Question

A person wears a hearing aid that uniformly increases the sound level of all audible frequencies of sound by $30.0{\text{ dB}}$. The hearing aid picks up sound having a frequency of $250{\text{ Hz}}$ at an intensity of $3.0 \times {10^{ - 11}}{W\m^2}$. What is the intensity delivered to the eardrum?

Answer 1

Use the relationship between the sound intensities given in decibels ${\text{dB}}$ and given in in watts per meter squared $W/{m^2}$ to find the resultant intensity of the sound after passing through the hearing aid. The difference between the sound intensities is given as $30{\text{ dB}}$.

Complete step by step answer:
Intensity of sound is defined to be power per unit area carried by a wave. Also, power is the rate at which sound energy is transferred by the wave. This can be represented in the form of an equation as intensity $I = \dfrac{P}{A}$, where P is the power through an area A. The SI unit for I is $W/{m^2}$.

Generally, sound intensity levels are quoted in terms of decibels (dB) much more often than sound intensities in watts per meter squared. Most commonly, decibels are the unit of choice in the scientific literature as well as in the popular media. The main reasons for this choice as units are related to how one perceives sounds. How our ears perceive sound can be more detaily described by the logarithm of the intensity rather than directly referring to the intensity. We have the sound intensity level $\beta$ in decibels of a sound with an intensity $I$ in watts per meter squared as $\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$, where ${I_0} = {10^{ - 12}}\,W/{m^2}$ is known as the reference intensity, which is the lowest or threshold intensity of sound, with a frequency of 1000 Hz, a person with normal hearing can perceive.

Let ${\beta _1}$ and ${\beta _2}$ be decibel levels of the sound before and after passing the hearing aid. Also, let ${I_1}$ and ${I_2}$ be the intensities of the sound before and after passing through the hearing aid. From the question, we have the values ${\beta _2} - {\beta _1} = 30.0{\text{ dB}}$ and ${I_1} = 3.0 \times {10^{ - 11}}\,W/{m^2}$
We know from definition that
${\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)$ and ${\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right)$
$\Rightarrow {\beta _2} - {\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) - 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)$
$\Rightarrow {\beta _2} - {\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}} \times \dfrac{{{I_0}}}{{{I_1}}}} \right)$ [Using $\log a - \log b = \log \left( {\dfrac{a}{b}} \right)$]
$\Rightarrow {\beta _2} - {\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)$
$\Rightarrow \dfrac{{{\beta _2} - {\beta _1}}}{{10}} = {\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)$
$\Rightarrow {10^{\dfrac{{{\beta _2} - {\beta _1}}}{{10}}}} = \dfrac{{{I_2}}}{{{I_1}}}$
$\Rightarrow {I_2} = {I_1}{10^{\dfrac{{{\beta _2} - {\beta _1}}}{{10}}}}$
Thus, we have ${I_2} = 3.0 \times {10^{ - 11}}\,W/{m^2} \times {10^{\dfrac{{30.0}}{{10}}}}$
$\Rightarrow {I_2} = 3.0 \times {10^{ - 11}} \times {10^3}\,W/{m^2}$
$\Rightarrow {I_2} = 3.0 \times {10^{ - 8}}\,W/{m^2}$

Hence, the intensity delivered to the eardrum is $3.0 \times {10^{ - 8}}\,W/{m^2}$.

Note: The intensity of a sound wave increases with increase in the amplitude of the wave, as amplitude denotes the energy carried by a sound wave. It tells about the loudness of a sound when heard. But intensity has nothing to do with the frequency of the sound wave as the same frequency wave can be amplified to have different energy levels.