Transverse and Longitudinal Waves — Question 10

Question 10

A transverse string deformation stretches the string at second order

A string segment has small uniform slope $s=\partial y/\partial x$, with fixed horizontal span $dx$. Its actual length is $d\ell=\sqrt{1+s^2}\,dx$.

1. Find the leading fractional increase in length and evaluate it for $s=.10$.

2. Explain why transverse displacement does not mean exactly zero length change, and why linear wave theory can neglect this geometric correction.

Question 10 - Solution

A transverse string deformation stretches the string at second order

Method and mathematical working

Use the geometric length before expanding. The first-order term in slope is absent because reversing the sign of the slope cannot change the segment length.$$\begin{aligned}\frac{d\ell}{dx}&=\sqrt{1+s^2}=1+\frac{s^2}{2}-\frac{s^4}{8}+\cdots.\\[4pt]\frac{d\ell-dx}{dx}&=\sqrt{1+s^2}-1,\\s=0.10&:\quad \sqrt{1.01}-1=0.00498756\ldots\simeq s^2/2=0.0050.\end{aligned}$$

(a) Calculation and conclusion

Expanding gives $(d\ell-dx)/dx\approx s^2/2$. At $.10$ the approximation is $.0050$; the exact value is $\sqrt{1.01}-1=.004988$, about $0.499\%$.

(b) Interpretation and limits

The motion direction is transverse, but the inclined segment is longer than its projection. The change is quadratic in small slope, whereas linear wave theory retains first-order terms. Large slopes invalidate this particular neglect and can make tension changes important.

Physical check

The quadratic leading term explains both the positive extension and why it is omitted in a first-order displacement equation. Its numerical size still matters when checking the model accuracy.