Asked by Anonymous
A source of sound (1000 Hz) emits uniformly in all directions. An observer 3.19 m from the source measures a sound level of 39.2 dB. Calculate the average power output of the source.
Answers
Answered by
drwls
Use the decibel level definition to compute the intensity in units of Watts per area. Multiply that number by 4 pi R^2 (the area of a sphere of radius R) to get the total power output.
Answered by
Anonymous
I = 39.2/1.0 x 10^-12
= 3.92 x 10^-11 W/m^2 x 4 pi (3.19^2m^2)
= 5.01e-9 W
i got the wrong answer
= 3.92 x 10^-11 W/m^2 x 4 pi (3.19^2m^2)
= 5.01e-9 W
i got the wrong answer
Answered by
drwls
You did not use the correct logarithmic formula that relates decibels to intensity.
To review it, see:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/sound/u11l2b.html
At 39.2 dB, the sound intensity level is 10^3.92 = 8318 times above the reference (hearing threshold) value of 10^-12 W/m^2. So the value of intensity you must use is 8.32*10^-9 W/m^2
To review it, see:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/sound/u11l2b.html
At 39.2 dB, the sound intensity level is 10^3.92 = 8318 times above the reference (hearing threshold) value of 10^-12 W/m^2. So the value of intensity you must use is 8.32*10^-9 W/m^2
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