(Power/area)*(area)*(time)= Energy
Convert 90 dB to power/area. Use SI units. See
http://www.sengpielaudio.com/TableOfSoundPressureLevels.htm
if you need help with that.
A 90-dB sound is absorbed by an eardrum 0.75 cm in diameter for 2 hours. How much energy in joules does the eardrum absorb at that time?
2 answers
I figured it out already, thanks. I was just having trouble with converting dB to W/m^2.
Anyway, here's my answer. Feel free to correct it.
I - Intensity
P - Power
E - Energy
t - time
A - Area
I = P/A ; I = E/t / A ; E = IAt
Io = 1x10^-12 W/m^2 (treshold of hearing)
Conversion of dB to W/m^2 :
B = 10 log I/Io
90 = 10log I/Io
9 = log I/Io
I = 10^9 Io
I = 10^9 (10^-12)
I = 1x10^-3 W/m^2
E = IAt
= 1x10^-3 W/m^2 [1/4 pi (0.0075m)^2] (7200 sec)
= 3.18086 x 10^-4 Joules
Anyway, here's my answer. Feel free to correct it.
I - Intensity
P - Power
E - Energy
t - time
A - Area
I = P/A ; I = E/t / A ; E = IAt
Io = 1x10^-12 W/m^2 (treshold of hearing)
Conversion of dB to W/m^2 :
B = 10 log I/Io
90 = 10log I/Io
9 = log I/Io
I = 10^9 Io
I = 10^9 (10^-12)
I = 1x10^-3 W/m^2
E = IAt
= 1x10^-3 W/m^2 [1/4 pi (0.0075m)^2] (7200 sec)
= 3.18086 x 10^-4 Joules
2 answers