A stereo speaker is placed between two observers who are 35 m apart, along the line connecting them. If one observer records an intensity level of 64 dB, and the other records an intensity level of 85 dB, how far is the speaker from each observer

   x = 2,864 m ,       Ra = 32.1 m                       

Explanation:

Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer

observer A β = 64 db

             β = 10 log Iₐ / I₀

where I₀ = 1 10⁻¹² W / m²

              Iₐ = I₀ 10 (β/ 10)

let's calculate

              Iₐ = 1 10⁻¹² (64/10)

              Iₐ = 2.51 10⁻⁶ W / m²

Observer B β = 85 db

             I_b = 1 10-12 10 (85/10)

             I_b = 3.16 10⁻⁴ W / m²

now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed

              P = I A

therefore for the two observers

              P = Ia Aa = Ib Ab

the area of a sphere is

               A = 4π R²

we substitute

               Ia 4pi Ra2 = Ib 4pi Rb2

               Ia Ra2 = Ib Rb2

Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute

             Ia (35 -x) 2 = Ib x2

we develop and solve

           35-x = Ra (Ib / Ia) x

           35 = [Ra (Ib / Ia) +1] x

           x (11.22 +1) = 35

           x = 35 / 12.22

            x = 2,864 m

This is the distance of observer B

The distance from observer A

            Ra = 35 - x

            Ra = 35 - 2,864

            Ra = 32.1 m

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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