Question

A jet engine emits sound uniformly in all directions, radiating an acoustic power of 2.35 × 10 5 2.35×105 W. Find the intensity of the sound at a distance of 51.3 51.3 m from the engine and calculate the corresponding sound intensity level.

Answer 1

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 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  = $\frac{P}{A}$            -----------------(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 = $\frac{2.35*10^{5}}{33075.08}$

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[$\frac{I}{I_{0} }$]               ----------------------(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[$\frac{7.11}{10^{-12}}$]

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