tarjetaroja2t

2022-09-23

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB.
What is the ratio of the final sound intensity to the original sound intensity?

elilsonoulp2l

Given
${\beta }_{1}=23db$
${\beta }_{2}=61db$
The sound intensity level is given by
$\beta =10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)$ where ${I}_{0}$ is intensity at reference level
Therefore the change in sound intensity level can be determined as
${\beta }_{2}-{\beta }_{1}=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{0}}\right)-10\mathrm{log}\left(\frac{{I}_{1}}{{I}_{0}}\right)$
$61-23=10\mathrm{log}\left(\frac{\frac{{I}_{2}}{{I}_{0}}}{\frac{{I}_{1}}{{I}_{0}}}\right)since\mathrm{log}a-\mathrm{log}b=\mathrm{log}\frac{a}{b}$
$38=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$3.8=\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$\frac{{I}_{2}}{{I}_{1}}={10}^{3.8}$
$\approx 6310$
Hence
The ratio of final sound intensity to original is approximately 6310
Result:
The ratio of final sound intensity to original is approximately 6310

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