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Unformatted text preview: Many Coupled Oscillators A V IBRATING S TRING Say we have n particles with the same mass m equally spaced on a string having tension . Let y k denote the vertical displacement if the k th mass. Assume the ends of the string are fixed; this is the same as having additional particles at the ends, but with zero displacement: y = 0 and y n + 1 = 0. Let k be the angle the segment of the string between the k th and k+1 st particle makes with the horizontal. Then Newtons second law of motion applied to the k th mass asserts that m y k = sin k sin k 1 , k = 1 ,..., n . (1) If the particles have horizontal separation h , then tan k = ( y k + 1 y k ) / h . For the case of small vibrations we assume that k 0; then sin k tan k = ( y k + 1 y k ) / h so we can rewrite (1) as y k = p 2 ( y k + 1 2 y k + y k 1 ) , k = 1 ,..., n , (2) where p 2 = / mh . This is a system of second order linear constant coefficient differential equations with the boundary conditions y ( t ) = 0 and y n + 1 ( t ) = 0. As usual, one seeks special solutions of the form y k ( t ) = v k e t . Substituting this into (2) we find 2 v k = p 2 ( v k + 1 2 v k + v k 1 ) , k = 1 ,..., n , that is, 2 is an eigenvalue of the matrix p 2 ( T 2 I ) , where T = 0 1 0 ... 0 0 0 1 0 1 ... 0 0 0 0 1 0 1 ... 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 ... 0 1 0 0 0 0 ... 1 0 1 0 0 0 ... 0 1 0 . (3) From the work in the next section (see (9)), we conclude that 2 k = 2 p 2 ( 1 cos k n + 1 ) = 4 p 2 sin 2 k 2 ( n + 1 ) , k = 1 ,..., n , so k = 2 ip sin k 2 ( n + 1 ) , k = 1 ,..., n . The corresponding eigenvectors V k are the same as for T (see (10)). Thus the special solutions are Y k ( t ) = V k e 2 ipt sin k 2 ( n + 1 ) , k = 1 ,..., n , where Y ( t ) = ( y 1 ( t ) ,..., y n ( t )) ....
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This note was uploaded on 03/06/2012 for the course MATH 260 taught by Professor Staff during the Spring '12 term at UPenn.
 Spring '12
 STAFF

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