314_pdfsam_math 54 differential equation solutions odd

314_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 3. We define the displacements of masses from equilibrium, x, y , and z , as in Example 2. For each mass, there are two forces acting on it due to Hook’s law. For the mass on the left, F11 = −kx for the mass in the middle, F21 = −k (y − x) finally, for the mass on the right, F31 = −k (z − y ) and F32 = −kz. and F22 = k (z − y ); and F12 = k (y − x); Applying Newton’s second law for each mass, we obtain the following system mx = −kx + k (y − x), my = −k (y − x) + k (z − y ), mz = −k (z − y ) − kz, or, in operator form, mD 2 + 2k [x] − ky = 0, −kx + mD 2 + 2k [y ] − kz = 0, −ky + mD 2 + 2k [z ] = 0. From the first equation, we express y= 1 mD 2 + 2k [x] k (5.29) and substitute this expression into the other two equations to get 1 mD 2 + 2k [x] − kz = 0, k − mD 2 + 2k [x] + mD 2 + 2k [z ] = 0. −kx + mD 2 + 2k 310 ...
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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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