(b) A student at a rock concert has a seat close to the speaker system, where th
ID: 2035023 • Letter: #
Question
(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 118 dB. By what factor does that sound intensity exceed the 2- Hours/day intensity limits from the graph?
Please indicate which answer is A and B
nu40@msu.edu Exam 3 Notes Google Docs LON-CAPA Standing So E Chegg Study TEXTBOOK SOLUTIONS EXPERT Os The graph shows the US Department of Labor noise regulation for workingv machinist is in an environment where the ambient sound level is of 85 dB, i.e Hours/day noise level. The machinist likes to listen to music, and plays a Boor 84.0 dB. 4 0 85 90 95 100 105 Sound Lavel (dB) (A) Calculate the INCREASE in the sound level from the ambient work environm (b) A student at a rock concert has a seat close to the speaker system, where the corresponds to a sound level of 118 dB. By what factor does that sound intensity intensity limits from the graph? Please indicate which answer is A and B Hide comments (1)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)
84 = 10log(I_2/10^-12)
=> I_2 = 2.51 x 10^-4
Add I_1 and I_2 together, let's denote this I_3.
I_3 = 5.67 x 10^-4
Solve for B_3,
B_3=10log(I_3I_0)
= 10log(5.67 x 10^-4/10^-12)
= 87.5
B_3-B_1=INCREASE= 87.5 – 85 = 2.54 (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)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.