Question

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 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 coustic output of 63 µW. What is the ratio of the acoustic intensity produced by the juvenile howler to the reference intensity I0, at a distance of 210 m?

Answer 1

`(I)/(I_0)=113.68`

Step-by-step explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

`P=IA\\\Rightarrow I=(P)/(A)\\\Rightarrow I=(63* 10^(-6))/(4* \pi * 210^2)`

Threshold intensity =
`I_0=1* 10^(-12)\ W/m^2`

Ratio

`(I)/(I_0)=((63* 10^(-6))/(4* \pi * 210^2))/(1* 10^(-12))\\\Rightarrow (I)/(I_0)=113.68`

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68