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

Beginner2022-10-01Added 6 answers

Consider sounds with intensities ${I}_{1}$ and ${I}_{2}$ with intensity levels ${\beta}_{1}$ and ${\beta}_{2}$ respectively. We have

${\beta}_{1}=10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{0}})$

and

${\beta}_{2}=10{\mathrm{log}}_{10}(\frac{{I}_{2}}{{I}_{0}})$

Now subtracting the second equation from first,

${\beta}_{1}-{\beta}_{2}=10[{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{0}})-{\mathrm{log}}_{10}(\frac{{I}_{2}}{{I}_{0}})]$

Fo

${\beta}_{1}-{\beta}_{2}=10{\mathrm{log}}_{e}(\frac{{I}_{1}/{I}_{0}}{{I}_{2}/{I}_{0}})$

${\beta}_{1}-{\beta}_{2}=10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{2}})$

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

${\beta}_{2}={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{2}})={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{2{I}_{1}})=90-10{\mathrm{log}}_{10}(\frac{1}{2})$

${\beta}_{2}=90-10\times ({\mathrm{log}}_{10}(1)-{\mathrm{log}}_{10}(2))=90-10\times (-0.301)=90+3.01=93.01dB$

(b) When ${I}_{2}=\frac{{I}_{1}}{5}$, the sound intensity level ${\beta}_{2}$ is given by

${\beta}_{2}={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{1}/5})=90-10{\mathrm{log}}_{10}(5)=83.01dB$

Result:

(a)93.01dB (b) 83.01 dB

${\beta}_{1}=10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{0}})$

and

${\beta}_{2}=10{\mathrm{log}}_{10}(\frac{{I}_{2}}{{I}_{0}})$

Now subtracting the second equation from first,

${\beta}_{1}-{\beta}_{2}=10[{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{0}})-{\mathrm{log}}_{10}(\frac{{I}_{2}}{{I}_{0}})]$

Fo

${\beta}_{1}-{\beta}_{2}=10{\mathrm{log}}_{e}(\frac{{I}_{1}/{I}_{0}}{{I}_{2}/{I}_{0}})$

${\beta}_{1}-{\beta}_{2}=10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{2}})$

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

${\beta}_{2}={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{2}})={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{2{I}_{1}})=90-10{\mathrm{log}}_{10}(\frac{1}{2})$

${\beta}_{2}=90-10\times ({\mathrm{log}}_{10}(1)-{\mathrm{log}}_{10}(2))=90-10\times (-0.301)=90+3.01=93.01dB$

(b) When ${I}_{2}=\frac{{I}_{1}}{5}$, the sound intensity level ${\beta}_{2}$ is given by

${\beta}_{2}={\beta}_{1}-10{\mathrm{log}}_{10}(\frac{{I}_{1}}{{I}_{1}/5})=90-10{\mathrm{log}}_{10}(5)=83.01dB$

Result:

(a)93.01dB (b) 83.01 dB

aurelegena

Beginner2022-10-02Added 2 answers

Sound intensity level in decibels is:

$\beta =10\ast \mathrm{log}(\frac{I}{{I}_{0}})$

$9=\mathrm{log}(\frac{I}{{I}_{0}})$

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:

${\beta}_{a}=93dB$

${\beta}_{b}=83dB$

$\beta =10\ast \mathrm{log}(\frac{I}{{I}_{0}})$

$9=\mathrm{log}(\frac{I}{{I}_{0}})$

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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