Sound Intensity Fraction Calculation Between Two Hockey Games
Introduction
The world of sound and acoustics is filled with fascinating phenomena, and one of the most intriguing aspects is the measurement of sound intensity. Sound intensity, often measured in decibels (dB), is a logarithmic scale that quantifies the amount of sound energy passing through a given area. Understanding sound intensity is crucial in various fields, from music and audio engineering to environmental noise control and medical diagnostics. In this article, we will delve into a practical problem involving sound intensity measurements taken during two hockey games, exploring the relationship between decibel levels and sound intensity ratios. The specific problem we will address is: If the loudest sound measured during a hockey game on one night was 112 dB, and the loudest sound measured during the game the following night was 118 dB, what fraction of the sound intensity of the second game was the sound intensity of the first? This problem provides an excellent opportunity to apply the fundamental principles of sound intensity and decibel calculations, shedding light on the practical implications of these concepts.
Background: Decibels and Sound Intensity
Before we dive into the specific problem, it is important to understand the fundamental concepts of decibels and sound intensity.
Sound Intensity
Sound intensity (I) is defined as the power (P) of a sound wave per unit area (A). Mathematically, it is represented as:
I = P/A
Sound intensity is typically measured in watts per square meter (W/m²). It quantifies the amount of sound energy that flows through a specific area per unit of time. However, due to the vast range of sound intensities that humans can perceive, a logarithmic scale called the decibel scale is more commonly used.
Decibel Scale
The decibel (dB) scale is a logarithmic scale used to measure sound intensity level. It is based on the ratio of the intensity of a sound to a reference intensity. The formula for calculating sound intensity level (L) in decibels is:
L = 10 * log₁₀(I/I₀)
Where:
- L is the sound intensity level in decibels (dB).
- I is the sound intensity of the sound being measured.
- I₀ is the reference intensity, which is the threshold of human hearing (10⁻¹² W/m²).
The logarithmic nature of the decibel scale means that an increase of 10 dB corresponds to a tenfold increase in sound intensity. For example, a sound at 60 dB is ten times more intense than a sound at 50 dB, and 100 times more intense than a sound at 40 dB. This property makes the decibel scale a convenient way to represent the wide range of sound intensities encountered in everyday life.
Key Relationships
Understanding the relationship between decibels and sound intensity is crucial for solving problems involving sound measurements. Here are some key relationships to keep in mind:
- An increase of 10 dB corresponds to a tenfold increase in sound intensity.
- An increase of 20 dB corresponds to a hundredfold increase in sound intensity.
- An increase of 3 dB corresponds to approximately a doubling of sound intensity.
These relationships provide a quick way to estimate the relative sound intensities of different sounds based on their decibel levels.
Problem Statement
Now, let's restate the problem we aim to solve: The loudest sound measured during a hockey game one night was 112 dB. The loudest sound measured during the 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?
This problem requires us to compare the sound intensities of two sounds based on their decibel levels. To solve it, we will need to use the formula relating decibels and sound intensity, and then calculate the ratio of the sound intensities.
Solution
To solve this problem, we will use the formula that relates sound intensity level in decibels (L) to sound intensity (I):
L = 10 * log₁₀(I/I₀)
where I₀ is the reference intensity (10⁻¹² W/m²).
Step 1: Determine the sound intensities (I₁ and I₂) corresponding to the given decibel levels (L₁ and L₂).
Let L₁ = 112 dB be the sound intensity level of the first game, and L₂ = 118 dB be the sound intensity level of the second game. Let I₁ and I₂ be the respective sound intensities.
For the first game:
112 = 10 * log₁₀(I₁/10⁻¹²)
Divide both sides by 10:
11.2 = log₁₀(I₁/10⁻¹²)
To remove the logarithm, take the antilog (10 to the power of both sides):
10¹¹.² = I₁/10⁻¹²
Multiply both sides by 10⁻¹²:
I₁ = 10¹¹.² * 10⁻¹²
I₁ = 10⁻⁰.⁸ W/m²
For the second game:
118 = 10 * log₁₀(I₂/10⁻¹²)
Divide both sides by 10:
11.8 = log₁₀(I₂/10⁻¹²)
To remove the logarithm, take the antilog (10 to the power of both sides):
10¹¹.⁸ = I₂/10⁻¹²
Multiply both sides by 10⁻¹²:
I₂ = 10¹¹.⁸ * 10⁻¹²
I₂ = 10⁻⁰.² W/m²
Step 2: Calculate the fraction of the sound intensity of the second game (I₂) to the sound intensity of the first game (I₁).
