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 82.0 dB.

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

B. 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 2- Hours/day intensity limits from the graph?

Explanation / Answer

part a )

Boom Box at an average level of 82.0 dB

82 = 10^(8.2)

85 = 10^(8.5)

The sound level due to ambient work environment and music is

10log(10^(8.5) + 10^(8.2)) = 86.76 dB

part b )

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

95 dB = 0.003162 W/m^2

118 dB = 0.630957 W/m^2

ratio = 199

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.