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488 chapter 18that the density and tension are proportional:ρ(x)=ρ 0 e αx , T(x)=T 0 e αx . (18.28)While we would expect the tension to be greater in regions of higher density (moremass to move and support), being proportional is clearly just an approximation.Substitution of these relations into (18.27) yields the new wave equation:∂ 2 y(x, t) ∂y(x, t)∂x 2 + α = 1 ∂ 2 y(x, t)∂x c 2 ∂t 2 , c 2 = T 0. (18.29)ρ 0Here c is a constant that would be the wave velocity if α =0. This equation issimilar to the wave equation with friction, only now the first derivative is withrespect to x and not t. The corresponding difference equation follows from usingcentral-difference approximations for the derivatives:y i,j+1 =2y i,j − y i,j−1 + αc2 (∆t) 22∆x[y i+1,j − y i,j ]+ c2c ′2 [y i+1,j + y i−1,j − 2y i,j ],y i,2 = y i,1 + c2c ′2 [y i+1,1 + y i−1,1 − 2y i,1 ]+ αc2 (∆t) 22∆x [y i+1,1 − y i,1 ]. (18.30)18.4.1 Waves on a CatenaryUp until this point we have been ignoring the effect of gravity upon our string’sshape and tension. This is a good approximation if there is very little sag in thestring, as might happen if the tension is very high and the string is light. Even ifthere is some sag, our solution for y(x, t) could be used as the disturbance aboutthe equilibrium shape. However, if the string is massive, say, like a chain or heavycable, then the sag in the middle caused by gravity could be quite large (Figure 18.5),and the resulting variation in shape and tension needs to be incorporated into thewave equation. Because the tension is no longer uniform, waves travel faster nearthe ends of the string, which are under greater tension since they must support theentire weight of the string.18.4.2 Derivation of a Catenary ShapeConsider a string of uniform density ρ acted upon by gravity. To avoid confusionwith our use of y(x) to describe a disturbance on a string, we call u(x) the equilibriumshape of the string (Figure 18.5). The statics problem we need to solve is todetermine the shape u(x) and the tension T (x). The inset in Figure 18.5 is a freebodydiagram of the midpoint of the string and shows that the weight W of thissection of arc length s is balanced by the vertical component of the tension T . The−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 488