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An aircraft maintenance technician walks past a tall hangar door that acts as 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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Answer:

Angle will be [tex]\Theta =45.09^{\circ}[/tex]

Explanation:

We have given frequency of the sound f = 600 Hz

Vertical opening is of 0.8 meter

So width D = 0.8 m

Velocity of sound v = 340 m/sec

We know that wavelength [tex]\lambda =\frac{v}{f}=\frac{340}{600}=0.566m[/tex]

Now for the first minimum [tex]Dsin\Theta =1\times \lambda[/tex]

[tex]0.8\times sin\Theta =1\times 0.566[/tex]

[tex]sin\Theta =0.7083[/tex]

[tex]\Theta =45.09^{\circ}[/tex]

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