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?
[tex]L=10 \log \left(\frac{1}{f_0}\right)[/tex]
[tex]L=[/tex] loudness, in decibels
[tex]I=[/tex] sound intensity, in watts / [tex]m^2[/tex]
[tex]I_0=10^{-12}[/tex] watts / [tex]m^2[/tex]
A. 0.25
B. 0.78
C. 0.95
D. 0.99
Answer (2)
Express the loudness on the first night as L 1 = 10 lo g ( I 0 I 1 ) = 112 and solve for I 1 .
Express the loudness on the second night as L 2 = 10 lo g ( I 0 I 2 ) = 118 and solve for I 2 .
Find the fraction I 2 I 1 = I 0 ⋅ 1 0 11.8 I 0 ⋅ 1 0 11.2 = 1 0 − 0.6 .
Calculate 1 0 − 0.6 to find the fraction: 0.25 .
Explanation
Understanding the Problem We are given the loudness of a hockey game on two consecutive nights, L 1 = 112 dB and L 2 = 118 dB. We want to find the fraction of the sound intensity of the first game to the sound intensity of the second game, which is I 2 I 1 . We are also given the formula 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 .
Finding Sound Intensity on the First Night We can express the loudness on the first night as L 1 = 10 lo g ( I 0 I 1 ) = 112 . Dividing both sides by 10, we get lo g ( I 0 I 1 ) = 11.2 . Then, we can rewrite this in exponential form as I 0 I 1 = 1 0 11.2 , so I 1 = I 0 ⋅ 1 0 11.2 .
Finding Sound Intensity on the Second Night Similarly, we can express the loudness on the second night as L 2 = 10 lo g ( I 0 I 2 ) = 118 . Dividing both sides by 10, we get lo g ( I 0 I 2 ) = 11.8 . Then, we can rewrite this in exponential form as I 0 I 2 = 1 0 11.8 , so I 2 = I 0 ⋅ 1 0 11.8 .
Calculating the Fraction of Sound Intensities Now we want to find the fraction I 2 I 1 . We have I 1 = I 0 ⋅ 1 0 11.2 and I 2 = I 0 ⋅ 1 0 11.8 , so I 2 I 1 = I 0 ⋅ 1 0 11.8 I 0 ⋅ 1 0 11.2 = 1 0 11.8 1 0 11.2 = 1 0 11.2 − 11.8 = 1 0 − 0.6 .
Final Calculation and Answer We calculate 1 0 − 0.6 ≈ 0.25118864315 . Rounding to two decimal places, we get 0.25.
Examples
Sound intensity, measured in decibels, is used in many real-world applications. For example, city planners use decibel measurements to assess noise pollution levels near airports or highways. Similarly, audio engineers use decibel measurements to optimize sound systems in concert halls and recording studios, ensuring that the sound is loud enough to be heard clearly without causing damage to listeners' ears. Understanding decibels and sound intensity is crucial for creating safe and enjoyable sound environments.
The fraction of sound intensity of the first hockey game to the second is approximately 0.25. This was calculated using the relationship between loudness in decibels and sound intensity. Therefore, the answer is A. 0.25.
;
