Wave Speed on a Stretched String — Question 6

Question 6

A hanging string has a travel-time gradient from its own weight

A uniform vertical string of length $L$ and density $\mu$ supports a mass $M>0$ at its bottom. Let $z$ measure upward from the bottom. In a local-wave approximation, $F(z)=(M+\mu z)g$ and $v(z)=\sqrt{F(z)/\mu}$.

1. Integrate the one-way travel time from bottom to top.

2. Find the formal limit as $M\to0$, and state why the local-wave description near the unloaded bottom requires care.

Question 6 - Solution

A hanging string has a travel-time gradient from its own weight

Method and mathematical working

At height $z$, the supported mass is $M+\mu z$. Integrate the reciprocal of the resulting local speed over the full height.$$\begin{aligned}F(z)&=(M+\mu z)g,\qquad v(z)=\sqrt{g(M/\mu+z)},\\t&=\frac1{\sqrt g}\int_0^L\frac{dz}{\sqrt{M/\mu+z}}.\\[4pt]t&=\frac2{\sqrt g}\left[\sqrt{M/\mu+L}-\sqrt{M/\mu}\right],\\M\to0&:\quad t\to2\sqrt{L/g}.\end{aligned}$$

(a) Calculation and conclusion

Writing $a=M/\mu$, $t=\int_0^L dz/\sqrt{g(a+z)}=2[\sqrt{a+L}-\sqrt a]/\sqrt g$. The upper part has larger tension and larger local speed.

(b) Interpretation and limits

The formal limit is $t=2\sqrt{L/g}$, finite despite $v(0)=0$. The reciprocal-square-root singularity is integrable. However, near the unloaded end the tension vanishes and the slowly varying local-wave approximation and taut-string idealization are delicate; the integral is a model result, not a complete boundary analysis.

Physical check

For $M\gg\mu L$, expansion of the square roots gives $t\approx L\sqrt{\mu/(Mg)}$, the nearly uniform-tension result. The massless-load limit has an integrable endpoint singularity in travel time.