US Department of Labor noise requlation for working without ear protection. A ma

Question

US Department of Labor noise requlation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB The graph shows the 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. 85 90 95 100 105 Sound Level (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 Suomit Angwer You have entered that answer before Incorrect. Tries 5/12 Previous Tries 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? Submit Answer Incorrect. Tries 1/12 Previous Tries

Explanation / Answer

Sound level B1= 85= 10log(I1/Io) ; where Io= 10^-12 W/m^2

So, I1= 3.16*10^-4

Similarly, B2= 82= 10log(I2/Io)

So, I2= 1.58*10^-4

Total intensity I3= I1+ I2= 4.74*10^-4 W/m^2

So, B3= 10log(I3/Io)

So, B3= 86.76dB

So, the increase in the sound level from the ambient is 86.76-85= 1.76dB

Now, 2hrs/day level is 95dB. So, B1= 95= 10log(I1/Io)

So, I1= 3.16*10^-3

B2= 118= 10log(I2/Io)

So, I2= 0.631

So, the sound intensity exceeds by the given sound intensity by a factor of I2/I1= 0.631/ 3.16*10^-3= 199.67

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