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.
3 answers
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.
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
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
0 answers