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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?

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Answer:

[tex]\frac{I}{I_0}=113.68[/tex]

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

[tex]P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}[/tex]

Threshold intensity = [tex]I_0=1\times 10^{-12}\ W/m^2[/tex]

Ratio

[tex]\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68[/tex]

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

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