Answer:
d2 = 4080 m
Explanation:
The intensity level of sound is given by
Intensity level = 10*log(I/Iā)
where Iā = 10ā»Ā¹Ā²W/mĀ² is the threshold of hearing
60 = 10*log(I/Iā) Ā eq. 1
The intensity of sound decreases with the increase in distance squared
I ā 1/dĀ²
Let d1 is the distance where intensity of sound is 64 dB and d2 is the distance where intensity of sound is 0 dB (barely audible) so
Iā/I = (d1/d2)Ā²
or Ā I/Iā = (d2/d1)Ā² put it in eq. 1
60 = 10*log(d2/d1)Ā²
60 = 20*log(d2/d1)
60/20 = log(d2/d1)
3 = log(d2/d1)
10Ā³ Ā = d2/d1
d1*10Ā³ Ā = d2
d2 = 4.08*10Ā³
d2 = 4080 m
Therefore, at a distance of 4080 m the sound of music will barely be audible to a person with normal hearing.
