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A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 52 dB by an observer 10 km away. (a) What is the average power generated by the horn? (b) What intensity level of the horn's sound is observed by someone waiting at an intersection 52 m from the train? Treat the horn as a point source and neglect any absorption of sound by the air.

Respuesta :

Answer:

a) 199 W

b) 5.86×10⁻³ ( = 97.7 dB ) W/m²

Explanation:

a) 52 dB is converted to W/m² using  the formula

[tex]\beta(dB) = 10 log(\frac{I}{10^{-12}})[/tex]

Here 10⁻¹² W/m² is the threshold intensity and I is the intensity

52 dB in W/m².

Solving for I from the above formula, the sound intensity in W/m² is

I=52 dB = 1.585×10⁻⁷ W/m².

Average power = Intensity x Area = I (4 π R²)

⇒ P = (1.585×10⁻⁷ )(4π)(10000)² = 199 W.

⇒ Power at a distance of 10 km = 199 W

b) intensity at 10 km be I₁ = 199 W.

 Let Intensity at 52 m = I₂

Intensity is inversely proportional to square of the distance.

I₁ / I₂ = R₂²/R₁²

⇒ I₂ = (R₁²/R₂²) I₁ = (10000²/52²)(1.585×10⁻⁷)

 ⇒Intensity at 52 m distance = 5.86×10⁻³ W/m² (= 97.7 dB)