A person has a hearing threshold \({\rm{10 dB}}\) above normal at \({\rm{100 Hz}}\) and \({\rm{50 dB}}\) above normal at \({\rm{4000 Hz}}\). How much more intense must a \({\rm{100 Hz}}\) tone be than a \({\rm{4000 Hz}}\) tone if they are both barely audible to this person?

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

First, make a vertical line for all given frequencies 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 vertical lines and \(0{\rm{ }}phon\) curve.

Now make horizontal lines corresponding to these points that are crossings of vertical lines and \(0{\rm{ }}phon\) curve, this horizontal line will meet on the y-axis at some value of sound level.

You got the following from the figure:

For \(100{\rm{ }}Hz\) frequency, horizontal line cut y-axis around\(40{\rm{ }}dB\)(threshold hearing)This \(40{\rm{ }}dB\) sound is for normal hearing but person hears a sound\(10{\rm{ }}dB\)above normal then an effective sound for hearing will be,

\(40{\rm{ }}dB + 10{\rm{ }}dB = 50{\rm{ }}dB\)

For \(4000{\rm{ }}Hz\) frequency, horizontal line cut y-axis around\( - 7{\rm{ }}dB\)(threshold hearing)

This \( - 7{\rm{ }}dB\) sound is for normal hearing but person hears a sound\(50{\rm{ }}dB\)above normal then an effective sound for hearing will be,

\( - 7{\rm{ }}dB + 50{\rm{ }}dB = 43{\rm{ }}dB\)

Now calculate how much \(100{\rm{ }}Hz\)is more intense than then\(4000{\rm{ }}Hz\)frequency of sound.

To do this let’s take the difference between these two sound levels as below.

\(50{\rm{ }}dB - 43{\rm{ }}dB = 7{\rm{ }}dB\)

So this is our final sound level \(\left( {7{\rm{ }}dB} \right)\) for the calculation of intensity.

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\({I_0}\) is the threshold intensity of hearing

Sound intensity level is,

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

Threshold intensity of hearing is,

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

Determine the intensity \(I\)for\(\beta = 7{\rm{ }}dB\) by using below formula.

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

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

\(\begin{aligned}{}I &= {10^{ - 12}}{\left( {10} \right)^{\frac{7}{{10}}}}\\ &= {10^{ - 12}}{\left( {10} \right)^{0.7}}\\ &= 5 \times {10^{ - 5}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)

Hence, a \(100{\rm{ }}Hz\) tone will be \(5 \times {10^{ - 5}}{\rm{ }}W \cdot {m^{ - 2}}\) more intense than \(4000{\rm{ }}Hz\) a tone.