You hear a sound at 65 dB. What is the sound intensity level if the intensity of the sound is doubled?

Jayvion Caldwell
2022-07-23
Answered

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Bradley Sherman

Answered 2022-07-24
Author has **17** answers

Use the equation given by:

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

If l is doubled, use a property of logarithms to get the new value of the intensity level, ${\beta}^{\prime}$. We have:

${\beta}^{\prime}=(10dB){\mathrm{log}}_{10}(\frac{2I}{{I}_{0}})(10dB)[{\mathrm{log}}_{10}(2)+{\mathrm{log}}_{10}(\frac{I}{{I}_{0}})]$

$=(10dB){\mathrm{log}}_{10}2+\beta =3dB+65dB=68dB$

Result:

68dB

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

If l is doubled, use a property of logarithms to get the new value of the intensity level, ${\beta}^{\prime}$. We have:

${\beta}^{\prime}=(10dB){\mathrm{log}}_{10}(\frac{2I}{{I}_{0}})(10dB)[{\mathrm{log}}_{10}(2)+{\mathrm{log}}_{10}(\frac{I}{{I}_{0}})]$

$=(10dB){\mathrm{log}}_{10}2+\beta =3dB+65dB=68dB$

Result:

68dB

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