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.

95 100 105 Sound Level (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 116 dB. By what factor does that sound intensity exceed the 2- Hours/day intensity limits from the graph?

Answer 1

a)

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

So, I1= 3.16*10^-4

Similarly, B2 = 86 = 10log(I2/Io)

So, I2 = 3.98*10^-4

Total intensity I3 = I1+ I2= 7.14*10^-4 W/m^2

So, B3= 10log(I3/Io)

So, B3= 88.539 dB

So, the increase in the sound level from the ambient is 88.539 - 85 = 3.539 dB

b)

Now, 2hrs/day level is 95 dB.

So, B1= 95= 10log(I1/Io)

So, I1= 3.16*10^-3

B2= 116= 10log(I2/Io)

So, I2 = 0.398

So, the sound intensity exceeds by the given sound intensity by a factor of

I2/I1 = 0.398/ (3.16*10^-3)

I2/I1 = 125.949