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An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Outside the door, on a line perpendicular to the opening in the door, a jet engine makes a 600-Hz sound. At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is 0.800 m wide and the speed of sound is 340 m/s?
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$45.1^\circ$
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OpenStax College Physics for AP® Courses Solution, Chapter 27, Problem 56 (Problems & Exercises) (2:18)

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Video Transcript
This is College Physics Answers with Shaun Dychko. An aircraft maintenance person walks past a open door in a hangar and outside the hangar, there's a jet engine projecting sound directly towards this open doorway and the question is at what angle with respect to this perpendicular line... this perpendicular to the wall at what angle can the worker position themselves such that they are at a minimum for the noise due to the diffraction pattern as this sound wave goes through this opening. So the frequency of this sound we are told is 600 hertz and the doorway has a size of 0.800 meters and we are also told that sound on this day with this humidity and temperature and so on is traveling at a speed of 340 meters per second. We are also looking for the very first minimum— the smallest angle minimum— and so that means m is 1... the order is 1. Here's our formula for finding the minima due to a diffraction pattern through a single slit: this is the size of the slit D times sin of the angle and that equals the order times the wavelength. So we can solve for sin Θ by dividing both sides by the size of the doorway D so sin Θ is over D. Now we are not told what the wavelength is but we can use the wave equation to figure it out knowing the speed of the wave and its frequency. So the speed of a wave is its frequency times its wavelength and we can divide both sides by f and then switch the sides around and we get λ is v over f and we can make a substitution for λ then and replace it with v over f. Take the inverse sin of both sides here to solve for the angle and we get that the angle then is the inverse sin of m times wave speed divided by frequency times door way width. So that's the inverse sin of 1 times 340 meters per second divided by 600 hertz times 0.800 meters and that is 45.1 degrees. So this worker— who has a smile on their face because they are in a nice, quiet position— is angled at 45.1 degrees, with respect to this perpendicular.