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Question: A window whose area is \[2{m^2}\] opens on the street where the street noise results in an intensity...

A window whose area is 2m22{m^2} opens on the street where the street noise results in an intensity level at the window of 60db60db. How much acoustic power enters the widow via sound waves? Now, if an acoustic absorber is fitted at the window?
A. 8μW8\mu W
B. 10μW10\mu W
C. 4μW4\mu W
D. 2μW2\mu W

Explanation

Solution

Intensity of sound is given as I=10log⁡10(ii0)I = 10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right), where iiis the intensity of the sound expressed in watts per meter and i0{i_0}is the reference intensity.
In the question, street noise intensity is given; hence we calculate the intensity of the sound entering through a window whose area is given hence; by using the intensity and the area, we calculate the power of the sound wave entering through the window.

Complete step by step solution:
The intensity level of street noise at the window =60db = 60db
As we know, the intensity of sound is given as I=10log⁡10(ii0)−−(i)I = 10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right) - - (i)
So we can write this formula as
10log⁡10(ii0)=60−−(ii)10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right) = 60 - - (ii)
Here i0{i_0}is the reference intensity whose value=10−12wattm2 = {10^{ - 12}}\dfrac{{watt}}{{{m^2}}}
So equation (ii) can be further written as

log⁡10(i10−12)=6 ⇒i10−12=antilog⁡10(6) ⇒i=10−12×antilog⁡10(6)  {\log _{10}}\left( {\dfrac{i}{{{{10}^{ - 12}}}}} \right) = 6 \\\ \Rightarrow \dfrac{i}{{{{10}^{ - 12}}}} = anti{\log _{10}}\left( 6 \right) \\\ \Rightarrow i = {10^{ - 12}} \times anti{\log _{10}}\left( 6 \right) \\\

As we know antilog⁡ax=axanti{\log _a}x = {a^x}, hence by using this above equation can be further written as

i=10−12×106 =10−6wattm  i = {10^{ - 12}} \times {10^6} \\\ = {10^{ - 6}}\dfrac{{watt}}{m} \\\

Hence the intensity of the sound =10−6wattm = {10^{ - 6}}\dfrac{{watt}}{m}
As we know the sound power is the intensity of the sound passing through a given area, given by the formula
I=PAI = \dfrac{P}{A}
Where the area of the window is given as 2m22{m^2}
The intensity of the sound passing through the window =10−6wattm = {10^{ - 6}}\dfrac{{watt}}{m}
Hence the acoustic power entering through the widow via sound waves will be

P=I.A =2×10−6 =2×10−6watt =2μW  P = I.A \\\ = 2 \times {10^{ - 6}} \\\ = 2 \times {10^{ - 6}}watt \\\ = 2\mu W \\\

Therefore the acoustic power entering through the window =2μW = 2\mu W

Option A is correct.

Note:
Students must note that the power is directly proportional to the frequency, so the intensity will also be proportional to frequency, but in case of noise, the loudness of the sound does not depend on frequency.