mod-matr - 11 Normal Modes - Matrix Methods Fall 2003...

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11 Normal Modes -- Matrix Methods Fall 2003 Matrix methods provide a very powerful tool for analyzing normal modes of coupled oscillator systems. We'll introduce this general approach by means of an example, the same system we discussed in detail in Section 8 of these notes. Our development will parallel the previous treatment, but we'll use matrix language. Example For the system discussed in Section 8, the equations of motion (from Σ F = ma ) are mx k k x k x k x k k x && ( ' ) ' , ' ( ' ) . 1 1 2 2 1 2 = - + + = - + (1) The equations of motion can be written as a single matrix equation: m m x x k k k k k k x x 0 0 1 2 1 2 F H G I K J F H G I K J = - + - - + F H G I K J F H G I K J ' ' ' ' . (2) Each side of this equation is a column matrix. You should verify that when the matrix products are carried out, equating the first (top) elements on the two sides gives the first of Eqs. (1), and equating the second (bottom) elements gives the second. We define the matrices M, K, and x as follows: M K x = F H G I K J = + - - + F H G I K J = F H G I K J m m k k k k k k x x 0 0 1 2 , ' ' ' ' , (3) Then Eq. (2) can be written simply as Mx Kx . = - (4) For other systems with different arrangements of masses and springs, the M K matrices will be different. But if the spring forces are linear functions of the coordinates, the equations of motion can always be written in the form of Eq. (4). So the following development is general, and is not restricted to the specific example cited above. Note that the (1,1) element of K represents the negative of the force on mass 1 when it is displaced a distance x 1 The (1,2) element is the negative of the force on mass when mass 2 is displaced a distance x 2 , and so on. This gives an alternative way to determine the elements of K instead of working out Eqs. (1) from Σ F = ma . Caution : Some books define K with the opposite sign, so there is no minus in Eq. (4). (See, for example, Edwards and Penney, p. 321.) We prefer to retain the minus sign, so that Eq. (4) has the same form as the equation of motion kx = - for a single particle.
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11-2 11 Normal Modes -- Matrix Methods General Formulation For a normal mode, each x must vary sinusoidally, and all x 's must have the same frequency. Therefore we try to find a solution of Eq. (4) in the form x x C a a t 1 2 1 2 F H G I K J = F H G I K J + cos , ϖ ϕ b g or x a = + C t cos . ϖ ϕ b g (5) In this expression, C is a scalar amplitude factor (determined by initial conditions), and a is a column matrix (or vector) whose elements a 1 , a 2 give the ratios of the amplitudes of the x for each normal mode. For example, if a 2 = 2 a 1 , the motion of the mass with coordinate x 2 is in phase with that of x 1 but with amplitude twice as great. Taking the second time derivative of Eq. (5), we get && cos cos , . x a a = - + = - + = ϖ ϖ ϕ λ ϖ ϕ λ ϖ 2 2 C t C t b g b g where (6) To test whether Eq. (5) really is a solution of Eq. (4), we substitute Eqs. (5) and (6) into Eq. (4). After dividing out the common factor C cos ( ϖ t + ϕ ), we get - = - λ Ma Ka , or K M a - = λ b g 0.
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This note was uploaded on 12/08/2011 for the course PHYSICS 340 taught by Professor Clarke during the Fall '08 term at University of Michigan.

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mod-matr - 11 Normal Modes - Matrix Methods Fall 2003...

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