How many times more intense is a 60 dB sound | vedclass

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(C) The loudness of two sounds is given as $L_2 = 60 \ dB$ and $L_1 = 30 \ dB$.The formula for the loudness of sound is $L = 10 \log_{10} \left( \frac

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$1000$

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(C) The loudness of two sounds is given as $L_2 = 60 \ dB$ and $L_1 = 30 \ dB$.The formula for the loudness of sound is $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.The difference in loudness is given by $L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.Substituting the values: $60 - 30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.$30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.$\log_{10} \left( \frac{I_2}{I_1} \right) = 3$.Therefore,$\frac{I_2}{I_1} = 10^3 = 1000$.Thus,the $60 \ dB$ sound is $1000$ times more intense than the $30 \ dB$ sound.

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

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