Analytical Mech Homework Solutions 173

Analytical Mech Homework Solutions 173 - 10 2V k12 = 1 2 1...

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10 22 12 3 12 21 0 00 2 VV k r θθ θ == ∂∂ 2 k l = () 2 2 2 2 02 cos sin sin sin kl V mgl rl l 1 =+ +− 2 2 23 2 2 021 sin 2 cos cos sin sin sin sin kl kl V mgl l l 1 =− 22 20 0 2 Vk km r l g l Thus, from Equation 11.3.37a or b, we have 11 1 12 1 2 22 2 1 2 2 k k  =++  But from Equation 11.3.9, for the coupled oscillator, we have 2 11 1 2 2 1 2 2 x k xx x x k x −+ + 2 1 The forms of the potential energy function are similar with … 11 22 12 k k k k and k k ′′ =+ = = In the case here …
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