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Show all your work clearly on the scratch paper) Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 1.00 m parallel to the wall, she experiences destructive interference for the first time. What is the frequency of the sound

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Rau7star

Answer:

Explanation:

The speed of sound in air to be 343 m/s.

Given:

distance 'd' = 5 m

L = 12 m

It can be concluded that path difference must be equal to half of the wavelength when person is observing destructive interference'y' at 1 m distance from the equidistant position

Since

λ/2 = yd/L

λ/2 = (1 x 5)/12

λ = 0.833m

Frequency of the sound is given by,

f = v / λ => 343 / 0.833

f=411.6 Hz