In figure, sound with a 4.40 cm wavelength travels rightward from a source and through a tube that consists of a straight portion and half-circle. Part of the sound wave travels through the half-circle and then rejoins the rest of the wave, which goes directly through the straight portion. This rejoining results in interference. What is the smallest radius $\mathbf{r}$that result in an intensity minimum at the detector?

Wavelength of sound$\mathrm{\lambda }=40.00\text{cm}=40\times {10}^{-2}\text{m}$.

The difference between the distance traveled by the sound wave, when it goes around a half circle and across the diameter of the circle, should satisfy the condition for destructive interference. Using this concept, find the radius of the circle.

Formula is as follow:

$\mathrm{f}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}△\mathrm{L}$

where, L is length, $\mathrm{\lambda }$ is wavelength and f is frequency.

Phase difference between two sound waves can be calculated by equation,

$\mathrm{\varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\text{Δ}\mathrm{L}$

Here, $\text{Δ}\mathrm{L}$is the path difference between two sound waves,

$\text{Δ}\mathrm{L}=\mathrm{\pi r}-2\mathrm{r}$

role="math" localid="1661493583341" $\mathrm{\varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{\pi r}-2\mathrm{r})$

But phase difference for destructive interference $\mathrm{\varphi }=\mathrm{\pi }$which is the smallest value for destructive interference is,

$\mathrm{\pi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{\pi r}-2\mathrm{r})$

$1=\frac{2}{\mathrm{\lambda }}(\mathrm{\pi }-2)\mathrm{r}$

$\mathrm{r}=\frac{\mathrm{\lambda }}{2(\mathrm{\pi }-2)}$

${\mathrm{r}}_{\min }=\frac{40\times {10}^{-2}\text{\hspace{0.17em}m}}{2\left(1.14\right)}$

${\mathrm{r}}_{\min }=\frac{40\times {10}^{-2}\text{\hspace{0.17em}m}}{2.28}$

${\mathrm{r}}_{\min }=17.5\text{cm}$

Hence, the smallest radius$\mathrm{r}$ that results in minimum intensity at the detector is ${\mathrm{r}}_{\min }=17.5\text{cm}$.