Sound Waves — Question 5

Question 5

Curved wavefronts become locally planar over a small receiver

A point source emits spherical sound fronts in a uniform medium. A circular receiving aperture has radius $b=0.50\,\mathrm m$ and its center is distance $R=10.0\,\mathrm m$ from the source on the aperture axis. Take $c=340\,\mathrm{m/s}$.

1. Find the edge-to-center path and arrival-time differences.

2. At $f=1000\,\mathrm{Hz}$, calculate the phase difference. State a quantitative criterion for treating the incident front as planar across the aperture.

Question 5 - Solution

Curved wavefronts become locally planar over a small receiver

(a) Method and calculation

The center path is $R$, whereas an edge path is $\sqrt{R^2+b^2}$. Hence$$\delta r=\sqrt{100.25}-10=0.012492\,\mathrm m,\quad\delta t=\frac{\delta r}{340}=3.674\times10^{-5}\,\mathrm s.$$For $b\ll R$, binomial expansion gives $\delta r\simeq b^2/(2R)=0.0125\,\mathrm m$.

(b) Interpretation and further calculation

The geometric delay corresponds to$$\delta\phi=2\pi f\delta t=0.231\,\mathrm{rad}.$$A useful phase-uniformity criterion is $2\pi\delta r/\lambda\ll1$, or $b^2/(2R)\ll\lambda/(2\pi)$. Small aperture angle $b/R$ alone does not fix phase error; the wavelength matters too.

Physical check

The edge arrives later because its path is longer. As $R\to\infty$ at fixed $b$, curvature delay vanishes.