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.
