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Mech5Sols

# Mech5Sols - Physics 105B Problem Set 5 May 6 2008 Jeff...

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Physics 105B Problem Set 5 May 6, 2008 Jef Schonert: schonert (at) physics.ucsb.edu Taylor 11.5 a) The easiest way to solve this problem is to see that it is equivalent to the example discussed in section 11.1 with k 3 = 0. Plugging k 3 = 0, k 1 = k 2 and m 1 = m 2 into eq. (11.5) gives that spring-constant matrix is K = ± 2 k - k - kk ² The mass matrix is still given by M = ± m 0 0 m ² To Fnd the normal ±requencies ω , we need to calculate det( K - ω 2 M ) = 0. ²irst, simpli±y the notation by deFning ω 0 such that k = 2 0 . Then this determinant is ³ ³ ³ ³ 2 2 0 - 2 - 2 0 - 2 0 2 0 - 2 ³ ³ ³ ³ = m 2 ( ω 4 - 3 ω 2 0 ω 2 + ω 4 0 )=0 Viewing the expression inside the parentheses as a quadratic equation in ω 2 , we can easily solve it to Fnd the (positive) ±requencies ω 1 = ω 0 2 ´ 3 - 5 , ω 2 = ω 0 2 ´ 3+ 5 b) To Fnd the normal modes a 1 ,a 2 , we must solve the equation ( K - ω 2 M ) a =0 , (1) where a = ± a 1 a 2 ² (2) Substituting ω = ω 1 , the Frst equation gives 2 2 0 a 1 - 2 0 2 (3 - 5) a 1 - 2 0 a 2 , (3) which implies that a 2 = a 1 2 (1 + 5) . (4) There±ore, the Frst normal mode is (plug the results ±or a 1 and a 2 into eq. (11.10)) ± x 1 x 2 ² = ± A cos( ω 1 t - δ ) A (1+ 5) 2 cos( ω 1 t - δ ) ² 1

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The constants A and δ must be determined by the boundary conditions. So we see that in this frst normal mode, the two carts oscillate in phase, but the second cart oscillates with a larger amplitude than the frst. Now do the same procedure with ω 2 instead oF ω 1 . Setting up the same matrix equation For a gives that a 2 = a 1 2 (1 - 5) So now the second normal mode is ± x 1 x 2 ² = ± A ± cos( ω 1 t - δ ± ) A ± (1 - 5) 2 cos( ω 1 t - δ ± ) ² Since 1 - 5 < 0, this means that the two carts oscillate 180 out oF phase, as x 1 always has the opposite sign oF x 2 . In addition to the phase di±erence, the second cart also oscillates with a smaller amplitude.
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Mech5Sols - Physics 105B Problem Set 5 May 6 2008 Jeff...

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