The Speed of a Traveling Wave — Question 7

Question 7

Position-dependent speed must be integrated along the path

A localized wave feature follows $dx/dt=v_0(1+x/L)$ for $0\le x\le L$, with $v_0>0$. Assume this ray description is applicable.

1. Find the time to traverse the region.

2. Compare with using the arithmetic average of the endpoint speeds. Does the assumed feature retain a constant speed?

Question 7 - Solution

Position-dependent speed must be integrated along the path

Method and mathematical working

Express each small contribution to travel time as $dx/v(x)$. With $u=1+x/L$, the integral becomes logarithmic rather than an arithmetic average.$$\begin{aligned}t&=\frac{1}{v_0}\int_0^L\frac{dx}{1+x/L}\\&=\frac{L}{v_0}\int_1^2\frac{du}{u}=\frac{L}{v_0}\ln2.\\[4pt]\bar v&=v_0/\ln2\simeq1.4427v_0,\quad t_{\mathrm{endpoint\ mean}}=\frac{2L}{3v_0}<\frac{L\ln2}{v_0}.\end{aligned}$$

(a) Calculation and conclusion

$dt=dx/[v_0(1+x/L)]$, so $t=(L/v_0)\ln2$. The path-time average speed is $L/t=v_0/\ln2\approx1.4427v_0$.

(b) Interpretation and limits

The endpoint speeds are $v_0$ and $2v_0$; their arithmetic mean would predict $2L/(3v_0)$, shorter than $(L/v_0)\ln2$. Speed increases continuously and time accumulates as the reciprocal of local speed. A single constant-speed translation is not the stated model.

Physical check

Since the speed stays between $v_0$ and $2v_0$, the time must lie between $L/(2v_0)$ and $L/v_0$. The logarithmic result satisfies both bounds.