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480 chapter 18Because (18.2) is second-order in time, a second initial condition (beyond initialdisplacement) is needed to determine the solution. We interpret the “gentleness”of the pluck to mean the string is released from rest:∂y(x, t =0)=0, (initial condition 2). (18.4)∂tThe boundary conditions have both ends of the string tied down for all times:y(0,t) ≡ 0, y(L, t) ≡ 0, (boundary conditions). (18.5)18.2.1 Solution via Normal-Mode ExpansionThe analytic solution to (18.2) is obtained via the familiar separation-of-variablestechnique. We assume that the solution is the product of a function of space and afunction of time:y(x, t)=X(x)T (t). (18.6)We substitute (18.6) into (18.2), divide by y(x, t), and are left with an equation thathas a solution only if there are solutions to the two ODEs:d 2 T (t)dt 2+ ω 2 T (t)=0,d 2 X(x)dt 2 + k 2 X(x)=0, k def= ω c . (18.7)The angular frequency ω and the wave vector k are determined by demanding thatthe solutions satisfy the boundary conditions. Specifically, the string being attachedat both ends demandsThe time solution isX(x =0,t)=X(x = l, t)=0 (18.8)⇒ X n (x)=A n sin k n x, k n =π(n +1), n=0, 1,.... (18.9)LT n (t)=C n sin ω n t + D n cos ω n t, ω n = nck 0 = n 2πcL , (18.10)where the frequency of this nth normal mode is also fixed. In fact, it is the singlefrequency of oscillation that defines a normal mode. The initial condition (18.3) ofzero velocity, ∂y/∂t(t =0)=0, requires the C n values in (18.10) to be zero. Puttingthe pieces together, the normal-mode solutions arey n (x, t) = sin k n x cos ω n t, n =0, 1,.... (18.11)Since the wave equation (18.2) is linear in y, the principle of linear superpositionholds and the most general solution for waves on a string with fixed ends can be−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 480