Question

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?

Answer 1

Answer:

Angle will be $\Theta =45.09^{\circ}$

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 $\lambda =\frac{v}{f}=\frac{340}{600}=0.566m$

Now for the first minimum $Dsin\Theta =1\times \lambda$

$0.8\times sin\Theta =1\times 0.566$

$sin\Theta =0.7083$

$\Theta =45.09^{\circ}$