When wearing proper ear protection, a jack hammer operator experiences a sound intensity of 1x10^-6 W/m2. The ear protection purpots to reduce the sound intensity by 100 times. What is the actual sound level of the jack hammer?
I = 10^-4 W/m^2
10 log(10^-4/10^-12)
10 log 10^8
80 log 10
80 dB
between busy street traffic and a subway train
To determine the actual sound level of the jack hammer, we need to first understand the concept of sound level. Sound level is measured in decibels (dB) and is a logarithmic scale that compares the sound intensity to a reference level. Mathematically, it can be calculated using the formula:
Sound level (dB) = 10 * log10 (I/I0)
Where I is the sound intensity being measured and I0 is the reference sound intensity.
Given that the operator experiences a sound intensity of 1x10^-6 W/m2 and the ear protection claims to reduce the sound intensity by 100 times, we can calculate the actual sound intensity by dividing the initial sound intensity by 100.
Actual sound intensity = 1x10^-6 W/m2 / 100 = 1x10^-8 W/m2
Now, let's calculate the sound level using the formula mentioned earlier.
Sound level (dB) = 10 * log10 (1x10^-8 / I0)
However, we still need the reference sound intensity (I0) to complete the calculation. In sound level measurements, the reference intensity is often taken as the threshold of human hearing, which is approximately I0 = 1x10^-12 W/m2.
Substituting this value into the formula, we get:
Sound level (dB) = 10 * log10 (1x10^-8 / 1x10^-12)
Simplifying the expression:
Sound level (dB) = 10 * log10 (1x10^4)
Using logarithmic rules, log10 (1x10^4) = 4, so we can further simplify:
Sound level (dB) = 10 * 4 = 40 dB
Therefore, the actual sound level of the jack hammer, when wearing the recommended ear protection, is 40 decibels.