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 acoustic output of 63 ȝW. 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

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

distance (r) = 210 m

intensity (I) = power/area

where we assume the area of a sphere due to the uniformity of the output in all directions

area = 4π$r^{2}$ =  4π x $210^{2}$ = 554,176.9 m^{2}

intensity (I) = $\frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}$

therefore the desired ratio I/I0 = $\frac{113.7 x 10^{-12}}{1 x 10^{-12}}$ = 113.7