At a distance of 7.5 m from a point sound source, the sound intensity level is 105 dB.
At what distance is the intensity level 90 dB?
To find the distance at which the sound intensity level is 90 dB, we need to use the inverse square law for sound intensity.
The inverse square law states that the sound intensity (I) is inversely proportional to the square of the distance (d) from the source. Mathematically, it can be expressed as:
I1 / I2 = (d2 / d1)^2
Where:
I1 = initial sound intensity level (105 dB)
I2 = final sound intensity level (90 dB)
d1 = initial distance (7.5 m)
d2 = final distance (unknown)
Let's substitute the given values into the equation and solve for d2:
(I1 / I2) = (d2 / d1)^2
(105 dB / 90 dB) = (d2 / 7.5 m)^2
(105 / 90) = (d2 / 7.5)^2
1.16 ≈ (d2 / 7.5)^2
Taking the square root of both sides:
sqrt(1.16) ≈ d2 / 7.5
1.08 ≈ d2 / 7.5
Now, let's solve for d2 by multiplying both sides by 7.5:
1.08 * 7.5 ≈ d2
8.1 m ≈ d2
Therefore, at a distance of approximately 8.1 meters from the sound source, the sound intensity level will be 90 dB.