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.

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 1.5- Hours/day intensity limits from the graph?

Answer 1

a )

addition of non coherent sounds is

10 log ( 10^8.5 + 10^8.6 ) = 88.54 dB

b )

given and as have to calculate at 1.5 hours/day sound level is 97.1 dB ( 0.005012 W/m² )

now

112 - 97.1 = 14.9 dB

10^1.49 = 30.90