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# CDIM12 - CHAPTER 12 Coupled Oscillations 12-1 m1 = M k1 x1...

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CHAPTER 12 Coupled Oscillations 12-1. m 1 = M k 1 x 1 k 12 k 2 m 2 = M x 2 The equations of motion are ( ) ( ) 1 1 12 1 12 2 2 2 12 2 12 1 0 0 Mx x x Mx x x κ κ κ κ κ κ + + = + + = ±± ±± (1) We attempt a solution of the form ( ) ( ) 1 1 2 2 i t i t x t B e x t B e ω ω = = (2) Substitution of (2) into (1) yields ( ) ( ) 2 1 12 1 12 2 2 12 1 2 12 2 0 0 M B B B M B κ κ ω κ κ κ κ ω + = + + = (3) In order for a non-trivial solution to exist, the determinant of coefficients of and must vanish. This yields 1 B 2 B ( ) ( ) 2 2 1 12 2 12 12 M M κ κ ω κ κ ω κ + + = 2 (4) from which we obtain ( ) 2 1 2 12 1 2 12 2 1 4 2 2 M M κ κ κ 2 2 ω κ κ κ + + = ± + (5) This result reduces to ( ) 2 12 12 M ω κ κ κ = + ± for the case 1 2 κ κ κ = = (compare Eq. (12.7)]. 397

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398 CHAPTER 12 If were held fixed, the frequency of oscillation of m would be 2 m 1 ( 2 01 1 12 1 M ) ω κ κ = + (6) while in the reverse case, would oscillate with the frequency 2 m ( 2 02 2 12 1 M ) ω κ κ = + (7) Comparing (6) and (7) with the two frequencies, ω + and ω , given by (5), we find ( ) 2 2 2 1 2 12 1 2 12 1 2 4 2 M κ κ κ κ κ + = + + + + ω κ ( ) ( ) 2 1 2 12 1 2 1 12 0 1 1 2 2 M M κ κ κ κ κ κ κ ω + + = + = 1 > + (8) so that 01 ω ω + > (9) Similarly, ( ) 2 2 2 1 2 12 1 2 12 1 2 4 2 M κ κ κ κ κ = + + + ω κ ( ) ( ) 2 1 2 12 1 2 2 12 0 1 1 2 2 M M κ κ κ κ κ κ κ ω + = + = 2 < + (10) so that 02 ω ω < (11) If , then the ordering of the frequencies is 1 κ κ > 2 01 02 ω ω ω ω + > > > (12) 12-2. From the preceding problem we find that for 12 1 2 , κ κ κ ² 1 12 2 12 1 2 ; M M κ κ κ ω ω + + κ (1) If we use 1 01 02 ; M M 2 κ κ ω ω = = (2) then the frequencies in (1) can be expressed as
COUPLED OSCILLATIONS 399 ( ) ( ) 12 1 01 01 1 1 12 2 02 02 2 2 1 1 1 1 κ ω ω ω ε κ κ ω ω ω ε κ = + + = + + (3) where 12 12 1 2 1 2 ; 2 2 κ κ ε ε κ κ = = (4) For the initial conditions [Eq. 12.22)], ( ) ( ) ( ) ( ) 1 2 1 2 0 , 0 0, 0 0, 0 0 x x x = = = ± ± , = x D (5) the solution for ( ) 1 x t is just Eq. (12.24): ( ) 1 2 1 2 1 cos cos 2 2 D t t x t ω ω ω ω + = (6) Using (3), we can write ( ) ( ) 1 2 01 02 1 01 2 02 2 2 ω ω ω ω ε ω ε ω ε + + + = + + + + (7) ( ) ( ) 1 2 01 02 1 01 2 02 2 2 ω ω ω ω ε ω ε ω ε = + + (8) Then, ( ) ( ) ( ) 1 cos cos x t D t t t t ε ε + + = + + (9) Similarly, ( ) ( ) ( 1 2 1 2 2 sin sin 2 2 sin sin x t D t t D t t t + + + ) t = = + + ω ω ω ω ε ε (10) Expanding the cosine and sine functions in (9) and (10) and taking account of the fact that ε + and ε are small quantities, we find, to first order in the ε ’s, ( ) 1 cos cos sin cos cos sin D t t t t t t t t ε ε + + + + x t (11) ( ) 2 sin sin cos sin sin cos D t t t t t t t t ε ε + + + + + + x t (12) When either ( ) 1 x t or ( ) 2 x t reaches a maximum, the other is at a minimum which is greater than zero. Thus, the energy is never transferred completely to one of the oscillators.

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400 CHAPTER 12 12-3.
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