Explanation Given: The spring mass system has, k 1 = 3 Nm -1 , k 2 = 4 Nm -1 , m 1 = 1 kg , m 2 = 4 3 kg . Approach: Write the general form of the differential equation for the system is, m 1 d 2 x d t 2 = − k 1 x + k 2 ( y − x ) , m 2 d 2 y d t 2 = − k 2 ( y − x ) . Substitute the values of k 1 , k 2 , m 1 , and m 2 . Consider the initial conditions x ( 0 ) = 1 , d x d t ( 0 ) = 0 , y ( 0 ) = 1 , and d y d t ( 0 ) = 0 for the new variables. Write the characteristic polynomial of matrix A . Find eigenvalues and eigenvector. Make the augmented matrix of the system and write that in reduced echelon form. Write the general solution in combined vector form. Put x = x 1 and y = x 3 . Write the motion of the coupled spring-mass system. Calculation: The general form of the differential equation for the system is, m 1 d 2 x d t 2 = − k 1 x + k 2 ( y − x ) ... ( 1 ) , And m 2 d 2 y d t 2 = − k 2 ( y − x ) ... ( 2 ) . Put the values of k 1 , k 2 , m 1 , and m 2 in above equations as follows. d 2 x d t 2 = − 3 x + 4 ( y − x ) , 4 3 d 2 y d t 2 = − 4 ( y − x ) . Consider the initial conditions x ( 0 ) = 1 , d x d t ( 0 ) = 0 , y ( 0 ) = 1 , and d y d t ( 0 ) = 0 for the new variables as follows. x 1 = x , x 2 = x ′ , x 3 = y , x 4 = y ′ Write, x ′ 1 = x 2 , x ′ 2 = − 7 x 1 + 4 x 3 , x ′ 3 = x 4 , x ′ 4 = 3 x 1 − 3 x 3 . And, x 1 ( 0 ) = 1 , x 2 ( 0 ) = 0 , x 3 ( 0 ) = 1 , x 4 ( 0 ) = 0 . Write initial value problem for x 1 , x 2 , x 3 , and x 4 as x ′ = A x , x ( 0 ) = x 0 , x = [ x 1 x 2 x 3 x 4 ] , A = [ 0 1 0 0 1 − m 1 ( k 1 + k 2 ) 0 k 2 m 1 0 0 0 0 1 k 2 m 2 0 − k 2 m 2 0 ] , x 0 = [ a 1 a 2 a 3 a 4 ] . Put values in the above system, A = [ 0 1 0 0 − 7 0 1 0 0 0 0 1 3 0 − 3 0 ] , x 0 = [ 1 0 1 0 ] . The characteristic polynomial of matrix A is det ( A − λ I ) , and I is the identity matrix. A − λ I = [ 0 1 0 0 − 7 0 1 0 0 0 0 1 3 0 − 3 0 ] − λ [ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] = [ − λ 1 0 0 − 7 − λ 1 0 0 0 − λ 1 3 0 − 3 − λ ] . The det ( A − λ I ) is, det A = [ − λ 1 0 0 − 7 − λ 1 0 0 0 − λ 1 3 0 − 3 − λ ] = λ 4 + 10 λ 2 + 9 = ( λ 2 + 1 ) ( λ 2 + 9 ) . Apply the condition that the system has non-trivial solution if and only if | A − λ I | = 0 for the characteristic equation. Put | A − λ I | = 0 . Then, | − λ 1 0 0 − 7 − λ 1 0 0 0 − λ 1 3 0 − 3 − λ | = 0 λ 4 + 10 λ 2 + 9 = 0 ( λ 2 + 1 ) ( λ 2 + 9 ) = 0 λ = ± i , ± 3 i . The eigenvector corresponding to eigenvalue λ = i is as follows. Consider the system ( A − λ I ) v 1 = 0 . The augmented matrix of the system is, [ − i 1 0 0 0 − 7 − i 4 0 0 0 0 − i 1 0 3 0 − 3 − i 0 ]

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Differential Equations and Linear Algebra (4th Edition)

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