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?
(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.
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
To calculate the energy absorbed by the eardrum, we need to use the formula for sound energy:
Energy = Power × Time
First, let's find the power absorbed by the eardrum. The power of sound is given by the formula:
Power = 10^(L/10)
Where L is the sound level in decibels (dB). In this case, the sound level is 90 dB, so the power can be calculated as:
Power = 10^(90/10) = 10^9
The power is then 10^9 times the reference power level of 0 dB, which is approximately 1 watt.
Now, we can calculate the energy absorbed by the eardrum over 2 hours. However, we need to convert the time from hours to seconds, as power is usually measured in watts per second (joules).
2 hours = 2 × 60 × 60 = 7200 seconds
Finally, we can calculate the energy absorbed:
Energy = Power × Time
Energy = 1 watt × 7200 seconds = 7200 joules
Therefore, the eardrum absorbs approximately 7200 joules of energy over 2 hours.