I am trying to understand the derivation of the normal modes of a dangling chain given here[pdf].

The author considers a chain with density per-unit-length $\rho$ hanging from a fixed point and adopts a coordinate system with the $x$ axis pointing vertically upwards from the chain's loose end in its equilibrium position and the $u$ axis pointing the the right. The horizontal displacements are considered to be small so that distances along the chain can be approximated as distances along $x$.

The tension in the chain at position $x$ is $w(x) = \rho g x$ and the accelerating force is due to the difference in the horizontal components of the tension at the ends of a small interval of chain, $\Delta x$.

If the segment at $x$ is displaced from the vertical by an angle $\alpha$, the horizontal component of the tension is $F(x) = w(x)\sin\alpha \approx Wu_x$. I think I get this – if the notation $u_x$ means $\mathrm{d}u/\mathrm{d}x$. The author then says:

The difference in force between the points on the change[sic – I think "chain" is meant] at $x$ and $x+\Delta x$ is thus $\Delta F = \Delta x(w u_x)_x$.

This bit I don't get. How does $\Delta F = w(x+\Delta x) u_x - w(x)u_x$ become the above expression?


1 Answer 1


Using Taylor expansion,

$$w(x+\Delta x)=w(x)+w_x(x) \Delta x+O(\Delta x^2)$$

$$u_x(x+\Delta x)=u_x(x)+u_{xx}(x) \Delta x+O(\Delta x^2)$$

Replace the above in the following, $$w(x+\Delta x) u_x(x+\Delta x) - w(x) u_x(x) \approx \Delta x w_x(x) u_x(x) + \Delta x w(x) u_{xx}(x) = \Delta x (w u_x)_x$$


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