soln_final

# soln_final - Massachusetts Institute of Technology Physics...

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Massachusetts Institute of Technology Physics 8.03 Spring 2004 Final Exam Solutions Thursday, December 16, 2004 Solution for Problem 1 Coupled Oscillators (a) If m A = , then ω A = 0 and ω B = k + k . m B (b) If k = 0, then m A ,m B and k ± form an isolate system and the center of mass of this system doesn’t move. Therefore, m A x A = m B x B (1) The force applied on m A is given by m A ¨ x A = k ± ( x A x B ) ( 2 ) Solving them gives ± ¨ x A + x A k ± ² m A + m B m A m B ³ =0 ω = k ± M (3) where M = m A m B m A + m B is called the “reduced mass”. (c) If k ± = 0, then ω A = k m A and ω B = k m B . (d) For the general situation, the coupled equations of motion are m A x A = kx A + k ± ( x B x A ) ( 4 ) ¨ m B x B = kx B k ± ( x B x A

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² ³ ³ ³ ³ ³ (e) Assume that x A = A cos ωt and x B = B cos ωt . The equations of motions become k + k ± k ± ω 2 A + A B =0 (6) m A m A k + k ± k ± ω 2 B + B A =0 (7) m B m B rewrite them with A and B as the variables ± k + k ± ² k ± A ω 2 + B =0 (8) m A m A ± k + k ± ² k ± A + ω 2 B =0 (9) m B m B for A and B have non-zero root, the determinant should be equal to zero, that is ± k + k ± ²± ² k ± 2 k + k ± ω 2 ω 2 = 0 (10) m A m A m A m B ± 1 ² 1 ω 4 ω 2 ( k + k ± ) + + ( k + k ± ) 2 k ± 2 = 0 (11) m A m B m A m B The solutions for ω are ´ 2 ω 2 p 1 p = ± p 2 4 q = p ± q (12) 1 , 2 2 2 2 4 1 where p =( k + k ± ) m A + 1 and q = ( k + k ) 2 k 2 . m B m A m B Solution for Problem 2 Dispersive String (a) The phase velocity is ω T v p = = + αk 2 (13) k µ (b) The group velocity is T +2 αk 2 T αk 2 µ v g = = + αk 2 + µ = µ (14) dk µ T + αk 2 T + αk 2 µ µ (c) There are three normal oscillation modes of the string 2 π The wave numbers k n = L and the frequencies ω n = v p n k n = v p n . Therefore λ
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soln_final - Massachusetts Institute of Technology Physics...

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