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.

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

Explanation / Answer

A) Take as new sound level- I' ; ambient sound level- Ib

in decibel, 86=10log(I'/Io) : I'=3.98x108 Io ---(1)

85= 10log(Ib/Io) ; Ib=3.16 x 108 Io---(2)

; (1)-(2) ; dI =8.2 x 107 Io

increase in dB = 10log(dI/Io) =69.14 dB

B) intensity in decibels = 10log(I/Io)

2 hours per day intensity(I1) ; 100=10*log(I1/Io) ---(1) {there should be the plot}

for the speaker system, 114=10log(I2/Io)------(2)

(2)-(1) ; 10log(I2/I1) = 114-100

I2=(25.12 )* I1

factor=25.12

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