The graph shows the us department of labor noise regulation for working ear prot
ID: 1418721 • Letter: T
Question
The graph shows the us department of labor noise regulation for working ear protection. A machinist is in an environment where the ambient sound level is of 85 db, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a boom box at an average level of 84.0 dB Calculate the INCREASE in the sound level from the ambient work environment level (in dB). A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 112 dB. By what factor does that sound intensity exceed the 1- Hours/day intensity limits from the graph?Explanation / Answer
dE = 10^(dB/10)
dE1 = 10^(85/10) = 3.16*10^8
dE2 = 10^(84/10) = 2.511*10^8
dE = dE1 + dE2 = (3.16 + 2.511)*10^8
dE = 5.671*10^8
dB = 10*log (dE)
dB = 10*log (5.671*10^8)
dB = 87.54 dB
increase in the sound level will be = 87.54 - 85 = 2.54 dB
2.
permissible noise exposure for 1 hours/day = 100 dB
average sound intentsity = 112 dB
difference = 112 - 100 = 12 dB
dE = 10^(dB/10)
dE = 10^(1.2) = 15.85 times greater than permissible
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