Types of Waves — Question 9

Question 9

Two travel times reveal a delay at the source

A triggered source sends identical pulses along a uniform line. Arrival times measured from the trigger are $.70\,\mathrm{s}$ at $x=2\,\mathrm{m}$ and $1.30\,\mathrm{s}$ at $x=5\,\mathrm{m}$. Assume an unknown fixed launch delay $\tau$ and one constant propagation speed.

1. Infer the speed and launch delay.

2. Why does using $x/t$ at either detector give an inconsistent answer, and what prediction tests the model at $x=8\,\mathrm{m}$?

Question 9 - Solution

Two travel times reveal a delay at the source

Method and mathematical working

Write one arrival-time equation per detector. Subtraction removes the unknown launch delay, after which either detector equation determines that delay.$$\begin{aligned}t_1&=\tau+\frac{x_1}{v},\qquad t_2=\tau+\frac{x_2}{v},\\ t_2-t_1&=\frac{x_2-x_1}{v}.\\[4pt]v&=\frac{5-2}{1.30-0.70}=5.0\,\mathrm{m/s},\\\tau&=0.70-2/5=0.30\,\mathrm s,\quad t(8)=0.30+8/5=1.90\,\mathrm s.\end{aligned}$$

(a) Calculation and conclusion

With $t=\tau+x/v$, subtraction gives $.60=3/v$, hence $v=5.0\,\mathrm{m/s}$. The launch delay is $.70-2/5=.30\,\mathrm{s}$. The slope determines speed independently of the common delay.

(b) Interpretation and limits

The trigger time is not the launch time, so $x/t$ wrongly includes the delay in the travel interval. At $8\,\mathrm{m}$ the prediction is $.30+8/5=1.90\,\mathrm{s}$. A third detector can test whether the same delay and speed explain all arrivals.

Physical check

The fitted time at $x=0$ must be a delay rather than a propagation time. Substituting the inferred speed and delay into both original arrival equations checks the inverse calculation before extrapolating to the third location.