the loudest sound measured one night during a hockey game was 112 db. the loudest sound measured during a…

the loudest sound measured one night during a hockey game was 112 db. the loudest sound measured during a hockey game the next night was 118 db. what fraction of sound intensity of the second game was the sound intensity of the first game?\n\nl = 10logleft(\frac{i}{i_0}\right)\n\n$l =$ loudness, in decibels\n$i =$ sound intensity, in watts/m²\n$i_0 = 10^{-12}$ watts/m²\n\n0.25\n0.78\n0.95\n0.99

the loudest sound measured one night during a hockey game was 112 db. the loudest sound measured during a hockey game the next night was 118 db. what fraction of sound intensity of the second game was the sound intensity of the first game?\n\nl = 10logleft(\frac{i}{i_0}\right)\n\n$l =$ loudness, in decibels\n$i =$ sound intensity, in watts/m²\n$i_0 = 10^{-12}$ watts/m²\n\n0.25\n0.78\n0.95\n0.99

Answer

Explanation:

Step1: Find intensity for first game

Given $L = 10\log(\frac{I}{I_0})$, for $L_1=112$ dB and $I_0 = 10^{- 12}\text{ watts/m}^2$. $112 = 10\log(\frac{I_1}{10^{-12}})$ $\frac{112}{10}=\log(\frac{I_1}{10^{-12}})$ $11.2=\log(\frac{I_1}{10^{-12}})$ By the definition of logarithm, $I_1 = 10^{11.2}\times10^{-12}=10^{-0.8}$

Step2: Find intensity for second game

For $L_2 = 118$ dB, $118=10\log(\frac{I_2}{10^{-12}})$ $\frac{118}{10}=\log(\frac{I_2}{10^{-12}})$ $11.8=\log(\frac{I_2}{10^{-12}})$ By the definition of logarithm, $I_2 = 10^{11.8}\times10^{-12}=10^{-0.2}$

Step3: Find the fraction

The fraction is $\frac{I_1}{I_2}=\frac{10^{-0.8}}{10^{-0.2}}=10^{-0.8 + 0.2}=10^{-0.6}\approx0.25$

Answer:

0.25