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 = -----------------(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 =
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[] ----------------------(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[]
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