Decibels and intensity of police car siren

[Aphrael]

I am struggling with understanding the difference between decibels and intensity. I know that intensity can be measure in decibels but they are not always the same thing.

I have a problem where there is a police car 200 meters from an accident runing sirens at 80 decibels. I am asked to find how many decibels higher the sound intensity will be when the police car is only 50 meters away. Then I am asked to find how many cars are needed to match the sound of a rock concert (120 dB).

I tried using (I1/I2)=(R2/R1)^2 but I ended up getting 1200 dB which I knew could not be correct.

What should I do?

 

[mathman]

Decibels use a logorithmic scale. 10 decibel increment means intensity increased by a factor of 10.

 

[FredGarvin]

You have the right idea in thinking about the inverse square relationship. You just didn't put it into a mathematical statement properly. If the distance is doubled, the intensity goes down by \frac{1}{2^2}. If the distance is tripled, the intensity goes down by \frac{1}{3^2}, etc...

Does that help?

EDIT: You'll need the hint from mathman too.

 

[Aphrael]

Okay I'm more confused now. I know about about the logorithmic scale thing. But I also know that since dB=10 log(I2/I1) Going down 3dB is equivalent to decreasing by a factor of two since 10 log(2)=dB. I also know that every 10dB the intensity doubles. So I don't understand how 10dB increases the intensity by 10. Please don't think I'm trying to say your wrong. I am just confused and want to show you my train of thought so you can tell me where I am wrong.

Also with the doubling, tripling distance thing...still confused. Are you saying I need to find the ratio between my two distances and then the total "R" equals the ratio squared?

 

[FredGarvin]

Perhaps you can take a look at this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/roomi.html#c1

Scroll down to where it talks about doubling the distance. You'll notice in your problem that you are doubling the distance twice, i.e. 50 to 100 ft and then 100 ft to 200 ft. For every doubling of the distance, the intensity decreases by 6 db. So your problem is simply going the other way. Instead of subtracting dB's from the source, you are adding.

Double check your calcs here:
http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/isprob.html#c3