Wave Speed on a Stretched String — Question 7

Question 7

A support angle can amplify string tension above a supported weight

A mass $M$ is supported symmetrically by two identical straight string segments, each making angle $\theta$ above horizontal. Each segment has density $\mu$, and its own weight is negligible. Consider small transverse disturbances away from the junction.

1. Find the tension and local wave speed.

2. Compare speeds for $\theta=30^\circ$ and $\theta=90^\circ$ with the same $M,\mu$. Explain the small-angle limitation.

Question 7 - Solution

A support angle can amplify string tension above a supported weight

Method and mathematical working

Balance the vertical components of the two equal tensions. Horizontal components cancel by symmetry; each supporting string then has its own wave speed.$$\begin{aligned}2F\sin\theta&=Mg,\\v&=\sqrt{\frac{Mg}{2\mu\sin\theta}}.\\[4pt]\frac{v(30^\circ)}{v(90^\circ)}&=\sqrt{\frac{\sin90^\circ}{\sin30^\circ}}=\sqrt2.\end{aligned}$$

(a) Calculation and conclusion

Vertical balance gives $2F\sin\theta=Mg$. Thus $F=Mg/(2\sin\theta)$ and $v=\sqrt{Mg/(2\mu\sin\theta)}$. Each tension supplies only its vertical component to support the load.

(b) Interpretation and limits

The speed ratio is $\sqrt{\sin90^\circ/\sin30^\circ}=\sqrt2$. The formula diverges as $\theta\to0$, signaling the unbounded tension required by the ideal geometry. Real stretching and finite strength invalidate extrapolation to arbitrarily small angles.

Physical check

As the supports approach horizontal, increasingly large tensions are required. The mathematical divergence flags the failure of ideal perfectly flexible, unbreakable supports in that limit.