Question

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 610 Hz sound. At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is 0.840 m wide and the speed of sound is 340 m/s

Answer 1

Answer:

The first minimum would be observed at 41.57°

Explanation:

v = 340m/s = speed of sound

f = 610Hz

d = 0.840m

λ = ?

Mλ = wsinθ

m = mth order minima

λ = wavelength incident on the single slit

θ = angular position of the mth minima

But, λ = v / f

λ = 340 / 610 = 0.557m

θ = sin⁻(mλ/d)

θ = sin⁻ [(1 * 0.557) / 0.840]

θ = sin⁻ 0.6635

θ = 41.57°

The first minimum would be observed at 41.57°