Sound Waves — Question 9

Question 9

Diffraction distinguishes acoustic wavelengths at one opening

A wall opening has width $D=0.50\,\mathrm m$. Compare tones at $170\,\mathrm{Hz}$ and $3400\,\mathrm{Hz}$ in air with $c=340\,\mathrm{m/s}$. Use the qualitative diffraction criterion: spreading is strong when $D/\lambda\lesssim1$ and weak when $D/\lambda\gg1$.

1. Calculate the two wavelength-to-opening comparisons and identify which tone spreads more strongly.

2. For the high-frequency tone, use the supplied far-field slit relation $\sin\theta_1=\lambda/D$ to locate the first minimum. Why does this formula yield no first minimum for the lower tone?

Question 9 - Solution

Diffraction distinguishes acoustic wavelengths at one opening

(a) Method and calculation

The wavelengths are$$\lambda_{\mathrm{low}}=340/170=2.00\,\mathrm m,\qquad\lambda_{\mathrm{high}}=340/3400=0.100\,\mathrm m.$$$$D/\lambda_{\mathrm{low}}=0.25,\qquad D/\lambda_{\mathrm{high}}=5.0.$$The lower tone spreads broadly; the higher tone has a narrower forward pattern.

(b) Interpretation and further calculation

For the high tone,$$\theta_1=\sin^{-1}(0.100/0.50)=11.54^\circ.$$For the low tone, the equation would require $\sin\theta_1=4$, which is impossible. It predicts no first zero within the forward angular range, consistent with broad spreading. The slit formula assumes a uniformly driven aperture and far-field observation.

Physical check

The relevant comparison is $D/\lambda$, so changing medium speed at fixed frequency also changes diffraction.