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 point P is located 5.4 m from one loudspeaker and 4.6 m from the other. The speed of sound is 344 m/s.

The answer to the question is 2.2khz

Explanation:

Let z₁ = 5.4m

Let z₂ = 4.6m

The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m

For the interference= Δz λ, 2λ, 3λ

The wavelength λ = 0.8m

The speed of sound v = 344m/s

The frequency f = v/λ = 344/0.8 = 430hz

Now,

f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,

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

f5 = 2120Hz = 2.200Hz

we will convert to two significant figures =2.2kHz

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