Past exam of the mathematical methods II course of the University of Oxford / 2019 / 9 / a

Assume small transverse slopes, a flexible string, constant tension $T$ and uniform linear mass density $\rho$. For an element of equilibrium length $dx$, the vertical force is the difference of the vertical tension components at its ends. To first order in slope,Its mass is $\rho dx$ and its transverse acceleration is $y_{tt}$. Newton's second law therefore gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$, orThis is the one-dimensional wave equation for small-amplitude transverse waves. Changes in string length and horizontal acceleration are second order in slope and are neglected in this linear approximation.

At distance $x$ below the support, the cable must carry the weight of the remaining length $L-x$. Thus its equilibrium tension is $T(x)=\mu g(L-x)$, and the local transverse wave speed isThe linear mass density cancels. More precisely the small-displacement equation with variable tension is $\mu y_{tt}={\partial }_x[T(x)y_x]$; its local characteristic speed is the expression above, even though a first-spatial-derivative term is also present.

The wave travel time on a hanging uniform cable is obtained from $dt=dx/c(x)$:Although the local speed tends to zero at the free end, the endpoint singularity in the travel-time integral is integrable. This finite result belongs to the ideal flexible-cable model, with bending stiffness neglected.