Respuesta :
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,
[tex]L = 10log_{10}\frac{I}{I_{o}}[/tex] Â Â Â Â Â
[tex]I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}[/tex] Â Â Â
[tex]L' = 10log_{10}\frac{I'}{I_{o}}[/tex]
[tex]I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}[/tex] Â Â Â Â
where
[tex]I_{o} = 10^{- 12} W/m^{2}[/tex]
I = Intensity of sound
Now,
[tex]I = \frac{P}{4\pi R^{2}}[/tex]
Similarly,
[tex]I' = \frac{P}{4\pi (R + 11.0)^{2}}[/tex]
Now,
[tex]\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}[/tex]
[tex]\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}[/tex]
[tex]R ^{2} + 22R + 121 = 11.64R^{2}}[/tex]
[tex]10.64R ^{2} - 22R - 121 = 0[/tex]
Solving the above quadratic eqn, we get:
R = 4.56 m
