The graph shows the US Department of Labor noise regulation for working without

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 118 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?

Explanation / Answer

E1 = 10 ^ (85/10)= 3.2 * 10 ^ 8
E2 = 10^(84/10) = 2.4 * 10 ^ 8

E = E1 + E2 = 5.7 * 10 ^ 8
New sound level = log(E) = 87.5 dB

Increase = new - old.= 87.5 - 85 = 2.5 dB increase


For part 2 the 1.5 hour level is 97 dB
118- 97 = 21dB
21 db -> factor of 10 ^ (21/10) = 125.89 times greater than the allowed level.

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