Bridger Holden

2022-09-30

What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

Tristin Durham

Consider sounds with intensities ${I}_{1}$ and ${I}_{2}$ with intensity levels ${\beta }_{1}$ and ${\beta }_{2}$ respectively. We have
${\beta }_{1}=10{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{{I}_{0}}\right)$
and
${\beta }_{2}=10{\mathrm{log}}_{10}\left(\frac{{I}_{2}}{{I}_{0}}\right)$
Now subtracting the second equation from first,
${\beta }_{1}-{\beta }_{2}=10\left[{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{{I}_{0}}\right)-{\mathrm{log}}_{10}\left(\frac{{I}_{2}}{{I}_{0}}\right)\right]$
Fo
${\beta }_{1}-{\beta }_{2}=10{\mathrm{log}}_{e}\left(\frac{{I}_{1}/{I}_{0}}{{I}_{2}/{I}_{0}}\right)$
${\beta }_{1}-{\beta }_{2}=10{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{{I}_{2}}\right)$
Let, for intensity ${I}_{1}$, the sound intensity level ${\beta }_{1}=90dB$
(a) For ${I}_{2}=2{I}_{1}$, the sound intensity level ${\beta }_{2}$ is given by
${\beta }_{2}={\beta }_{1}-10{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{{I}_{2}}\right)={\beta }_{1}-10{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{2{I}_{1}}\right)=90-10{\mathrm{log}}_{10}\left(\frac{1}{2}\right)$
${\beta }_{2}=90-10×\left({\mathrm{log}}_{10}\left(1\right)-{\mathrm{log}}_{10}\left(2\right)\right)=90-10×\left(-0.301\right)=90+3.01=93.01dB$
(b) When ${I}_{2}=\frac{{I}_{1}}{5}$, the sound intensity level ${\beta }_{2}$ is given by
${\beta }_{2}={\beta }_{1}-10{\mathrm{log}}_{10}\left(\frac{{I}_{1}}{{I}_{1}/5}\right)=90-10{\mathrm{log}}_{10}\left(5\right)=83.01dB$
Result:
(a)93.01dB (b) 83.01 dB

aurelegena

Sound intensity level in decibels is:
$\beta =10\ast \mathrm{log}\left(\frac{I}{{I}_{0}}\right)$
$9=\mathrm{log}\left(\frac{I}{{I}_{0}}\right)$
Where ${I}_{0}={10}^{-12}W/{m}^{2}$, the limit of audiable sound. Ihe expression in the brackets can be written as:
${10}^{9}=y$
Where y is just a substitute symbol for the expression in the brackets. If we double the expression in the y and put it back in the equation, we get solution for the first part of the problem:
${\beta }_{a}=10\ast \mathrm{log}2\ast {10}^{9}$
${\beta }_{a}=93dB$
Doing the same process for the second part of the problem, but this time instead of doubling y, we divide it with 5 and we get:
${\beta }_{b}=10\ast \mathrm{log}{10}^{9}/5$
${\beta }_{b}=833dB$
Result:
${\beta }_{a}=93dB$
${\beta }_{b}=83dB$

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