Two identical loudspeakers are located at points A and B, 2.00 m apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of 784 Hz. Take the speed of sound in air to be 344 m/s. A small microphone is moved out from point Balong a line perpendicular to the line connecting Aand B(line BC in Fig. P16.65). (a) At what distances from Bwill there be destructiveinterference? (b) At what distances from Bwill there be constructiveinterference? (c) If the frequency is made low enough, there will be no positions along the line BCat which destructive interference occurs. How low must the frequency be for this to be the case?

Distance between the speakers is$2.00\text{\hspace{0.17em}m}$

Sound’s speed in air is $344\text{\hspace{0.17em}m}/\text{s}$

Frequency emitted$784\text{\hspace{0.17em}Hz}$

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,

$\sqrt{x^2+h^2}-x=\beta \lambda$

Solving for x,

$x=\frac{h^2}{2\beta \lambda }-\frac{\beta }{2}\lambda \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

Similarly, for $\beta =\frac{3}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\frac{3}{2}0.439\text{\hspace{0.17em}m}}-\frac{\frac{3}{2}}{2}0.439\text{\hspace{0.17em}m}\\ =2.71\text{\hspace{0.17em}m}\end{array}$

for $\beta =\frac{5}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\frac{5}{2}0.439\text{\hspace{0.17em}m}}-\frac{\frac{5}{2}}{2}0.439\text{\hspace{0.17em}m}\\ =1.27\text{\hspace{0.17em}m}\end{array}$

For $\beta =\frac{7}{2}$

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

For, $\beta =\frac{9}{2}$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\frac{9}{2}0.439\text{\hspace{0.17em}m}}-\frac{\frac{9}{2}}{2}0.439\text{\hspace{0.17em}m}\\ =0.026\text{\hspace{0.17em}m}\end{array}$

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 $\beta =1$

$\begin{array}{c}x=\frac{h^2}{2\beta \lambda }-\frac{\beta }{2}\lambda \\ x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\times 1\times 0.439\text{\hspace{0.17em}m}}-\frac{1}{2}\times 0.439\text{\hspace{0.17em}m}\\ =4.34\text{\hspace{0.17em}m}\end{array}$

Similarly, for $\beta =2$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\times 2\times 0.439\text{\hspace{0.17em}m}}-\frac{2}{2}0.439\text{\hspace{0.17em}m}\\ =1.84\text{\hspace{0.17em}m}\end{array}$

for $\beta =3$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\times 3\times 0.439\text{\hspace{0.17em}m}}-\frac{3}{2}0.439\text{\hspace{0.17em}m}\\ =0.86\text{\hspace{0.17em}m}\end{array}$

For $\beta =4$

$\begin{array}{c}x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\times 4\times 0.439\text{\hspace{0.17em}m}}-\frac{4}{2}0.439\text{\hspace{0.17em}m}\\ =0.26\text{\hspace{0.17em}m}\end{array}$

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$h=\lambda /2$.

The path difference can never be more than or even as big as $\lambda /2$.

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 $86\text{\hspace{0.17em}Hz}$.