The graph shows the US Department of Labor noise regulation for working without
ID: 1403891 • Letter: T
Question
The graph shows the US Department of Labor noise regulation for working without 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 114 dB. By what factor does that sound intensity exceed the 1- Hours/day intensity limits from the graph?
Tries 0/12 95 100 105 Sound Level (dB)Explanation / Answer
a)
The sound intensity in dB is defined as 10*log10(I/I0).
where Io = 1*10-12W/m2 (Threshold Intensity)
Calculating Intensity of Each Sound -
85 dB = 10*log10(I/I0).
I = (1*10^-12)*10^8.5 = 3.16 * 10^-4 W/m^2
84 dB = 10*log10(I/I0)
I = (1*10^-12)*10^8.4 = 2.51 * 10^-4 W/m^2
Adding them = 5.67 * 10^-4 W/m^2
Converting them back to dB = 10*log10( 5.67 * 10^-4 / 1 * 10^-12 ) = 87.53 dB
Increase in sound level = 87.53 dB - 85 dB = 2.53 dB
Increase in the sound level from the ambient work environment level = 2.53 dB
b)
From Graph as we can see 1-Hour/Day Intensity Limits Corresponds to 100 Db of Sound.
Level difference = 114 dB - 100 dB = 14 dB.
14dB = 10*log10(I/I0)
(I/I0) = 10^(14/10)
Factor by which sound intensity exceeds = 10 ^ 1.4
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