kakvoglq

Answered

2022-09-24

A biologist builds a device to detect and measure the size of insects. The device emits sound waves. If an insect passes through the beam of sound waves, it produces a diffraction pattern on an array of sound detectors behind the insect What is the lowest-frequency sound that can be used to detect a fly that is about 3 mm in diameter? The speed of sound is 340 m/s.

Answer & Explanation

vyhlodatis

Expert

2022-09-25Added 14 answers

We would like to determine the lowest frequency sound that can be used to detect a fly that is about 3 mm in diameter.

Lowest frequency means the largest wavelength since the speed of sound is constant and the wavelength is inversely proportional to the frequency.

So, our goal is to determine the largest wavelength of a sound wave that can be Used to detect a fly that is about 3 mm in diameter. The equation that describes the diffraction in this case, is

$\mathrm{sin}(\theta )=\frac{1.22\lambda}{D}$

$\theta $ , the angular deflection of the first minimum region from the fly.

$D=3\times {10}^{-3}m$, the diameter of the fly.

$\lambda $ , the maximum wavelength at which the first minimum region will still be detectable.

In order to detect a fly, the first minimum region must be detectable. After a certain wavelength, the first minimum region will not be detectable anymore meaning that there is no first minimum region and the central maximum is spread over all the detector. We are interested in that certain wavelength at which the angle of deflection of the first minimum will reach its maximum value, and this maximum value is $\theta ={90}^{\circ}$

So, let's substitute into (1) with $3\times {10}^{-3}m$ for D and ${90}^{\circ}$ for $\theta $

$\mathrm{sin}({90}^{\circ})=\frac{1.22\lambda}{3\times {10}^{-3}m}$

$\lambda =\frac{3\times {10}^{-3}m}{1.22}$

$\lambda =2.46\times {10}^{-3}m$

Now that we have the maximum wavelength of the needed sound wave, we will use it to determine the lowest frequency as follows

${f}_{Min}=\frac{{v}_{sound}}{{\lambda}_{Max}}$

Where ${v}_{sound}=340m/s$ and ${\lambda}_{Max}=2.46\times {10}^{-3}m$

${f}_{Min}=\frac{340m/s}{2.46\times {10}^{-3}m}$

${f}_{Min}=1.3\times {10}^{5}Hz$

Result:

${f}_{Min}=1.3\times {10}^{5}Hz$

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