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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? The speed of sound in air is 343 m/s

User Nicksweet
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Answer:

Step-by-step explanation:

This is a case of interference of sound , akin to YDSE in optics .

Here, like interference dark and bright fringes, region of silence and intense sound will be formed due to destructive and constructive interference respectively.

Here d = distance between two sources = 5 m

D = distance of source and screen = 12m

position of first destructive interference

= λ D /2d

1 = λ 12 /2x 5

λ = 5 / 6 m

frequency = v / λ

= 343 x 6/ 5

= 411.6 Hz

User Rajohan
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