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 84.0 dB Calculate the INCREASE in the sound level from the ambient work environment level (in dB). 87 5 Compute the intensities from the levels, add them to get the total intensity, then find the total sound level. A student at a rock concert has a seat close to the speaker system where the average sound intensity corresponds to the sound level of 116 dB By what factor does that sound intensity exceed the 1- Hours/day intensity limits from the graph?

Explanation / Answer

a] i = i1+i2

85 = 10 log (i1/10^-12)

8.5 = 12 + log i1

i1 = 10^(8.5-12) = 0.0003162

i2 = 10^(8.4-12) = 0.0002512

i = 0.0002511+0.0003162 = 0.0005673

sound level = 10* log ( 0.0005673 /10^-12)

= 87.538 dB

increase = 87.538-85

= 2.538 dB answer

b] intensity in dB = 10*log (i/io)

116 = 10*log (i/io)

11.6 = log [i*10^12]

log i = 11.6 -12 = -0.4

i = 10^-0.4

intensity of 1 hr limit = 100 dB

log i2 = 100/10 - 12 = -2

i2 = 10^-2

ratio = i/i2 = 10^-0.4/10^-2

= 39.81 times

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