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 acoustic 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? A) 110 B) 230 C) 76 D) 170 E) 300

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Quizplus · Accepted Answer

The intensity of sound is given by:$$I = \frac{P}{4\pi r^2}$$where P is the acoustic output and r is the distance from the source.The reference intensity is:$$I_0 = 1.0 	imes 10^{-12} 	ext{ W/m}^2$$The ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is:$$\frac{I}{I_0} = \frac{P}{4\pi r^2 I_0}$$Substituting the given values, we get:$$\frac{I}{I_0} = \frac{63 	imes 10^{-6} 	ext{ W}}{4\pi (210 	ext{ m})^2 (1.0 	imes 10^{-12} 	ext{ W/m}^2)} = 110$$Therefore, the correct answer is A.

The Howler Monkey Is the Loudest Land Animal And, Under