Question

A child has a hearing loss of 60 dB near 5000Hz, due to noise exposure, and normal hearing elsewhere. How much more intense is a 5000 Hz tone than a 400 Hz tone if they are both barely audible to the child?

Short answer

A \(5000{\rm{ }}Hz\) tone will be \(7.9 \times {10^{ - 8}}\) more intense than \(400{\rm{ }}Hz\) a tone.

Step-by-step solution

Threshold of hearing:

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 phon.

Figure for sound level, intensity, and loudness at given frequencies

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.

Analysis of this figure:

Here the points that are crossings of vertical lines and 0 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 6000 Hz frequency, horizontal line cut y-axis around -1 dB (threshold hearing)

This -1 dB sound is not a normal sound level, person hears a sound 60 dB above normal then an effective sound for hearing will be

\(6{\rm{ }}dB - 1{\rm{ }}dB = 59{\rm{ }}dB\)

For 400 Hz frequency, horizontal line cut y-axis around 10 dB (threshold hearing)

This 10 dB sound is a normal hearable sound level.

Now calculate how much 400 Hzis more intense than then 6000 Hz frequency of sound

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

\(59dB - 10{\rm{ }}dB = 49{\rm{ }}dB\)

So this is our final sound level (49 dB) for the calculation of intensity.

Calculation:

Now change this sound intensity level (dB) \(W \cdot {m^{ - 2}}\)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,

\(\beta = 49{\rm{ }}dB\)

Threshold intensity of hearing is,

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

Determine the intensity I for \(\beta = 49{\rm{ }}dB\) by using below 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.

\(\begin{aligned}{}I &= {10^{ - 12}}{\left( {10} \right)^{\frac{{49}}{{10}}}}\\ &= {10^{ - 12}}{\left( {10} \right)^{4.9}}\\ &= 7.9 \times {10^{ - 8}}{\rm{ }}W \cdot {m^{ - 2}}\end{aligned}\)

Hence, a 5000 Hz tone will be \(7.9 \times {10^{ - 8}}\) more intense than 400 Hz a tone.