The problem asks for the fraction of the sound intensity of the first game to the sound intensity of the second game, which is I₁/I₂:
Fraction = I₁/I₂
Fraction = (10⁻⁰.⁸) / (10⁻⁰.²)
Using the properties of exponents, we can simplify this expression:
Fraction = 10^(-0.8 - (-0.2))
Fraction = 10^(-0.8 + 0.2)
Fraction = 10⁻⁰.⁶
To calculate this value, we can use a calculator:
Fraction ≈ 0.251188643
Step 3: Round the result to an appropriate number of significant figures.
Since the given decibel levels have three significant figures, we should round our result to three significant figures as well.
Fraction ≈ 0.251
Therefore, the sound intensity of the first game was approximately 0.251 times the sound intensity of the second game.
Alternative Approach Using Decibel Difference
An alternative approach to solving this problem involves using the difference in decibel levels directly. Since the decibel scale is logarithmic, the difference in decibels is related to the ratio of the sound intensities. If L₁ and L₂ are the sound levels in decibels and I₁ and I₂ are the corresponding sound intensities, then:
L₂ - L₁ = 10 * log₁₀(I₂/I₁)
In our case, L₁ = 112 dB and L₂ = 118 dB, so:
118 - 112 = 10 * log₁₀(I₂/I₁)
6 = 10 * log₁₀(I₂/I₁)
Divide by 10:
- 6 = log₁₀(I₂/I₁)
Take the antilog (10 to the power of both sides):
10⁰.⁶ = I₂/I₁
We want to find the fraction I₁/I₂, which is the inverse of I₂/I₁:
I₁/I₂ = 1 / 10⁰.⁶
I₁/I₂ = 10⁻⁰.⁶
This is the same expression we obtained in the previous method, so the result will be the same:
I₁/I₂ ≈ 0.251
This alternative approach highlights the logarithmic relationship between decibels and sound intensity, providing a more direct way to calculate the ratio of sound intensities based on the difference in decibel levels.
Conclusion
In summary, we have determined that the sound intensity of the first hockey game, which measured 112 dB, was approximately 0.251 times the sound intensity of the second hockey game, which measured 118 dB. This result demonstrates the significant impact that even a relatively small difference in decibel levels can have on sound intensity. The 6 dB difference between the two measurements translates to a considerable change in the amount of sound energy present. This is because the decibel scale is logarithmic, each 10 dB increase represents a tenfold increase in sound intensity. Therefore, a sound at 118 dB is significantly more intense than a sound at 112 dB.
Understanding the relationship between decibel levels and sound intensity is crucial in various real-world applications. For instance, in audio engineering, it is essential to manage sound levels to prevent hearing damage and ensure optimal listening experiences. In environmental noise control, assessing and mitigating noise pollution often involves measuring sound levels in decibels and comparing them to established standards. In medical diagnostics, sound intensity measurements are used in procedures like audiometry to assess hearing sensitivity and diagnose hearing disorders. The problem we solved here, while seemingly simple, underscores the importance of these fundamental concepts and their relevance in practical scenarios. By applying the formula relating decibels and sound intensity, we were able to quantitatively compare the sound levels of two events and gain a deeper understanding of the nature of sound and its measurement.
Moreover, the alternative approach we explored, using the decibel difference directly, highlights the elegance and efficiency of the logarithmic scale in handling sound intensity measurements. This method provides a quick and intuitive way to estimate the ratio of sound intensities, reinforcing the practical utility of the decibel scale. In conclusion, the problem of comparing sound intensities at hockey games serves as a valuable illustration of the principles of acoustics and the importance of understanding decibels and sound intensity in a variety of contexts.
