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

Angle will be
`\Theta =45.09^(\circ)`

Step-by-step 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 =(v)/(f)=(340)/(600)=0.566m`

Now for the first minimum
`Dsin\Theta =1* \lambda`

`0.8* sin\Theta =1* 0.566`

`sin\Theta =0.7083`

`\Theta =45.09^(\circ)`