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Q.68PE

Expert-verifiedFound in: Page 631

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Find the intensity in watts per meter squared of a \({\rm{60}}{\rm{.0 Hz}}\)**** sound having a loudness of \({\bf{60}}{\rm{ }}{\bf{phons}}\)****. (b) Find the intensity in watts per meter squared of a \({\rm{10,000 Hz}}\)**** sound having a loudness of \({\bf{60}}{\rm{ }}{\bf{phons}}\)****.**

(a) Intensity of \(60{\rm{ }}Hz\) will be \(3.17 \times {10^{ - 5}}{\rm{ }}W \cdot {m^{ - 2}}\).

(b) The intensity \(10000{\rm{ }}Hz\) will be \(1.5 \times {10^{ - 10}}{\rm{ }}W \cdot {m^{ - 2}}\).

**The greater the amplitude, the louder and more intense the sound.**

**Intensity is the number of incident photons per second per unit area (more precisely, it is energy per unit area per unit time), while frequency refers to the frequency of the photon when referred to as a wave, which is the number of waves in a second.**

First, make a vertical line for a given frequency \(\left( {60{\rm{ }}Hz} \right)\) and draw a horizontal line corresponding to these vertical lines then we will get our sound intensity levels in \(dB\) as shown in the figure.

Here, the points that are crossings of the vertical line \(\left( {60{\rm{ }}Hz} \right)\) and \(60{\rm{ }}phons\) curve.

Now make horizontal lines corresponding to these points that are crossings of vertical lines and \(60{\rm{ }}phons\), this horizontal line will meet on the y-axis at some value of sound level.

For \(60{\rm{ }}Hz\) frequency, \(60{\rm{ }}phons\) loudness - horizontal line cut y-axis around \(73{\rm{ }}dB\) (see only small blue line)

Now change this sound intensity level \(\left( {dB} \right)\) \(W \cdot {m^{ - 2}}\) by using the following formula.

\(\beta = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\)

Here, \(\beta \) is the sound intensity level in \(dB\), \(I\) is the intensity (in \(W \cdot {m^{ - 2}}\)) of ultrasound wave and \({I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\) is the threshold intensity of hearing.

Sound intensity level is given by,

\(\beta = 73{\rm{ }}dB\)

Threshold intensity of hearing is,

\({I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\)

Define the intensity \(I\) for \(\beta = 73{\rm{ }}dB\) by using following formula.

\(\begin{aligned}{}\beta &= 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{\beta }{{10}} &= {\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{\beta }{{10}} &= {\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{I}{{{I_0}}} &= {\left( {10} \right)^{\frac{\beta }{{10}}}}\end{aligned}\)

Now replace \({I_0}\) and \(\beta \) with the given data.

\(\begin{aligned}{}I &= {10^{ - 12}}{\left( {10} \right)^{\frac{{73}}{{10}}}}\\ &= {10^{ - 12}}{\left( {10} \right)^{7.3}}\\ &= 3.17 \times {10^{ - 5}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)

First, make a vertical line for a given frequency \(\left( {10000{\rm{ }}Hz} \right)\) and draw a horizontal line corresponding to these vertical lines then we will get our sound intensity levels in \(dB\) as shown in the figure.

Here, the points that are crossings of the vertical line \(\left( {10000{\rm{ }}Hz} \right)\) and \(60{\rm{ }}phons\) curve.

Now make horizontal lines corresponding to these points that are crossings of vertical lines and \(60{\rm{ }}phons\), these horizontal lines will meet on the y-axis at some value of sound level.

For \(10000{\rm{ }}Hz\) frequency, \(60{\rm{ }}phons\) loudness - horizontal line cut y-axis around \(70{\rm{ }}dB\).

Now change this sound intensity level \(\left( {dB} \right)\) \(W \cdot {m^{ - 2}}\) by using the following formula.

\(\beta = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\)

Here, \(\beta \) is the sound intensity level in \(dB\), \(I\) is the intensity (in \(W \cdot {m^{ - 2}}\)) of ultrasound wave and \({I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\) is the threshold intensity of hearing

Sound intensity level is,

\(\beta = 70{\rm{ }}dB\)

Threshold intensity of hearing is,

\({I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\)

Determine the intensity \(I\) for \(\beta = 70{\rm{ }}dB\) by using the following formula.

\(\begin{aligned}{}\beta &= 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{\beta }{{10}} &= {\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{\beta }{{10}} &= {\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\\\frac{I}{{{I_0}}} &= {\left( {10} \right)^{\frac{\beta }{{10}}}}\end{aligned}\)

Now replace \({I_0}\) and \(\beta \) with the given data.

\(\begin{aligned}{}I &= {10^{ - 12}}{\left( {10} \right)^{\frac{{70}}{{10}}}}\\ &= {10^{ - 12}}{\left( {10} \right)^7}\\ &= 1.5 \times {10^{ - 10}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)

Hence, the intensity \(10000{\rm{ }}Hz\) will be \(1.5 \times {10^{ - 10}}{\rm{ }}W \cdot {m^{ - 2}}\).

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