Answer:
The distance of the second sound intensity level is 28.12 m.
Explanation:
Given;
intensity of first sound level, I₁ = 110 dB
distance of first sound level, r₁ = 5 m
intensity of second sound level, I₂ = 95 dB
let the distance of second sound level = r₂
The intensity of the sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ][/tex]
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
for 110 dB to W/m²
[tex]110 = 10 Log[\frac{I}{1 \ \times \ 10^{-12}} ]\\\\11 = Log[\frac{I}{1 \ \times \ 10^{-12}} ]\\\\10^{11} = \frac{I}{1 \ \times \ 10^{-12}} \\\\I = 1 \ \times \ 10^{-12} \ \times \ 10^{11}\\\\I = 10^{-1} \ W/m^2\\\\I = 0.1 \ W/m^2[/tex]
for 95 dB to W/m²
[tex]95 = 10 Log[\frac{I}{1 \ \times \ 10^{-12}} ]\\\\9.5 = Log[\frac{I}{1 \ \times \ 10^{-12}} ]\\\\10^{9.5} = \frac{I}{1 \ \times \ 10^{-12}} \\\\I = 1 \ \times \ 10^{-12} \ \times \ 10^{9.5}\\\\I = 10^{-2.5} \ W/m^2\\\\[/tex]
The intensity of sound is related to distance as follows;
[tex]I = \frac{P}{A} = \frac{P}{\pi r^2} \\\\I = \frac{1}{r^2} \\\\I_1r_1^2 = I_2r_2^2 \\\\[/tex]
Now, determine the distance of the second sound intensity level
[tex]r_2 ^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2} }\\\\r_2 = \sqrt{\frac{0.1 \ \times \ 5^2}{10^{-2.5}}}\\\\r_2 = 28.12 \ m[/tex]
Therefore, the distance of the second sound intensity level is 28.12 m.
