Question

A 50-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.

Answer 1

Sound Level of 50 dB is given to us

So the intensity of sound for this level of sound is given as

$L = 10 Log \frac{I}{I_0}$

now we will have

$L = 50 dB$

$50 = 10 Log\frac{I}{1 \times 10^{-12}}$

by solving above equation we will have

$I = 10^{-7} W/m^2$

now energy received by ear drum per second will be

$P = intensity \times area$

$P = 10^{-7} \times 5 \times 10^{-5} = 5 \times 10^{-12} W$

Now total energy received in one year will be

$E = power \times time$

$E = 5 \times 10^{-12} \times 3.156 \times 10^7$

$E = 1.6 \times 10^{-4} J$

so above is the energy received by year in one year