A worker is exposed to 98 dB for five hours and 82 dB for a three hours, giving an eight hour working day. On average what noise level is this worker exposed to? (round to the nearest whole number)
first exposure = 98(5) dBhours
2nd exposure = 82(3) dBhours
average exposure = (490+246)dBhours/8 hours
= 92 dB
this would make "mathematical" sense, I don't know if it works that way in reality
Well, let me calculate that for you, but first let me put on my noise-canceling wig. *Puts on a brightly colored wig* Okay, here we go! *Pulls out calculator*
If the worker is exposed to 98 dB for 5 hours and 82 dB for 3 hours, we can calculate the average noise level by adding up the noise exposures and dividing by the total time:
(98 dB × 5 hours) + (82 dB × 3 hours) = 490 + 246 = 736
So, the worker is exposed to a total of 736 decibels over the 8-hour working day.
To find the average, we now divide by the total time:
736 dB ÷ 8 hours = 92 dB
So, on average, this worker is exposed to approximately 92 decibels. And remember, laughter is always the best noise-canceller! 🤡🎉
To calculate the average noise level the worker is exposed to, we need to calculate the equivalent continuous sound level (Leq). Leq is calculated using the following formula:
Leq = 10 × log10(1/T × ∑10(L/10))
Where:
T = Total time of exposure (in hours)
L = Sound level (in dB)
Let's calculate the equivalent continuous sound level for the given exposures:
For the 98 dB exposure:
Leq1 = 10 × log10(1/5 × 10^(98/10))
= 10 × log10(0.2 × 10^9.8)
= 10 × log10(0.2 × 63095734.62)
= 10 × log10(12619146.92)
≈ 10 × 7.1001
≈ 71.001
For the 82 dB exposure:
Leq2 = 10 × log10(1/3 × 10^(82/10))
= 10 × log10(0.333 × 10^8.2)
= 10 × log10(0.333 × 158489.319)
= 10 × log10(52811.999)
≈ 10 × 4.721
≈ 47.210
Now, we can calculate the total equivalent continuous sound level (Leq-total) for both exposures:
Leq-total = (Leq1 × T1 + Leq2 × T2) / T
= (71.001 × 5 + 47.210 × 3) / 8
= (355.005 + 141.63) / 8
= 496.635 / 8
≈ 62.079
Therefore, on average, the worker is exposed to approximately 62 dB of noise.
To find the average noise level, we need to calculate the total noise exposure over the entire eight-hour working day.
First, let's convert the decibel values to their corresponding sound pressure levels (SPL) using the formula:
SPL = 10 * log10(P / P0)
Where:
- SPL is the sound pressure level in decibels (dB)
- P is the sound pressure in pascals (Pa)
- P0 is the reference sound pressure, typically set to 20 micro pascals (0.00002 Pa)
Using this formula, we can calculate the sound pressure for each duration:
For the exposure to 98 dB for five hours:
SPL1 = 98 dB
For the exposure to 82 dB for three hours:
SPL2 = 82 dB
Now, let's calculate the corresponding sound pressure levels (SPL) for each duration:
P1 = P0 * 10^(SPL1 / 10) = 0.00002 * 10^(98 / 10) Pa
P2 = P0 * 10^(SPL2 / 10) = 0.00002 * 10^(82 / 10) Pa
Next, we will calculate the total sound pressure over the entire eight-hour working day:
TotalP = (5 hours * P1) + (3 hours * P2)
Finally, we'll convert the total sound pressure back into decibels to find the average noise level:
AverageSPL = 10 * log10(TotalP / (P0 * 8))
Round the AverageSPL to the nearest whole number to get the final answer.
Note: The reference sound pressure (P0) value used in the calculation can vary depending on the context. In this case, we used the standard reference sound pressure value of 20 micro pascals (0.00002 Pa).