By how many decibels does the sound intensity from a point source decrease if you increase the distance to it by a factor 6?
Surface area is proportional to distance^2
so if you increase distance by 6, then sound goes down by a factor of 1/36.
db= 10log(1/6)
so db goes down by a factor of 7.78
so if the original were 40 db, the new is 32.22 db
To determine the decrease in sound intensity in decibels as the distance to a point source increases by a factor of 6, we need to use the inverse square law formula. According to the inverse square law, sound intensity decreases with the square of the distance.
First, let's understand the relationship between sound intensity and decibels. The sound intensity level in decibels (dB) can be calculated using the formula:
L = 10 * log10(I/I0)
Where:
L = sound intensity level in decibels (dB)
I = sound intensity
I0 = reference sound intensity (usually taken as the threshold of hearing, which is 10^(-12) watts per square meter)
Since we want to find the decrease in sound intensity, we can compare the initial sound intensity (I1) to the final sound intensity (I2) after the distance is increased by a factor of 6.
According to the inverse square law, the sound intensity is inversely proportional to the square of the distance. Mathematically, this can be expressed as:
I2 = (d1^2 / d2^2) * I1
Where:
I2 = final sound intensity
I1 = initial sound intensity
d1 = initial distance
d2 = final distance
Given that the initial distance (d1) is 1 unit, and the final distance (d2) is 6 units (since it increases by a factor of 6), we can substitute these values into the equation:
I2 = (1^2 / 6^2) * I1
I2 = (1/36) * I1
Now, to find the decrease in decibels, we will calculate the ratio of the final sound intensity to the initial sound intensity:
dB decrease = 10 * log10(I2/I1)
dB decrease = 10 * log10((1/36) * I1 / I1)
dB decrease = 10 * log10(1/36)
dB decrease = 10 * log10(1) - log10(36)
dB decrease = 10 * 0 - log10(36)
dB decrease = 0 - log10(36)
dB decrease ≈ -17.8 dB
Therefore, the sound intensity from a point source will decrease by approximately 17.8 decibels if you increase the distance to it by a factor of 6.