Answer:
The source is at a distance of 4.56 m from the first point.
Solution:
As per the question:
Separation distance between the points, d = 11.0 m
Sound level at the first point, L = 66.40 dB
Sound level at the second point, L'= 55.74 dB
Now,
![L' = 10log_(10)(I')/(I_(o))](https://img.qammunity.org/2020/formulas/physics/high-school/ryun65oqfkeam94d5nxrepxnk4hivdinkt.png)
where
![I_(o) = 10^(- 12) W/m^(2)](https://img.qammunity.org/2020/formulas/physics/high-school/46ue64z7pljvv6siu8d94xifn4koahxdoy.png)
I = Intensity of sound
Now,
![I = (P)/(4\pi R^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/hoa6inc4d4ygczietistbhkmcfbjp9kdax.png)
Similarly,
![I' = (P)/(4\pi (R + 11.0)^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/7pxj3j4qebomqhs6vaiwti186liui6xq1k.png)
Now,
![(I)/(I') = ((R + 11.0)^(2))/(R^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/bh1t1og6q8fcm5177ihi0yftxz4rwxh4bz.png)
![(10^(- 5.36))/(10^(- 6.426)) = ((R + 11.0)^(2))/(R^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/p4sdiceygmgimtz34n78udsrxkhe665j7q.png)
![R ^(2) + 22R + 121 = 11.64R^(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/c18csfmq0ls567o048dgvqiku884jdvhye.png)
![10.64R ^(2) - 22R - 121 = 0](https://img.qammunity.org/2020/formulas/physics/high-school/alzmovuqbhkdmwqt8a5sdmyliuxsp00wuk.png)
Solving the above quadratic eqn, we get:
R = 4.56 m