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.

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 118 dB. By what factor does that sound intensity exceed the 2- Hours/day intensity limits from the graph?

Explanation / Answer

(A)

use the equation B=10log(II_0)

I_0=10^(-12) W/m^2.

Let B_1 and B_2 correspond to the ambient sound level and the sound level of the boom box.

Find I_1 and I_2 from B_1=10log(I_1I_0) and B_2=10log(I_2I_0).

B_1=10log(I_1I_0)

85 = 10log(I_1/10^-12)

=> I_1 = 3.16 x 10^-4

B_2=10log(I_2I_0)

82 = 10log(I_2/10^-12)

=> I_2 = 1.58 x 10^-4

Add I_1 and I_2 together, let's denote this I_3.

I_3 = 4.74 x 10^-4

Solve for B_3,

B_3=10log(I_3I_0)

= 10log(4.74 x 10^-4/10^-12)

= 86.76

B_3-B_1=INCREASE= 86.76 – 85 = 1.76 (answer to part A)

(B)

2 hrs per day corresponds to 95 dB from the graph.

So the question becomes " what is the ratio of the intensities of 118dB to 95 dB"

= 10 ^ 2.3    ( 118 -95 = 23 dB = 2.3 Bell )

or 199.5 times greater than the 2 hr per day limit. (answer to part B)

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