(a) What is the intensity in watts per meter squared of a just barely audible \({\rm{200 Hz}}\) sound? (b) What is the intensity in watts per meter squared of a barely audible \({\rm{4000 Hz}}\) sound?

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 \(\left( {200{\rm{ }}Hz} \right)\) and draw a horizontal line corresponding to this vertical line then we will get our sound intensity levels in \(dB\) as shown in the figure below.

Threshold of hearing is minimum sound level below which person cann’t hear any sound and loudness for barely audible is \(0\)phons.

From the figure you can see that the horizontal line cross y-axis around \(23{\rm{ }}dB\).

So \(23{\rm{ }}dB\) intensity sound level is barely audible \(t\) is the person.

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,

\(\beta = {\rm{ }}23{\rm{ }}dB\)

Threshold intensity of hearing is,

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

Determine the intensity \(l\) for \(\beta = {\rm{ }}23{\rm{ }}dB\) by using following formula.

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

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

\(\begin{aligned}{}I = {10^{ - 12}}{\left( {10} \right)^{\frac{{23}}{{10}}}}\\ = {10^{ - 12}}{\left( {10} \right)^{2.3}}\\ = 1.3 \times {10^{ - 10}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)

First, make a vertical line for a given frequency \(\left( {4000{\rm{ }}Hz} \right)\) and draw a horizontal line corresponding to this vertical line then we will get our sound intensity levels in \(dB\) as shown in the figure

The threshold of hearing is a minimum sound level below which a person can't hear any sound. and loudness for barely audible is \(0\)phons.

From the figure you can see that this horizontal line cross y-axis around \( - 7{\rm{ }}dB\).

So \( - 7{\rm{ }}dB\) intensity sound level is barely audible t is the person.

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 \(dB\), \(l\) 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,

\(\beta = - 7{\rm{ }}dB\)

Threshold intensity of hearing is,

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

Calculate intensity \(I\) for \(\beta = - 7{\rm{ }}dB\) by using following formula.

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

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

\(\begin{aligned}{}I = {10^{ - 12}}{\left( {10} \right)^{\frac{{ - 7}}{{10}}}}\\ = {10^{ - 12}}{\left( {10} \right)^{ - 0.7}}\\ = 2 \times {10^{ - 13}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)