Q.71PE - College Physics (Urone)

Question

What is the ratio of intensities of two sounds of identical frequency if the first is just barely discernible as louder to a person than the second?

Step-by-Step Solution

Answer

The ratio of two intensities of two sounds will be 1.99.

Step 1: Just barely discernible sound

Two sounds are called just barely discernible sounds if difference in intensity sound level between them as 3 dB.

So write the relation as below.

\[{\beta _2} - {\beta _1} = 3{\rm{ }}dB\]

Here, \[{\beta _2}\] and \[{\beta _1}\] are two intensity sound levels that have a difference of 3 dB.

Step 2: Formula for sound intensity level in dB ( decibels)

The sound intensity level (SIL) or sound intensity level is the level (logarithmic quantity) of sound intensity relative to a reference value.

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.

Threshold intensity of hearing is, \[{I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\]

Step 3: Ratio of intensities

The sound intensity level is given as

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

 

Now to get the ratio of intensities we have to rearrange this relation like below.

\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}}}}\end{aligned}

This is a general formula for intensity now let’s make to the formula for intensities

For \[{\beta _2}\] intensity will be

\[{I_2} = {I_0}{\left( {10} \right)^{\frac{{{\beta _2}}}{{10}}}}\]                                                                                                          ….. (1)

 

For \[{\beta _1}\] intensity will be

\[{I_1} = {I_0}{\left( {10} \right)^{\frac{{{\beta _1}}}{{10}}}}\]                                                                                                           ….. (2)

 

The ratio can be written as

\begin{aligned}\frac{{{I_2}}}{{{I_1}}} &= \frac{{{I_0}{{\left( {10} \right)}^{\frac{{{\beta _2}}}{{10}}}}}}{{{I_0}{{\left( {10} \right)}^{\frac{{{\beta _1}}}{{10}}}}}}\\ &= \frac{{{{\left( {10} \right)}^{\frac{{{\beta _2}}}{{10}}}}}}{{{{\left( {10} \right)}^{\frac{{{\beta _1}}}{{10}}}}}}\\ &= {\left( {10} \right)^{\frac{{{\beta _2} - {\beta _1}}}{{10}}}}\end{aligned}

As know,

\begin{aligned}{\beta _2} - {\beta _1} = 3{\rm{ }}dB\\{\beta _2} = {\beta _1} + 3{\rm{ }}dB\end{aligned}

Now replace \[{\beta _2}\] as \[{\beta _1} + 3{\rm{ }}dB\] then you will get,

\begin{aligned}\frac{{{I_2}}}{{{I_1}}} &= {\left( {10} \right)^{\frac{{{\beta _1} + 3 - {\beta _1}}}{{10}}}}\\ &= {\left( {10} \right)^{\frac{3}{{10}}}}\\ &= 1.99\end{aligned}