Question

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

Answer 1

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,

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