91Ó°ÊÓ

Distance between the speakers is$2.00\text{″¾}$

Sound’s speed in air is $344\text{″¾}/\text{s}$

Frequency emitted$784\text{ H³ú}$

When the path difference between two coherent sources is a half-integer number of Wavelengths, Destructive interference occurs. If path difference is integral multiple of wavelength, constructive interference occurs.

Wavelength can be calculated as,

Take the distance between the speakers as h. The condition for destructive interference is,

$\sqrt{x^2+h^2}-x=\mathrm{β}\mathrm{λ}$

Solving for x,

$x=\frac{h^2}{2\mathrm{β}\mathrm{λ}}−\frac{\mathrm{β}}{2}\mathrm{λ}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

Substitute the values in the above equation and put $\mathrm{β}=\frac{1}{2}$

Similarly, for $\mathrm{β}=\frac{3}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2\frac{3}{2}0.439\text{″¾}}−\frac{\frac{3}{2}}{2}0.439\text{″¾}\\ =2.71\text{″¾}\end{array}$

for $\mathrm{β}=\frac{5}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2\frac{5}{2}0.439\text{″¾}}−\frac{\frac{5}{2}}{2}0.439\text{″¾}\\ =1.27\text{″¾}\end{array}$

For $\mathrm{β}=\frac{7}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{ }m\right)}^2}{2\frac{7}{2}0.439\text{ }m}−\frac{\frac{7}{2}}{2}0.439\text{ }m\\ =0.53\text{ }m\end{array}$

For, $\mathrm{β}=\frac{9}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2\frac{9}{2}0.439\text{″¾}}−\frac{\frac{9}{2}}{2}0.439\text{″¾}\\ =0.026\text{″¾}\end{array}$

Thus,$9.01m$ ,$2.71m$ , $1.27m$,$0.53m$ ,$0.026m$ are the positions of destructive interference.

Substitute the values in equation $\left(1\right)$and put $\mathrm{β}=1$

$\begin{array}{c}x=\frac{h^2}{2\mathrm{β}\mathrm{λ}}−\frac{\mathrm{β}}{2}\mathrm{λ}\\ x=\frac{{\left(2.00\text{″¾}\right)}^2}{2×1×0.439\text{″¾}}−\frac{1}{2}×0.439\text{″¾}\\ =4.34\text{″¾}\end{array}$

Similarly, for $\mathrm{β}=2$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2×2×0.439\text{″¾}}−\frac{2}{2}0.439\text{″¾}\\ =1.84\text{″¾}\end{array}$

for $\mathrm{β}=3$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2×3×0.439\text{″¾}}−\frac{3}{2}0.439\text{″¾}\\ =0.86\text{″¾}\end{array}$

For $\mathrm{β}=4$

$\begin{array}{c}x=\frac{{\left(2.00\text{″¾}\right)}^2}{2×4×0.439\text{″¾}}−\frac{4}{2}0.439\text{″¾}\\ =0.26\text{″¾}\end{array}$

Thus, the positions of constructive interference are 4.34 m, 1.84 m, 0.86 m, 0.26 m.

There will be destructive interference at speaker B when$h=\mathrm{λ}/2$.

The path difference can never be more than or even as big as $\mathrm{λ}/2$.

Therefore, the minimum frequency is then,

$\begin{array}{c}\frac{v}{2h}=\left(344\text{m/s}\right)/\left(4.0\text{m}\right)\\ =86\text{Hz}.\end{array}$

The lowest frequency at which destructive interference occurs is $86\text{​ H³ú}$.