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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 HzHz. You are 8.00 mm from speaker A. Take the speed of sound in air to be 344 m/sm/s. What is the closest you can be to speaker B and be at a point of perfectly destructive interference?

Respuesta :

Answer:

1.008 m

Explanation:

Wavelength of sound emitted by speakers

= velocity / frequency

= 344 / 172

= 2 m

For destructive interference , path difference should be equal to odd multiple of  half the wavelength .

path diff = 2 / 2 = 1 m

d₂ - d₁ = 1 m

d₂ = d₁ + 1 m

= .008 m + 1

= 1.008 m

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