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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.

Respuesta :

Answer:

a)   I = 3.63 W / m² , b)   I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

         I = P / A = ½ ρ v (2π f [tex]s_{max}[/tex])²

         I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

        cte = (½ ρ v 4π² s_{max}²)

         I₀ = cte s² f₀²

        I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

         I = constant (2.20 10³) 2

         I = cte 4.84 10⁶

let's find the relationship of the two quantities

        I / Io = 4.84

        I = 4.84 Io

        I = 4.84 0.750

        I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

        I = cte (f s)²

        I = constant (0.250 10³ 4)²

 

        I = cte 1 10⁶

         

the relationship

        I / Io = 1

        I = Io

        I = 0.750 W / m²

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