(Sound Intensity) = (Energy)/[(Area)*(Time)]
It can be converted to decibels using a logarithmic formula, if that is what they want.
A microphone has an area of 3.8 cm2. It receives during a 2.87 s time period a sound energy of 4.2x10-11 J. What is the intensity of the sound?
3 answers
How would I do that?
Do the indicated division first, to get 3.85*10^-8 W/m^2
Zero dB corresponds to 10^-12 W/m^2
Each factor of ten in intensity above the zero dB reference level corresponds to a 10 dB increase.
I get 46 dB for your case
See http://www.animations.physics.unsw.edu.au/jw/dB.htm
for a better explanation
Zero dB corresponds to 10^-12 W/m^2
Each factor of ten in intensity above the zero dB reference level corresponds to a 10 dB increase.
I get 46 dB for your case
See http://www.animations.physics.unsw.edu.au/jw/dB.htm
for a better explanation
0 answers