Analytical Mech Homework Solutions 181

# Analytical Mech Homework Solutions 181 - 18 2 0 Thus K =...

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Unformatted text preview: 18 2 0 Thus K = mω0 2 The above matrix equation is thus … 0 1 a 0 2 2 2 2 1 1 mω0 − mω a = 0 0 1 1 1 2 2 (ω0 2 − ω 2 ) −ω 2 a1 = 0 ω0 2 − ω 2 a2 −ω 2 From the top equation in the matrix equation above, we get … 2 (ω0 2 − ω 2 ) a1 − ω 2 a2 = 0 2 2 a2 2 (ω0 − ω ) = a1 ω2 The amplitudes for the two normal modes can be found by setting ω 2 = ω12 or ω2 2 . In each case we set a1 = 1 , which we are free to do since only ratios of the amplitudes can be determined. For ω 2 = ω12 = 2 − 2 ω0 2 we obtain … ( a2 = ) 2 (ω0 2 − ω12 ) ω 2 1 ( ) 2 1 − 2 − 2 = 2 = 2− 2 ( ) Thus, for mode 1, the lower frequency, symmetric mode … 1 iω t q1 = e 1 2 ( ) For ω 2 = ω2 2 = 2 + 2 ω0 2 we obtain … a2 = 2 ( ω0 2 − ω 2 2 ) ω 2 1 ( ) 2 1 − 2 + 2 =− 2 = 2+ 2 ( ) Thus, for mode 2, the higher frequency, anti-symmetric mode … 1 iω 2 t q2 = e − 2 (b) In this case, m << 1 or M >> m . We also assume that the spring is “slack”, m+M k g << , an assumption not stated in the problem (which needs to be rectified in M +m r the next edition, I suppose) k k g and Ω0 2 = Also, 2mg ≠ kr , so we let ω0 2 = ≈ M +m M r i.e., The kinetic energy is ...
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