At a distance of 30 m the noise from the engine of an jet has an intensity level of 130 dB.
At this level, you will be in pain and your ears will hurt. That's why this intensity is know as the "pain threshold".
How far do you have to be from a jet (total distance in meters), in order for the noise to drop down in intensity to 69 dB, a level comparable to that of a spoken conversation?
A decrease in sound level by 61 dB amounts to a factor of 10^6.1 = 1.26^10^6 in sound power per area. To achieve this, you need to be sqrt(1.25*10^6) = 1122 times farther away. That corresonds to 30*1122 = 33,660 m away. (about 21 miles)
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To determine the distance at which the noise intensity drops to 69 dB, we can use the inverse square law for sound propagation. According to the inverse square law, the sound intensity decreases by 6 dB for every doubling of the distance from the source.
Let's calculate:
1. Calculate the difference in intensity levels between the two scenarios:
ΔdB = 130 dB - 69 dB = 61 dB
2. Determine the number of times the sound intensity should be halved to decrease by 61 dB:
n = ΔdB / 6 dB = 61 dB / 6 dB ≈ 10.16
This means that the sound intensity needs to be halved approximately 10 times for it to decrease by 61 dB.
3. Find the distance by which the sound intensity should be halved 10 times from the initial distance of 30m:
Final distance = 30 m * (1/2)^n = 30 m * (1/2)^10 ≈ 0.293 m
So, you need to be approximately 0.293 meters away from the jet for the noise intensity to drop down to 69 dB, comparable to a spoken conversation.
To determine the distance from the jet where the noise drops down to an intensity level of 69 dB, we can use the inverse square law formula. The inverse square law states that the intensity of sound decreases inversely proportional to the square of the distance from the source.
The formula is:
I2 = I1 * (d1/d2)^2
Where:
I1 = initial intensity (130 dB) at distance d1 (30 m)
I2 = final intensity (69 dB) at distance d2 (unknown)
d1 = initial distance (30 m)
d2 = final distance (unknown)
Rearranging the formula to solve for d2:
d2 = √((I1/I2) * d1^2)
Substituting the given values:
I1 = 130 dB
I2 = 69 dB
d1 = 30 m
d2 = √((130/69) * 30^2)
d2 ≈ √(230.435 * 900)
d2 ≈ √207391.5
d2 ≈ 454.9 m
Therefore, you would need to be approximately 454.9 meters away from the jet for the noise intensity to drop down to 69 dB, comparable to the level of a spoken conversation.