Sound Waves — Question 7

Question 7

Reciprocal acoustic timing separates wind from sound speed

Two stationary stations are separated by $L=100\,\mathrm m$ along uniform air flow. Sound takes $t_+=0.250\,\mathrm s$ downstream and $t_-=0.3125\,\mathrm s$ upstream. Let $c$ be sound speed relative to air and $w<c$ the flow speed.

1. Infer $c$ and $w$ from these measurements.

2. Find the round-trip time and show why using it alone with still-air ranging overestimates the actual distance when $c$ is known.

Question 7 - Solution

Reciprocal acoustic timing separates wind from sound speed

(a) Method and calculation

The ground speeds are $c+w$ and $c-w$, so$$c=\frac L2\left(\frac1{t_+}+\frac1{t_-}\right)=360\,\mathrm{m/s},$$$$w=\frac L2\left(\frac1{t_+}-\frac1{t_-}\right)=40\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

The round-trip time is$$t_{\mathrm{rt}}=\frac L{c+w}+\frac L{c-w}=\frac{2Lc}{c^2-w^2}=0.5625\,\mathrm s.$$A still-air interpretation gives $L_{\mathrm{app}}=ct_{\mathrm{rt}}/2=101.25\,\mathrm m$. More generally $L_{\mathrm{app}}/L=1/[1-(w/c)^2]>1$: the upstream delay exceeds the downstream saving.

Physical check

The round-trip correction starts at order $(w/c)^2$; a one-way timing difference is first order in flow.