Ph12a_s4_2009

Ph12a_s4_2009 - Ph12a Solution Set 4 October 28, 2009 3.19...

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Unformatted text preview: Ph12a Solution Set 4 October 28, 2009 3.19 The absorptive and elastic amplitudes are given by A ab = F M 2 2 + ( 2- 2 ) 2 (1) A el = F M 2- 2 2 2 + ( 2- 2 ) 2 (2) The weak damping approximation is / 2 << . This implies that , so we have ( 2- 2 ) 2 = ( + )( - ) = 2 ( - ) and the above expressions become: A ab = F M " 2 2 (1 + 4( - ) 2 2 ) # (3) A el = F M " 2 ( - ) 2 2 (1 + 4( - ) 2 2 ) # (4) Defining x = - / 2 , and once again using we therefore have: A ab = F M 1 1 + x 2 (5) A el = F M - x 1 + x 2 (6) Therefore in units of F M , we have the desired forms of A ab and A el . 3.30 The equation of motion is Eq. (62) in the text: M d 2 n dt 2 =- Mg l n + K ( n +1- n )- K ( n- n- 1 ) . (7) We are to show that a solution of the form ( z,t ) = A ( z )cos t with A ( z ) = A e- z- e- L e- ( L- z ) 1- e- 2 L (8) 1 Ph12a Solutions Problem Set 4 October 28, 2009 satisfies the equations of motion, subject to the boundary conditions (0 ,t ) = A cos t and ( L,t ) = 0. We must check the solution satis- fies both the boundary conditions and the equation of motion. Boundary conditions Substituting z = 0 and z = L into Eq. 8, we find that A (0) = A , A ( L ) = 0. Equation of motion Substituting the solution into the RHS of the equa- tion of motion gives- Mg l n + K ( n +1- n )- K ( n- n- 1 ) = cos t- Mg l A ( z ) + K [ A ( z + a ) + A ( z- a )- 2 A ( z...
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Ph12a_s4_2009 - Ph12a Solution Set 4 October 28, 2009 3.19...

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