Question

A howler monkey is the loudest land animal and under some circumstances, can be heard up to a distance of 5.0 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/m^2. The acoustic power emitted by the howler is closest to

a) 3.2 mW b) 0.11 mW c) 11 mW d) 1.1 mW e) .31 mW

Answer 1

The threshold of hearing is

`I=1.0 \cdot 10^(-12)W/m^2`
and in order to hear the howler at a distance of 5 km, its sound should have at least this intensity.

The sound wave is emitted uniformly in all directions, so the area we should consider to calculate the power is the area of a sphere with radius
r=5 km=5000 m:

`A=4 \pi r^2 = 4 \pi (5000 m)^2 =3.14 \cdot 10^8 m^2`

And by using these data, we can calculate the power emitted by the howler:

`P=IA=(1.0 \cdot 10^(-12) W/m^2)(3.14 \cdot 10^8 m^2)=3.1 \cdot 10^(-4) W=0.31 mW`
So, the correct answer is E) 0.31 mW.