316_pdfsam_math 54 differential equation solutions odd

316_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 mD 2 + 2 k 2 k [ x 2 ] = 0 , (5.33) mD 2 + 2 k + 2 k [ x 3 ] = 0 . For normal modes, we find solutions y j ( t ) and z j ( t ), corresponding to x j , j = 1 , 2, and 3 by using (5.29), (5.30), and identities (5.33). ω 1 : y 1 = 1 k ( mD 2 + 2 k ) [ x 1 ] 0 , z 1 = 1 k ( mD 2 + 2 k ) 2 1 [ x 1 ] = x 1 ; ω 2 : y 2 = 1 k ( mD 2 + 2 k ) [ x 2 ] = 1 k mD 2 + 2 k 2 k + 2 [ x 2 ] = 2 x 2 , z 2 = 1 k ( mD 2 + 2 k ) 2 1 [ x 2 ] = 1 k ( mD 2 + 2 k ) 2 2 + 1 [ x 2 ] = x 2 ; ω 3 : y 3 = 1 k ( mD 2 + 2 k ) [ x 3 ] = 1 k mD 2 + 2 k + 2 k 2 [ x 3 ] = 2 x 3 , z 3 = 1 k ( mD 2 + 2 k ) 2 1 [ x 3 ] = 1 k ( mD 2 + 2 k ) 2 2 + 1 [ x 3 ] = x 3 ; 5. This spring system is similar to the system in Example 2 on page 282 of the text, except the middle spring has been replaced by a dashpot. We proceed as in Example 1. Let x and y represent the displacement of masses m 1 and m
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