Question

A 59-db sound wave strikes an eardrum whose area is 5.0×10−5m2. the intensity of the reference level required to determine the sound level is 1.0×10−12w/m2. 1yr=3.156×107s. part a how much energy is absorbed by the eardrum per second?

Answer 1

Sound intensity is defined as the power carried by sound waves per unit area in a direction perpendicular to that area. The formula for sound intensity level in decibels is:
$L=10log_{10}\frac{I}{I_0}dB$
Using this formula we can find sound intensity:
$59=10log_{10}\frac{I}{10^{-12}}\\ 10^{5.9}=I\cdot 10^{12}\\ I=\frac{10^{5.9}}{10^{12}}\\ I=7.94\cdot10^{-7}\frac{W}{m^2}$
Now we just multiply this number by the area to get the energy absorbed per second:
$P=I\cdot A=7.94\cdot10^{-7}\cdot 5\cdot 10^{-5}=3.97\cdot 10^{-11} W$