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 88.0 dB. Calculate the INCREASE in the sound level from the ambient work environment level (in dB). 4.76 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 116 dB. By what factor does that sound intensity exceed the 2- Hours/day intensity limits from the graph? 126x10^2

Explanation / Answer

Note that

I = 10^[(dB/10) - 12] W/m^2

Thus, the work sound is

Iwork = 3.1622E-4 W/m^2
Iboom = 6.3095E-4 W/m^2

Thus,

Itot = 9.47185E-4 W/m^2

Thus, converting this to dB,

dB = 10 log (I / [10^-12 W/m^2])

dBtot = 89.76 dB

Thus, the increase is

dBtot - dBwork = 4.76 dB   [ANSWER]

********************

For 116 dB, the intensity is

Irock = 0.3981 W/m^2

For 2 hours a day, the dB level is

dB2hrs = 95 dB

which has intensity

I2hrs = 3.1622E-3 W/m^2

Thus,

Irock / I2hrs = 126   [ANSWER]

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