The Speed of a Traveling Wave — Question 6

Question 6

Travel through unequal regions uses a harmonic mean

A pulse travels through two consecutive equal lengths, each $3\,\mathrm{m}$, at speeds $2$ and $6\,\mathrm{m/s}$. Ignore delays at the interface.

1. Find total travel time and overall average speed.

2. Compare with the arithmetic mean of the two speeds and explain which weighting is appropriate.

Question 6 - Solution

Travel through unequal regions uses a harmonic mean

Method and mathematical working

Accumulate travel time separately in each region. Divide the total distance by this total time only after the two segments have been combined.$$\begin{aligned}t&=\frac{L}{v_1}+\frac{L}{v_2},\\\bar v&=\frac{2L}{t}=\frac{2v_1v_2}{v_1+v_2}.\\[4pt]t_1&=3/2=1.5\,\mathrm s,\quad t_2=3/6=0.5\,\mathrm s,\\\bar v&=\frac{3+3}{1.5+0.5}=3.0\,\mathrm{m/s}\ne\frac{2+6}{2}.\end{aligned}$$

(a) Calculation and conclusion

The times are $3/2=1.5$ and $3/6=.5\,\mathrm{s}$. Total time is $2\,\mathrm{s}$ over $6\,\mathrm{m}$, giving average speed $3\,\mathrm{m/s}$.

(b) Interpretation and limits

The arithmetic mean is $4\,\mathrm{m/s}$ and would apply to equal times, not equal distances. Equal lengths give the harmonic mean $2/(1/v_1+1/v_2)$. The pulse spends more time in the slower region.

Physical check

The result must lie between the two speeds and be weighted toward the slower one. If either speed tends to zero, the crossing time diverges and the overall speed tends to zero.