Question

A speaker at an open-air concert emits 600 W of sound power, radiated equally in all directions.

Answer 1

Answer

Given,

Sound Power, P = 600 W

a) Intensity of sound,

    r = 5 m

   we know,

     $I = \dfrac{P}{A}$

     $I = \dfrac{P}{4\pi r^2}$

     $I = \dfrac{600}{4\pi\times 5^2}$

            I = 1.91 W/m²

b) Sound intensity

  $\beta = 10 log(\dfrac{I}{I_0})$

       I₀ = 10⁻¹²

  $\beta = 10 log(\dfrac{1.91}{10^{-12}})$

  $\beta = 123\ dB$

c) Sound experienced by the Phil

    β₁ = 123 - 23 = 100 \ dB

distance from the sound

again using sound intensity formula

  $\beta = 10 log(\dfrac{I}{I_0})$

  $100= 10 log(\dfrac{I}{10^{-12}})$

  $I = 10^{-12}\times 10^{10}$

where, I = P/A

  $\dfrac{P}{4\pi r^2} = 10^{-2}$

  $r^2 = \dfrac{600}{0.04\times \pi}$

        r = 69.09 m

Distance where Phil will experience same sound intensity without earplug is equal to r = 69.09 m