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?
L = loudness, in decibels
I = sound intensity, in watts/m²
I₀ = 10⁻¹² watts/m²
Options: 0.25, 0.78, 0.95, 0.99
Answer (2)
To compare the sound intensities of the two hockey games, we need to understand the relationship between sound intensity and decibels.
The loudness level in decibels (dB) is given by the formula:
L = 10 lo g 10 ( I 0 I )
where:
L is the loudness level in decibels,
I is the sound intensity in watts per square meter ( W/m 2 ),
I 0 is the reference sound intensity, typically 1 0 − 12 W/m 2 .
We are given:
The loudness of the first game ( L 1 ) is 112 dB,
The loudness of the second game ( L 2 ) is 118 dB.
We need to find the fraction I 2 I 1 where I 1 is the intensity at the first game, and I 2 is the intensity at the second game.
**Step 1: Express I 1 and I 2
Using the formula:
For the first game: 112 = 10 lo g 10 ( I 0 I 1 )
Simplifying for I 1 : I 0 I 1 = 1 0 11.2 I 1 = 1 0 11.2 × I 0
For the second game: 118 = 10 lo g 10 ( I 0 I 2 )
Simplifying for I 2 : I 0 I 2 = 1 0 11.8 I 2 = 1 0 11.8 × I 0
**Step 2: Find the fraction I 2 I 1
I 2 I 1 = 1 0 11.8 × I 0 1 0 11.2 × I 0
The I 0 cancels out, simplifying to: I 2 I 1 = 1 0 11.2 − 11.8
I 2 I 1 = 1 0 − 0.6
Calculate 1 0 − 0.6 : 1 0 − 0.6 ≈ 0.2512
Rounding this value to two decimal places, we get approximately 0.25.
Therefore, the fraction of the sound intensity of the first game to the second game is 0.25 .
The correct answer from the given options is 0.25.
The fraction of the sound intensity of the first hockey game to the second is 0.25. This calculation is based on the relationship between sound intensity and decibels using the formula L = 10 lo g 10 ( I 0 I ) . Therefore, the correct answer from the given options is 0.25.
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