By how many decibels does the sound intensity from a point source decrease if you increase the distance to it by a factor 6?
A rule of thumb: About 3 db each time one halves sound intensity. But sound intensity changes as 1/r^2, so, 6 is changing sound intensity by 1/36
db= 3*number doubles
but 8- 2*2*2
and 4=2*2
so, 36 is 8*4
log 36= log8+log4=3+2=5
so,going down by 3 db five times, or -15 db about...
Check: 10 log 1/36= -10 log(36)=down 15.56 db the sound goes down
To calculate the decrease in sound intensity, we can use the inverse square law formula. According to this law, the sound intensity decreases in proportion to the square of the distance from the sound source.
The formula for the inverse square law is:
I2 = I1 * (r1/r2)^2
Where:
I1 is the initial sound intensity
I2 is the final sound intensity
r1 is the initial distance from the sound source
r2 is the final distance from the sound source
In this case, the question states that the distance to the sound source is increased by a factor of 6. So we can assume that r2 = 6 * r1.
Now let's plug the values into the formula:
I2 = I1 * (r1/(6 * r1))^2
Simplifying the equation:
I2 = I1 * (1/6)^2
I2 = I1 * 1/36
I2 = I1 / 36
From the equation, we see that I2 is 1/36 times the original intensity I1. It means the sound intensity decreases by 36 times, which is equivalent to 36 decibels.
Therefore, the sound intensity decreases by 36 decibels when the distance to the point source is increased by a factor of 6.