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Unformatted text preview: ially displaced from its equilibrium position by a small vertical distance—say bead k is displaced by an amount ck at t = 0. The beads are then released so that they can vibrate freely. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 560 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 m m L Equilibrium Position A Typical Initial Position It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu Figure 7.6.1 Problem: For small vibrations, determine the position of each bead at time t > 0 for any given initial configuration. D E Solution: The small vibration hypothesis validates the following assumptions. • The tension T remains constant for all time. • There is only vertical motion (the horizontal forces cancel each other). • Only small angles are involved, so the approximation sin θ ≈ tan θ is valid. T H Let yk (t) = yk be the vertical distance of the k th bead from equilibrium at time t, and set y0 = 0 = yn+1 . IG y k +1 yk θk R Y θk–1 yk–1 k–1 P k θ k +1 k+1 Figure 7.6.2 If θk is the angle depicted in Figure 7.6.2, the diagram above, then the upward force on the k th bead at time t is Fu = T sin θk , while the downward force is Fd = T sin θk−1 , so the total force on the k th bead at time t is O C F = Fu − Fd = T (sin θk − sin θk−1 ) ≈ T (tan θk − tan θk−1 ) =T yk+1 − yk yk − yk−1 − L L = T (yk−1 − 2yk + yk+1 ). L Newton’s second law says force = mass × acceleration, so we set myk = T T (yk−1 − 2yk + yk+1 ) =⇒ yk + (−yk−1 + 2yk − yk+1 ) = 0 (7.6.1) L mL together with yk (0) = ck and yk (0) = 0 to model the motion of the k th bead. Altogether, equations (7.6.1) represent a system of n second-order linear differential equations, and each is coupled to its neighbors so that no single Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot7...
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