314_pdfsam_math 54 differential equation solutions odd

314_pdfsam_math 54 differential equation solutions odd -...

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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, F 11 = kx and F 12 = k ( y x ); for the mass in the middle, F 21 = k ( y x ) and F 22 = k ( z y ); finally, for the mass on the right, F 31 = k ( z y ) and F 32 = kz. 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,
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