x p ( t ) = c cos 10 t and so derivatives are written as, x ′ p ( t ) = − 10 c sin 10 t x ″ p ( t ) = − 100 c cos 10 t Substitute, the above periodic particular solution and its derivative in the matrix differential equation and solve for the vector c as, − 100 c cos 10 t = [ − 50 12.5 50 − 50 ] c cos 10 t + [ 0 200 ] cos 10 t − 100 c = [ − 50 12.5 50 − 50 ] c + [ 0 200 ] − [ 0 200 ] = [ − 50 12.5 50 − 50 ] c + 100 c [ 0 − 200 ] = [ 50 12.5 50 50 ] c Further, solve the above equation as, [ 0 − 200 ] = [ 50 12.5 50 50 ] [ c 1 c 2 ] [ 0 − 200 ] = [ 50 c 1 + 12.5 c 2 50 c 1 + 50 c 2 ] { 50 c 1 + 12.5 c 2 = 0 50 c 1 + 50 c 2 = − 200 Solve the system of equations by substitution method and obtain the value of c 1 and c 2 as, c 1 = 4 3 c 2 = − 16 3 Therefore, the period solution x p ( t ) = c cos 10 t of the equation is written as, x p ( t ) = c cos 10 t x p ( t ) = [ c 1 c 2 ] cos 10 t x p ( t ) = [ 4 3 − 16 3 ] cos 10 t Solve the homogeneous part of the non-homogeneous equation as follows: The matrix A for the homogeneous part is written below

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ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 5.4, Problem 15P

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Program Description: Purpose of the problem is to find the solution of the system Mx″=Kx+F for the initial conditions x(0)=x′(0)=0 .

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Chapter 5.4, Problem 14P

Chapter 5.4, Problem 16P

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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

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Program Description: Purpose of the problem is to find the solution of the system M x ″ = K x + F for the initial conditions x ( 0 ) = x ′ ( 0 ) = 0 .

Solution Summary: The author explains how the matrix differential equation for the given system is written as, Mx′′=Kx+F.

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Program Description: Purpose of the problem is to find the solution of the system Mx″=Kx+F for the initial conditions x(0)=x′(0)=0 .

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Chapter 5 Solutions