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Unformatted text preview: L = −T
∂x 2 L dx −
0 ∂u ∂ξ
∂x ∂x L
2 ∂ (u + 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
F (x), ξ (x) = −T
0 ∂x ∂x
Integration by parts yields
L F (x), ξ (x) = T
ξ (x)dx − T
ξ (L) − T
∂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
∂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
dx = ρdx 2 .
∂t If we divide by ρdx, we obtain the wav...
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- Winter '14