Answer:
Given Acousting power P=2.35*10^5W
Distance r=51.3m
Area=4πr^2=4*3.142*51.3^2=33075.08
Sound intenstity I= (2.35*10^5)/33075.08=6962.8W/m^Reference sound intenstity Io=10^-12w/m^2
Sound intenstity I1=10*log(I/Io)
10*log(6962.8/10^-12)db =24.5w/m^2
Answer:
Intensity of sound = 7.11W/m²
Corresponding sound intensity level = 128.5dB
Explanation:
(a) Intensity (I) of a sound is related to the power (P) of the sound and coverage area (A) of the sound as follows;
I = [tex]\frac{P}{A}[/tex] -----------------(i)
From the question;
P = acoustic power = 2.35 x 10⁵W
distance = radius of coverage = r = 51.3m
Let's first calculate the area of coverage (A) of the sound as follows;
A = 4 x π x r² [substitute the value of r = 51.3 and take π = 3.142]
A = 4 x 3.142 x 51.3²
A = 33075.08m²
Now, substitute the values of P and A into equation (i) as follows;
I = [tex]\frac{2.35*10^{5}}{33075.08}[/tex]
I = 7.11W/m²
Therefore, the intensity of the sound at that distance is 7.11W/m².
(b) To calculate the sound intensity level, β, (basically in decibels), the following relation is used;
β = log₁₀[[tex]\frac{I}{I_{0} }[/tex]] ----------------------(ii)
Where;
I₀ = reference sound intensity level (threshold of hearing) = 10⁻¹²W/m²
I = Intensity of sound = 7.11W/m² [as calculated above]
Substitute these values into equation (ii) as follows;
β = 10 x log₁₀[[tex]\frac{7.11}{10^{-12}}[/tex]]
β = 10 x log₁₀[7.11 x 10¹²]
β = 10 x 12.85dB
β = 128.5dB
Therefore, the corresponding sound intensity level is 128.5dB
