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Unformatted text preview: L = −T 0 T 2 ∂u ∂x 2 L dx − 0 ∂u ∂ξ dx + ∂x ∂x L 0 T 2 T 2 ∂ (u + ξ ) ∂x ∂ξ ∂x 2 dx 2 dx. We are imagining that the displacement ξ is inﬁnitesimally small, so terms containing the square of ξ or the square of a derivative of ξ can be ignored, and hence L ∂u ∂ξ F (x), ξ (x) = −T dx. 0 ∂x ∂x Integration by parts yields L F (x), ξ (x) = T 0 ∂2u ξ (x)dx − T ∂x2 ∂u ξ (L) − T ∂x ∂u ξ (0). ∂x Since ξ (0) = ξ (L) = 0, we conclude that L L F (x)ξ (x)dx = F (x), ξ (x) = T 0 0 ∂2u ξ (x)dx. ∂x2 Since this formula holds for all inﬁnitesimal displacements ξ (x), we must have F (x) = T ∂2u , ∂x2 for the force density per unit length. Now we apply Newton’s second law, force = mass × acceleration, to the function u(x, t). The force acting on a tiny piece of the string of length dx is F (x)dx, while the mass of this piece of string is just ρdx, where ρ is the density of the string. Thus Newton’s law becomes T ∂2u ∂2u dx = ρdx 2 . 2 ∂x ∂t If we divide by ρdx, we obtain the wav...
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