Homework 4

# 3660254037844380 0

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Unformatted text preview: in the lecture notes. The matrix was divided by the values of mass 1 = 1 and mass 2 = 2. >> B = [2/1 ‐1/1; ‐1/2 2/2] B = 2.000000000000000 ‐1.000000000000000 ‐0.500000000000000 1.000000000000000 >> [v e] = eig(B) v = 0.939070801588044 0.590690494568872 ‐0.343723769333440 0.806898221355074 e = 2.366025403784438 0 0 The first column of v shows that the two masses are moving opposite of each other (reverse of what the case is with equal masses) and with the highest mode. In the case of equal masses, the lowest mode corresponded to the first column, but here that is switched. The second column of v indicates that the masses are moving together, the second mass faster than the first, and the total mode of this motion is the lowest mode. 15.2) b) Calculations for the general matrix were supplied in the lecture notes. The matrix was divided by the values of mass 1 = 1, mass 2 = 2 amd mass 3 = 3. >> B = [2 ‐1 0; ‐1/2 2/2 ‐1/2; 0/3 ‐1/3 2/3] B = 2.000000000000000 ‐1.000000000000000 0 ‐0.500000000000000 1.000000000000000 ‐0.500000000000000 0 ‐0.3333333333333...
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## This note was uploaded on 01/15/2011 for the course MATH 345 taught by Professor Staff during the Spring '08 term at Ohio State.

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