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 86.0 dB.

1) Calculate the INCREASE in the sound level from the ambient work environment level (in dB).

2) 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 2- Hours/day intensity limits from the graph?

Explanation / Answer

(a). The ambient sound level is of 85 dB. 8 Hours/day noise level

Boom Box at an average level of 86.0 dB

Lets turn the dB levels in powers of 10. To do this simply divide the dB level by 10.

85 = 10^(8.5)
86 = 10^(8.6)

The sound level due to ambient work enviroment and music is

10log(10^(8.5) + 10^(8.6)) = 88.539 dB

Therefore, the increase in sound level from the ambient work environment level (in dB).

88.539 - 85 = 3.539 dB i.e. 3.54 rounded (2dp)

b.

The sound intensity for 2- Hours/day from the graph=95dB

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.

(B). 2 hours/day sound level is 95 dB (0.003162 W/m

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