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Unformatted text preview: Homework 6 Classical Mechanics (Phys701) Due: November 28, 2011 Problem 65 (a) In the linear triatomic molecule, suppose the initial conditions is that the center atom is at rest, but displaced by an amount a from equilibrium, the other two being at their equilibrium points. Find the amplitudes of the longitudinal small oscillations about the center of mass. Give the amplitudes of the normal modes (b) Repeat part (a) but with the center atom initially at its equilibrium position but with initial speed v . Solution: In terms of normal modes ζ 1 , 2 , 3 the solution always has the form ζ i ( t ) = ζ i (0) cos ω i t + ˙ ζ i (0) ω i sin ω i t, for ω i ̸ = 0 or ζ i ( t ) = ζ i (0) + ˙ ζ i (0) t, for ω i = 0 First, we find ζ i (0) and ˙ ζ i (0) from the initial velocities and displacements η i (0) and ˙ η i (0). For a triatomic molecule this can be done using expressions (6.59) of Chapter 6. For the case (a) we have η 1 = 0, η 2 = a , η 3 = 0, and ˙ η i (0) = 0. Using (6.59) we get ζ 1 (0) = a M √ 2 m + M ζ 2 (0) = ζ 3 (0) = − 2 a m √ 2 m (1 + 2 m/M ) and ˙ ζ i (0) = 0 for all i . This gives ζ 1 ( t ) = a M √ 2 m + M ζ 2 ( t ) = ζ 3 ( t ) = − 2 a m √ 2 m (1 + 2 m/M ) cos ω 3 t , ω 3 = √ k m ( 1 + 2 m M ) The next step is to find the time dependence of the original small displacements η i ( t ). This is done using the formula ⃗η = ˆ A ⃗ ζ , where the columns of matrix ˆ A are given by the eigenvectors (6.58 a,b,c). η 1 η 2 η 3 = 1 / √ 2 m + M 1 / √ 2 m 1 / √ 2 m (1 + 2 m/M ) 1 / √ 2 m + M − 2 / √ 2 M (2 + M/m ) 1 / √ 2 m + M − 1 / √ 2 m 1 / √ 2 m (1 + 2 m/M ) ζ 1 ζ 2 ζ 3 We obtain the final result η 1 ( t ) = a M 2 m + M − a M 2 m + M cos ω 3 t η 2 ( t ) = a M 2 m + M + 2 a m 2 m + M cos ω 3 t η 3 ( t ) = a M 2 m + M − a M 2 m + M cos ω 3 t 1 All atoms are on average displaced from the original position by the distance a M/ (2 m + M ) and oscillate around these new positions with frequency ω 3 . The amplitudes of the oscillations of the side atoms are equal to the average displacement, and the amplitude of the central atom is 2 a m/ (2 m + M ). (b) In this section we have initial conditions η i = 0 and ˙ η 1 = 0, ˙ η 2 = v , ˙ η 3 = 0. Using the same procedure as in the previous section one gets ζ 1 ( t ) = v Mt √ 2 m + M ζ 2 ( t ) = ζ 3 ( t ) = − 2 v m ω 3 √ 2 m (1 + 2 m/M ) sin ω 3 t and then η 1 ( t ) = v Mt 2 m + M − v M (2 m + M ) ω 3 sin ω 3 t η 2 ( t ) = v Mt 2 m + M + 2 v m (2 m + M ) ω 3 sin ω 3 t η 3 ( t ) = v Mt 2 m + M − v M (2 m + M ) ω 3 sin ω 3 t The center of mass of the molecule moves with constant velocity v M/ (2 m + M ), the atoms oscillate about their equilibrium positions with respect to the center of mass with amplitudes that have the same ratios as in Sec.(a) of the problem....
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 Fall '11
 Bazaliy
 mechanics, Energy, Kinetic Energy, Work, Characteristic polynomial, Normal mode

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