The decibel level beta is given by:
beta = 10log_10(I/I_0);
where I is the itensity at a given distance r from the sound source, and I_0 = 10^-12 W/m^².
Now, assuming a spherical sound wave, the itensity is dependent on the distance from the source according to the relation
I = P/(4pir^2),
where P indicates the source's power.
When the distance increases 190 times (that is, going from 1m to 190m), the intensity will decrease by 190^2; that is:
I_2 = I_1/190^2.
Our new decibel level, beta_2, is then given by:
beta_2 = 10log_10(I_2/I_0);
since I_2 = I_1/190^2, then
beta_2 = 10log_10(I_1/(190^2I_0)).
This can be rewritten as
beta_2 = 10[log_10(I_1) - log_10(190^2I_0)].
beta_2 = 10{log_10(I_1) - [log_10(190^2) + log_10(I_0)]}
beta_2 = 10{log_10(I_1) - [2log_10(190) + log_10(I_0)]}(E1)
Moreover, note that beta_1 = 10log_10(I_1/I_0) and, hence, log_10(I_1) = beta_1/10 + log_10(I_0).
Since beta_1 = 80 dB:
log_10(I_1) = 80/10 + log_10(10^-12) ;
log_10(I_1) = 8 - 12 = -4. (E2)
Now we can take the value of log_10(I_1) from (E2) and put it into (E1), yielding:
beta_2 = 10{-4 - [2*2.3 -12]}
beta_2 = 34 dB.
Note that we get a smaller value, as we would expect, since the sound source is know farther away from the house.
