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. Calculate the INCREASE in the sound level from the ambient work environment level (in dB) Compute the intensities from the levels, add them to get the total intensity, then find the total sound level. Note the question asks for the increase. 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-Hours/day intensity limits from the graph?

Explanation / Answer

Q1.

let intensity of ambient sound level is I1.

==>10*log10(I1/I0)=85

==>I1=I0*10^8.5...(1)

intensity of boom box be I2.

==>10*log10(I2/I0)=86

==>I2=I0*10^8.6...(2)

total intensity=I1+i2

=I0*(10^8.5+10^8.6)

=I0*10^8.8539

then decibel level=10*log10(I0*10^8.8539/I0)

=88.539 dB

hence increase=88.539-85=3.539 dB

question 2:

let intensity corresponding to 112 dB is I1.

==>10*log10(I1/I0)=112

==>I1=10^11.2*I0...(3)

from the graph, 1 hour/day intensity level is 99.5 dB

if intensity of this level is I2,

==>10*log10(I2/I0)=99.5

==>I2=I0*10^9.95...(4)

dividing equation 3 by equation 4

I1/I2=10^1.25=17.7828

hence the sound intensity exceed the 1 hours/day intensity limit by 17.7828 times

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