Question

A point source emits 50.0 W of sound isotropically. A small microphone intercepts the sound in an area of 0.30 cm", 85 m from the source. a) What is the the sound intensity at the microphone? b) What is the power intercepted by the microphone? c) What is the sound level in dB? d) How much closer would we need to place the mic for a 3dB gain in sound level? What difference in intensity does this difference correspond to?

Answer 1

Answer:

a)  I = 5,507 10⁻⁴ W / m² , b)    P = 1.65 10⁻⁸ W / m² , c)  R = 4.06 m,   I = 1,259 10⁻⁶ W / m²

Explanation:

A) the intensity is defined as the power per unit area

         I = P / A

The intensity for the radius of the sphere of 85 m is

         A = 4π R²

         A = 4π 85²

         A = 9,079 10⁴ m²

The intensity at this point is

           I = 50.0 / 9.079 10⁴

           I = 5,507 10⁻⁴ W / m²

           

b) The power collected by the microphone

     We reduce the area to the SI system

         a = 0.30 cm² (1 m / 10² cm)² = 0.3 10⁻⁴ m²

         

          P = I A

          P = 5,507 10-4 0.30 10-4

          P = 1.65 10⁻⁸ W / m²

c) Let's reduce dβ

      β = 10 log (I / I₀)

      β = 10 log (5,507 10⁻⁴ / 1 10⁻¹²)

      β = 59 dB

d) if the gain is 3 dB, the total intensity is

     β = 59 + 3

     β = 61 dB

We look for the intensity

       I = I₀ 10  β / 10

        I = 1 10⁻¹² 10 (61/10)

        I = 1,259 10⁻⁶ W / m²

Use

          A₁ I₁ = A₂ I₂

          A = 4pi R2

    We replace

          R₁² I₁ = R₂² I₂

          R₂ = R₁ √ I₁ / I₂

           R = 85 RA (1,259 10⁻⁶ / 5,507 10⁻⁴)

          R = 4.06 m