1. Mar 3, 2008

### Shiina-kun

1. The problem statement, all variables and given/known data
The sound intensity level 5.0m from a point source is 95 dB. At what distance will it be 75 dB?

2. Relevant equations
dB = 10log(I/Io)
I = P/4*pi*r^2

3. The attempt at a solution
I honestly don't know what to do with this problem. I started out by plugging 5.0m and 95 dB into the equation >>

dB = 10log((P/4*pi*r^2)/Io)
>> 95 = 10log((P/4*pi*25)/1*10^-12)
>> 95 =10log(P/4*pi*25)-10log(1*10^-12)
>> 95 + 10log(1*10^-12) = 10log(P/4*pi*25)
>> -25 = 10log(P)-10log(4*pi*25)
>> -25+10log(4*pi*25) = 10log(P)
>> -.0307 = 10log(P)
>> -.00307 = log(P)
>> P = 10^-.00307

After that, I used the P that I found in the equation >>

75 = 10log((10^-.00307/4*pi*r^2)/1*10^-12)
>> 75 + 10log(1*10^-12)=10log(10^-.00307/4*pi*r^2)
>> -45 = 10log(10^-.00307)-10log(4*pi*r^2)
>> -44.95 =-10log(4*pi)+10log(r^2)
>> -33.96 = 10log(r^2)
>> -3.39 = log(r^2)
>> r^2 = 10^-3.39
>> r = .0200

This doesn't make any sense at all, but I don't know what I'm doing wrong. Our physics teacher didn't teach us the lesson on this. He just gave us a bunch of formulas and told us to figure it out. Am I using the wrong formulas, or am I making a stupid math error? Please help!!!

2. Mar 3, 2008

deleted

3. Mar 3, 2008

### mysqlpress

r=50m
95=10log(I/Io)
75=10log(I'/Io)
20=10log(I/I') - log properties
2=log(I/I')
100=I/I'
I=100I'
I/I' = r^2/R^2
100=r^2/25
r= 50

in fact 10dB = 10 times , 20dB = 10^2 times.
=.= my unlucky day
always got mistakes, haha~

4. Mar 3, 2008

### Shiina-kun

Thank you so much for your help!

So... let me see if I understand this...
Instead of solving for P, you get the difference between the decibels and solve for r using the first distance given (R=5). Is that right?

5. Mar 3, 2008

### mysqlpress

Yes, that's.
A better approach is to "delete" as much "constants" as possible,
it makes the calculation easier..

6. Mar 3, 2008

### Shiina-kun

er... one more question...

would I add 25 to 50? r=50, which equals the difference between the two distances, right?

7. Mar 3, 2008

### mysqlpress

No

at distance = 5m 95dB
at distance = 50m 75dB
difference between two distance =50-5 = 45m

8. Mar 3, 2008

### Shiina-kun

Oh! I get it now! Thank you!!! ^^