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} watts / m^2[/tex]
A. 0.25
B. 0.78
C. 0.95
D. 0.99
Answer (1)
Express the relationship between loudness (L) and sound intensity (I) using the formula: L = 10 \t \t lo g ( I 0 I ) .
Calculate the ratio of sound intensities I 2 I 1 using the given loudness values L 1 = 112 dB and L 2 = 118 dB.
Simplify the expression to I 2 I 1 = 1 0 10 L 1 − L 2 = 1 0 − 0.6 .
Evaluate 1 0 − 0.6 to find the fraction of sound intensity: 0.25 .
Explanation
Understanding the Problem We are given the loudness of sound in decibels (dB) for two hockey games and asked to find the fraction of the sound intensity of the second game that was the sound intensity of the first game. We are given the formula:
L = 10 lo g ( I 0 I )
where: L = loudness, in decibels I = sound intensity, in watts / m 2 I 0 = 1 0 − 12 w a tt s / m 2
Let L 1 and I 1 be the loudness and sound intensity of the first game, and L 2 and I 2 be the loudness and sound intensity of the second game. We are given L 1 = 112 dB and L 2 = 118 dB. We want to find the ratio I 2 I 1 .
Expressing Sound Intensities Using the given formula, we have:
L 1 = 10 lo g ( I 0 I 1 ) L 2 = 10 lo g ( I 0 I 2 )
Dividing both sides of each equation by 10, we get:
10 L 1 = lo g ( I 0 I 1 ) 10 L 2 = lo g ( I 0 I 2 )
Using the property that 1 0 l o g x = x , we can rewrite the equations as:
1 0 10 L 1 = I 0 I 1 1 0 10 L 2 = I 0 I 2
Multiplying both sides of each equation by I 0 , we get:
I 1 = I 0 ⋅ 1 0 10 L 1 I 2 = I 0 ⋅ 1 0 10 L 2
Finding the Ratio Now we want to find the ratio I 2 I 1 . We have:
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 − 10 L 2 = 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
Calculating the Value We can calculate 1 0 − 0.6 :
1 0 − 0.6 ≈ 0.25118864315
Rounding to two decimal places, we get 0.25.
Final Answer The fraction of sound intensity of the second game that was the sound intensity of the first game is approximately 0.25.
Therefore, the answer is 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 from traffic, construction, and industrial activities. By understanding the logarithmic scale of decibels, they can determine the impact of different noise sources on residential areas and implement noise reduction strategies. Similarly, audio engineers use decibel measurements to optimize sound systems in concert halls and recording studios, ensuring that sound levels are safe and enjoyable for the audience. The relationship between sound intensity and decibels helps professionals manage and mitigate noise-related issues in various environments.
