The Speed of a Traveling Wave — Question 5

Question 5

A current makes upstream and downstream travel times unequal

A wave travels at speed $c$ relative to a uniform moving medium. The medium drifts right at speed $U<c$. Across a fixed laboratory distance $L=12\,\mathrm{m}$, travel times are $2\,\mathrm{s}$ downstream and $3\,\mathrm{s}$ upstream. Use Galilean velocity addition.

1. Infer $c$ and $U$.

2. Find the round-trip time and compare with $2L/c$. Explain the difference.

Question 5 - Solution

A current makes upstream and downstream travel times unequal

Method and mathematical working

Write one laboratory speed for each direction and solve their sum and difference. For a round trip, add the two times, since each leg covers the same distance.$$\begin{aligned}c+U&=\frac{12}{2}=6,\qquad c-U=\frac{12}{3}=4,\\t_{\mathrm{round}}&=\frac{L}{c+U}+\frac{L}{c-U}=\frac{2Lc}{c^2-U^2}.\\[4pt]c&=(6+4)/2=5,\quad U=(6-4)/2=1\quad(\mathrm{m/s}),\\t_{\mathrm{round}}&=2+3=5.0\,\mathrm s>24/5=4.8\,\mathrm s.\end{aligned}$$

(a) Calculation and conclusion

Laboratory speed magnitudes are $c+U=L/2=6$ and $c-U=L/3=4\,\mathrm{m/s}$. Adding and subtracting gives $c=5\,\mathrm{m/s}$ and $U=1\,\mathrm{m/s}$.

(b) Interpretation and limits

The round trip takes $5\,\mathrm{s}$, compared with $2L/c=4.8\,\mathrm{s}$. Exactly, $t=2Lc/(c^2-U^2)>2L/c$: the extra upstream time exceeds the downstream saving. The two directional speeds must not be averaged as travel times.

Physical check

The denominator is positive because $U<c$. As $U\to0$, the still-medium result is recovered; as $U\to c$, the upstream time diverges.