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3 years ago

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Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc

ate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Take the speed of sound in air to be 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter 'none' in all unused answer boxes.)

1 answer:

Juli2301 [7.4K]3 years ago

6 0

Answer:

0.2286 m, 0.686 m and 1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

Δr = (2n +1) λ/2

where r is the distance, λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

v = λ f

λ = v / f

we substitute

Δr = (2n + 1) v / 2f

let's calculate for difference values of n

Δr = (2n +1) 343/(2 750)

Δr = (2n + 1) 0.2286

we locate the different values for a minimum of interim

n Δr (m)

0 0.2286

1 0.686

2 1,143

therefore we see that there is respect even where the intensity is minimal

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