316_pdfsam_math 54 differential equation solutions odd

316_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 mD 2 + 2k − mD 2 + 2k + √ √ 2k [x2 ] = 0, (5.33) 2k [x3 ] = 0. For normal modes, we find solutions yj (t) and zj (t), corresponding to xj , j = 1, 2, and 3 by using (5.29), (5.30), and identities (5.33). ω1 : 1 mD 2 + 2k [x1 ] ≡ 0, k 1 2 z1 = mD 2 + 2k − 1 [x1 ] = −x1 ; k y1 = ω2 : √ √ √ 1 1 mD 2 + 2k [x2 ] = mD 2 + 2k − 2k + 2 [x2 ] = 2x2 , k k 1 1 2 2 mD 2 + 2k − 1 [x2 ] = mD 2 + 2k − 2 + 1 [x2 ] = x2 ; z2 = k k y2 = ω3 : √ √ √ 1 1 mD 2 + 2k [x3 ] = mD 2 + 2k + 2k − 2 [x3 ] = − 2x3 , k k 1 1 2 2 mD 2 + 2k − 1 [x3 ] = mD 2 + 2k − 2 + 1 [x3 ] = x3 ; z3 = k k y3 = 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 This spring system is similar to the system in Example 2 on page 282 of the text, except the middle spring has been replaced by a dashpot.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 m1 and m2 to the right of their respective equilibrium positions. The mass m1 has a force F1 acting on its left side due to the left spring and a force F2 acting on its right side due to the dashpot. Applying Hooke’s law, we see that F1 = −k1 x. Assuming as we did in Section 4.1 that the damping force due to the dashpot is proportional to the magnitude of the velocity, but opposite in direction, we have F2 = b (y − x ) , 312 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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