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LekaFEV [45]
2 years ago
14

Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A po

int P is located 5.4 m from one loudspeaker and 4.6 m from the other. The speed of sound is 344 m/s.
Physics
1 answer:
My name is Ann [436]2 years ago
8 0

Answer:

The answer to the question is 2.2khz

Explanation:

<em>Let z₁ = 5.4m</em>

<em>Let z₂ = 4.6m</em>

<em>The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m</em>

<em>For the interference= Δz λ, 2λ, 3λ......</em>

<em>The wavelength λ = 0.8m</em>

<em>The speed of sound v = 344m/s</em>

<em>The frequency f = v/λ = 344/0.8 = 430hz</em>

<em>Now,</em>

<em>f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,</em>

<em>f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz</em>

<em>f5 = 2120Hz = 2.200Hz </em>

<em>we will convert to two significant figures =2.2kHz</em>

<em> </em>

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<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>
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