00:01
Our question says, at a rock concert, a decibel meter registers 130 decibels when placed 2 .2 meters in front of a loudspeaker on the stage.
00:10
A, what was the power output of the speaker, assuming uniformed spherical spreading of the sound and neglecting absorption in the air? b, how far away would the sound level be a somewhat reasonable 35 decibels? okay.
00:23
Well, i wrote out what we were given.
00:24
We were given that the best decibels, which i have here is beta, 130.
00:28
And the distance is 2 .2 meters.
00:32
I wrote that as r.
00:34
Okay, so we're going to use the definition of decibels, which says that beta is equal to 10 times the log.
00:44
This is log base 10 of the intensity, which the intensity we can use to find the power divided by i0.
00:52
All right, where i not is just a constant.
00:56
Okay, so we know i not.
00:57
It's just a constant.
00:58
We know beta.
00:59
So we can find i from here.
01:01
So if we divide each side by 10, we'll have beta over 10 is equal to log i over i0, and then we raise each side to the 10th power to get rid of the log because it's log -based 10.
01:12
So that's going to give us i over i0 is equal to 10 to the beta over 10.
01:30
Then solving for i, we have i is equal to i0 beta over 10.
01:47
Well, i not is equal to, so if we plug in the values that we have, we find that i is equal to i not, which is 1 .0, which is 1 .0 times 10 to the minus 12 watts per meter squared.
02:08
It's 12.
02:13
Okay.
02:14
Multiplied by, oh, i'm sorry, i forgot up here to write a 10.
02:17
So this is supposed to be 10 to the beta over 10, not i not to the beta over 10.
02:21
So let's fix that just for clarity here.
02:23
Let's put the 10 here and then put the i -not on the other side just to make this simpler.
02:31
So this is 10 to the beta over 10 times i -0...