How many times more intense is a 60 dB sound than a 30 dB sound?

Question

TestHub · Accepted Answer

The loudness of sound in decibels is given by $L = 10 lo g_{10} (\frac{I}{I _{0}})$ .

For the first sound, $L_{1} = 60$ dB:

$60 = 10 lo g_{10} (\frac{I _{1}}{I _{0}}) \Rightarrow \frac{I _{1}}{I _{0}} = 1 0^{6}$

For the second sound, $L_{2} = 30$ dB:

$30 = 10 lo g_{10} (\frac{I _{2}}{I _{0}}) \Rightarrow \frac{I _{2}}{I _{0}} = 1 0^{3}$

Taking the ratio of the two intensities:

$\frac{I _{1}}{I _{2}} = \frac{1 0 ^{6}}{1 0 ^{3}} = 1000$

Thus, a 60 dB sound is 1000 times more intense than a 30 dB sound.

Answer: 1000

Physics - Sound Wave Question with Solution | TestHub

Options:

Doubts & Discussion