Q65P - University Physics with Modern Physics

Question

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 B along a line perpendicular to the line connecting A and B (line BC in Fig. P16.65). (a) At what distances from B will there be destructive interference? (b) At what distances from B will there be constructive interference? (c) If the frequency is made low enough, there will be no positions along the line BC at which destructive interference occurs. How low must the frequency be for this to be the case?

Step-by-Step Solution

Answer

1. Points of destructive interference with respect to B are $9.01m$ , 2.71 m, 1.27 m, 0.53 m, 0.026 m.
2. Points of constructive interference with respect to B are 4.34 m, 1.84 m, 0.86 m, 0.26 m.
3. The lowest frequency for the case mentioned is  $86\text{ Hz}.$

Step 1: Given Data

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 $784\text{\hspace{0.17em}Hz}$

Step 2: (a) Determination of the distance from B where destructive interference occur.

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,

$\begin{array}{c}\lambda =\frac{v}{f}\\ \text{=}\frac{\text{344\hspace{0.17em}m/s}}{\text{784\hspace{0.17em}Hz}}\\ \text{=0.439\hspace{0.17em}m}\end{array}$

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,

$\begin{array}{c}\sqrt{x^2+h^2}-x+x=\beta \lambda +x\\ \sqrt{x^2+h^2}=\beta \lambda +x\\ x^2+h^2={(\beta \lambda +x)}^2\end{array}$

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

 

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

$\begin{array}{c}x=\frac{h^2}{2\beta \lambda }-\frac{\beta }{2}\lambda \\ x=\frac{{\left(2.00\text{\hspace{0.17em}m}\right)}^2}{2\frac{1}{2}0.439\text{\hspace{0.17em}m}}-\frac{\frac{1}{2}}{2}0.439\text{\hspace{0.17em}m}\\ =9.01\text{\hspace{0.17em}m}\end{array}$

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

Step 3: (b) Determination of the distance from B where constructive destructive occur.

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.

Step 4: (c) Determination of the lowest frequency.

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