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Svetlanka [38]
3 years ago
6

By what factor does the sound intensity increase if the sound intensity level increases from 60 db to 61 db?

Physics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

\frac{I_{2}}{I_{1}} = 10^{0.1}

Explanation:

The sound intensity level β can be mathematically defined as

\beta(dB) = 10log_{10}(\frac{I}{I_{o} } )

where I_{o} is the reference intensity and is equal to 10^{-12} W/m^{2}.

Now,

\beta_{2} - \beta_{1} = 10 log\frac{I_{2} }{I_{o}} -  10 log\frac{I_{1} }{I_{o}}\\62-60= 10log(\frac{\frac{I_{2} }{I_{o} } }{\frac{I_{1} }{I_{o} } } )\\1=10log(\frac{I_{2} }{I_{1} } )\\\\\frac{I_{2} }{I_{1} }=10^{0.1}

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