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 the sound intensity of the second game was the sound intensity of the first game?
[tex]
\begin{array}{l}
L=10 \log \left(\frac{1}{t_0}\right) \\
L=\text { loudness, in decibels } \\
I=\text { sound intensity, in watts } / m^2 \\
t_0=10^{-12} \text { watts } / m^2
\end{array}
[/tex]
A. 0.25
B. 0.78
C. 0.95
D. 0.99
Answer (1)
We are given the formula for loudness in decibels and the loudness values for two hockey games.
We use the formula to express the sound intensities I 1 and I 2 in terms of the given loudness values L 1 and L 2 .
We find the fraction I 2 I 1 by simplifying the expression 1 0 10 L 1 − L 2 .
Substituting the given values, we calculate the fraction to be approximately 0.25, so the final answer is 0.25 .
Explanation
Understanding the Problem We are given the formula for loudness in decibels: L = 10 lo g ( I 0 I ) , where L is the loudness in decibels, I is the sound intensity in watts/ m 2 , and I 0 = 1 0 − 12 watts/ m 2 . We are given that the loudness on the first night is L 1 = 112 dB and the loudness on the second night is L 2 = 118 dB. We want to find the fraction I 2 I 1 , where I 1 is the sound intensity on the first night and I 2 is the sound intensity on the second night.
Applying the Loudness Formula From the formula L = 10 lo g ( I 0 I ) , we can write L 1 = 10 lo g ( I 0 I 1 ) and L 2 = 10 lo g ( I 0 I 2 ) . Dividing both sides of the equations by 10, we get 10 L 1 = lo g ( I 0 I 1 ) and 10 L 2 = lo g ( I 0 I 2 ) .
Isolating Sound Intensities Using the property that 1 0 l o g ( x ) = x , we get 1 0 10 L 1 = I 0 I 1 and 1 0 10 L 2 = I 0 I 2 . Multiplying both sides of the equations by I 0 , we get I 1 = I 0 ⋅ 1 0 10 L 1 and I 2 = I 0 ⋅ 1 0 10 L 2 .
Calculating the Fraction Now we find the fraction I 2 I 1 = I 0 ⋅ 1 0 10 L 2 I 0 ⋅ 1 0 10 L 1 = 1 0 10 L 2 1 0 10 L 1 = 1 0 10 L 1 − L 2 . Substituting the given values L 1 = 112 and L 2 = 118 , we get I 2 I 1 = 1 0 10 112 − 118 = 1 0 10 − 6 = 1 0 − 0.6 .
Final Calculation and Answer Calculating 1 0 − 0.6 , we find that I 2 I 1 ≈ 0.251188643150958 . Therefore, the fraction of sound intensity of the second game that was the sound intensity of the first game is approximately 0.25.
Examples
Understanding sound intensity is crucial in various fields, such as acoustics, environmental science, and audio engineering. For instance, when designing concert halls or recording studios, engineers need to carefully manage sound levels to ensure optimal listening experiences and prevent hearing damage. By using the decibel scale and understanding the relationship between sound intensity and loudness, they can make informed decisions about room acoustics, speaker placement, and soundproofing materials. This ensures that the sound is clear, balanced, and safe for the audience or performers.
