A jet engine produces a sound level of 150 dB from a distance of 30 m (assuming
spherical dissipation). The average human ear has an effective area of 12.0 cm2.
What is the sound level at 60 m from a jet engine is?
What would the power intercepted by an unprotected ear at 60 m
from a jet engine be?
To find the sound level at 60 m from a jet engine, we can use the inverse square law, which states that the intensity of sound decreases as the square of the distance from the source increases. The formula for calculating the sound level in decibels is:
L2 = L1 + 20 * log10(r2/r1)
Where L1 is the initial sound level, L2 is the final sound level, r1 is the initial distance, and r2 is the final distance.
Given that the initial sound level (L1) is 150 dB and the initial distance (r1) is 30 m, we can plug these values into the formula:
L2 = 150 + 20 * log10(60/30)
L2 = 150 + 20 * log10(2)
Using a calculator, we can find that log10(2) is approximately 0.3010:
L2 = 150 + 20 * 0.3010
L2 = 150 + 6.02
L2 ≈ 156.02 dB
Therefore, the sound level at 60 m from the jet engine is approximately 156.02 dB.
To calculate the power intercepted by an unprotected ear at 60 m from the jet engine, we need to use the formula for intensity:
I = P/A
Where I is the intensity, P is the power, and A is the effective area of the ear.
Given that the effective area of the ear (A) is 12.0 cm^2, we need to convert it to square meters:
A = 12.0 cm^2 = 12.0 * 10^-4 m^2 = 1.2 * 10^-3 m^2
We can now calculate the intensity:
I = P / 1.2 * 10^-3
To solve for P, we need to rearrange the formula:
P = I * 1.2 * 10^-3
Since we know the sound level (L2) at 60 m, we can find the intensity (I) using the formula:
I = (10^ (L2 / 10)) / (4πr2^2)
Where r2 is the final distance (60 m).
Plugging in the values, we have:
I = (10^ (156.02 / 10)) / (4π * 60^2)
Using a calculator, we find that (10^ (156.02 / 10)) ≈ 1.58*10^15.
I ≈ 1.58*10^10 / (4π * 3600)
I ≈ 1.39*10^7 W/m^2
Now we can calculate the power intercepted by the unprotected ear:
P = 1.39*10^7 * 1.2 * 10^-3
P ≈ 1.67*10^4 W
Therefore, the power intercepted by an unprotected ear at 60 m from the jet engine would be approximately 1.67*10^4 watts.