S1113bryant - 13a L= m1 m2(x1)2(x2)2 k(x1 x2)2 2 2(5 L L x1 x2 x1 x2 = 2k(x1 x2)x1 2k(x1 x2)x2 x L = = 2k(x1 x2)x 2k(x1 x2)x =0(6 where x1 = x2 = x

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13a) L = m 1 2 ( ˙ x 1 ) 2 + m 2 2 ( ˙ x 2 ) 2 - k ( x 1 - x 2 ) 2 (5) δ x L = ∂L ∂x 1 δx 1 + ∂L ∂x 2 δx 2 = - 2 k ( x 1 - x 2 ) δx 1 + 2 k ( x 1 - x 2 ) δx 2 = - 2 k ( x 1 - x 2 ) δx + 2 k ( x 1 - x 2 ) δx = 0 (6) where δx 1 = δx 2 = δx because both particles are being translated by the same amount. This shows that the sum of the forces in a closed system is 0. The force particle 1 exerts on
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This note was uploaded on 11/10/2011 for the course PHY 4241 taught by Professor Berg during the Spring '11 term at University of Florida.

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