Question

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

Answer 1

Hint:
Intensity of sound is given as \[I = 10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right)\], where \[i\]is the intensity of the sound expressed in watts per meter and \[{i_0}\]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 \[ = 60db\]
As we know, the intensity of sound is given as \[I = 10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right) - - (i)\]
So we can write this formula as
\[10{\log _{10}}\left( {\dfrac{i}{{{i_0}}}} \right) = 60 - - (ii)\]
Here \[{i_0}\]is the reference intensity whose value\[ = {10^{ - 12}}\dfrac{{watt}}{{{m^2}}}\]
So equation (ii) can be further written as
\[
  {\log _{10}}\left( {\dfrac{i}{{{{10}^{ - 12}}}}} \right) = 6 \\
  \Rightarrow \dfrac{i}{{{{10}^{ - 12}}}} = anti{\log _{10}}\left( 6 \right) \\
 \Rightarrow i = {10^{ - 12}} \times anti{\log _{10}}\left( 6 \right) \\
 \]

As we know \[anti{\log _a}x = {a^x}\], hence by using this above equation can be further written as
\[
  i = {10^{ - 12}} \times {10^6} \\
   = {10^{ - 6}}\dfrac{{watt}}{m} \\
 \]
Hence the intensity of the sound \[ = {10^{ - 6}}\dfrac{{watt}}{m}\]
As we know the sound power is the intensity of the sound passing through a given area, given by the formula
\[I = \dfrac{P}{A}\]
Where the area of the window is given as \[2{m^2}\]
The intensity of the sound passing through the window \[ = {10^{ - 6}}\dfrac{{watt}}{m}\]
Hence the acoustic power entering through the widow via sound waves will be
\[
  P = I.A \\
   = 2 \times {10^{ - 6}} \\
   = 2 \times {10^{ - 6}}watt \\
   = 2\mu W \\
 \]
Therefore the acoustic power entering through the window \[ = 2\mu W\]

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.