Question

A person has a hearing threshold $10\mathrm{d}\mathrm{B}$ above normal at $100\mathrm{H}\mathrm{z}$ and $50\mathrm{d}\mathrm{B}$ above normal at $4000\mathrm{H}\mathrm{z}$. How much more intense must a $100\mathrm{H}\mathrm{z}$ tone be than a $4000\mathrm{H}\mathrm{z}$ tone if they are both barely audible to this person?

Short answer

A $100Hz$ tone will be $5\times {10}^{-5}$ more intense than $4000Hz$ 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 $0\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{n}$.

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 $0phon$ curve.

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

You got the following from the figure:

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

$40dB+10dB=50dB$

For $4000Hz$ frequency, horizontal line cut y-axis around$-7dB$(threshold hearing)

This $-7dB$ sound is for normal hearing but person hears a sound$50dB$above normal then an effective sound for hearing will be,

$-7dB+50dB=43dB$

Now calculate how much $100Hz$is more intense than then$4000Hz$frequency of sound.

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

$50dB-43dB=7dB$

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

Calculation

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,

$\beta =7dB$

Threshold intensity of hearing is,

$I_0={10}^{-12}W\cdot m^{-2}$

Determine the intensity $I$for$\beta =7dB$ by using below formula.

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

Now replace$I_0$and$\beta$with the given data.

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

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