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.

a) How much energy is absorbed by the eardrum per second?

b) At this rate, how long would it take the eardrum to receive a total energy of 1.0 J?

Answer 1

Intensity level B = 10*log(I/Io) = 59 dB

Io = ntensity of the reference level = 10^-12 W/m^2

10*log(I/10^-12) = 59

I = 7.94*10^-7 W/m^2

(a)

energy absorbed by the ear drum per second P = I*A

A = area of ear drum

P = 7.94*10^-7*5*10^-5

P = 3.97*10^-11 W

------------------------------

b)

amount of energy asorbed E = P*t

t = time for which the energy is absorbed

E = energy absorbed

time t = E/P

time t = 1/(3.97*10^-11) = 0.252*10^11 s = (0.252*10^11)/(3.156*10^7) yr

time t = 798.5 yrs