The howler monkey is the loudest land animal and under some

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8 km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10^-12 W/m². A juvenile howler monkey has an acoustic output of 44 W. What is the ratio of the acoustic intensity produced by the juvenile howler to the reference intensity I₀, at a distance of 310 m?

Solution

This is an exercise of sound propagation, as the output is even, all energy must be distributed in a sphere of given radius

I = W/ A I sound intensity

Data

W = 44 W = 44 10^-6W

R= 310 m

A = 4 R²    A = 4 310²

A= 1.208 10⁶ m²

I = 44 10^-6/ 1.208 10⁶

I= 36.42 10^-12W/m²

Ratio = I /Io

Ratio = 36.42 10^-12/ 1.0 10^-12

Ratio = 36.42   

Your result is 36.

But this overall amount is given in decibels, the expression

= 10 log (Ratio)

= 10 log ( 36.42 )

= 10 1.132

= 11.32 d