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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 2.4 m. a person is standing 10.0 m away from the wall, equidistant from the loudspeakers. when the person moves 0.20 m parallel to the wall, she experiences destructive interference for the first time. what is the frequency of the sound? take the speed of sound in air to be 343 m/s.

Respuesta :

When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength

now we will have

[tex]\frac{\lambda}{2} = \frac{yd}{L}[/tex]

now we know that

y = 0.20 m

d = 2.4 m

L = 10 m

now here we have

[tex]\frac{\lambda}{2} = \frac{0.20\times 2.4}{10}[/tex]

[tex]\lambda = 0.096 m[/tex]

now frequency of wave is given as

[tex]f = \frac{v}{\lambda}[/tex]

[tex]f = \frac{343}{0.096} = 3573 Hz[/tex]

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