PHY.7:5.3 - Derive an interior law and change the endpoint condition

A taut string has uniform tension T and mass per unit length mu. Its transverse displacement u(x,t) is small enough that slopes can be treated to leading order; changes of tension and longitudinal motion are neglected. The kinetic energy per length is mu*u_t^2/2. Expanding the extra length to second order in slope gives the stored elastic contribution T*u_x^2/2. For length l, use

S[u]=integral over time and 0<=x<=l of (mu*u_t^2-T*u_x^2)/2 dx dt.

Take variations eta that vanish at the two temporal endpoints. Integration by parts gives the interior coefficient -mu*u_tt+T*u_xx and the spatial boundary contribution

integral over time of [-T*u_x*eta] at x=0,l dt.

Thus the interior equation is mu*u_tt=T*u_xx. At a fixed endpoint, eta is zero. At an unloaded endpoint free to move transversely in this model, eta is arbitrary and the corresponding slope must be zero.

For mu=0.01 kg/m, T=100 N and l=1 m, wave speed is sqrt(T/mu)=100 m/s. Two fixed endpoints admit the lowest nonzero spatial mode sin(pi*x/l), giving frequency 50 Hz. Keep the left endpoint fixed and free the right endpoint transversely while maintaining its axial tension; the lowest mode becomes sin(pi*x/(2*l)) and its frequency is 25 Hz. The interior equation did not change. Reusing the fixed-end spectrum would miss the changed boundary.

Loaded endpoint. Attach a massless transverse spring of stiffness kappa at x=l. Add -integral kappa*u(l,t)^2/2 dt to the action. With the right endpoint variation free, its coefficient gives T*u_x(l,t)+kappa*u(l,t)=0. The spring changes the boundary condition through its stored energy. If an endpoint mass is consequential, its kinetic term must be included too; the massless condition would no longer supply that boundary’s dynamics.

Retain the boundary contribution until the physical freedom or load determines its use.