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
Answer
Explanation:
Step1: Write the decibel - intensity formula for the first game
Given $L = 10\log\left(\frac{I}{I_0}\right)$, for the first game with $L_1=112$ dB, we have $112 = 10\log\left(\frac{I_1}{I_0}\right)$. Then $\log\left(\frac{I_1}{I_0}\right)=\frac{112}{10}=11.2$, and $\frac{I_1}{I_0}=10^{11.2}$. So, $I_1 = I_0\times10^{11.2}$.
Step2: Write the decibel - intensity formula for the second game
For the second game with $L_2 = 118$ dB, we have $118=10\log\left(\frac{I_2}{I_0}\right)$. Then $\log\left(\frac{I_2}{I_0}\right)=\frac{118}{10}=11.8$, and $\frac{I_2}{I_0}=10^{11.8}$. So, $I_2 = I_0\times10^{11.8}$.
Step3: Find the fraction of $I_1$ to $I_2$
We want to find $\frac{I_1}{I_2}$. Substitute $I_1 = I_0\times10^{11.2}$ and $I_2 = I_0\times10^{11.8}$ into the fraction: $\frac{I_1}{I_2}=\frac{I_0\times10^{11.2}}{I_0\times10^{11.8}}$. Since $I_0$ cancels out, and using the rule $\frac{a^m}{a^n}=a^{m - n}$, we get $\frac{I_1}{I_2}=10^{11.2-11.8}=10^{- 0.6}\approx0.25$.
Answer:
0.25