Problem 100 - Physics

Question

When one person shouts at a football game, the sound intensity level at the center of the field is \(60.0 \mathrm{~dB}\). When all the people shout together, the intensity level increases to \(109 \mathrm{~dB}\). Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game?

Step-by-Step Solution

Answer

Approximately 7943 people are at the game.

Step 1: Understanding Decibel Levels

The sound intensity level in decibels (dB) relates to the intensity of the sound by the formula: \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( \beta \) is the sound level in dB, \( I \) is the intensity of the sound, and \( I_0 \) is a reference intensity, typically \( 10^{-12} \text{ W/m}^2 \). Here, we have individual intensity level, \( \beta_1 = 60 \text{ dB} \), and total intensity level, \( \beta_2 = 109 \text{ dB} \).

Step 2: Calculate Individual Intensity

Recognize that the intensity \( I_1 \) for one person can be calculated from the sound level \( \beta_1 = 60 \text{ dB} \):\[ 60 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]Rewriting gives:\[ \frac{I_1}{I_0} = 10^{6} \]Thus,\[ I_1 = 10^{6} \cdot I_0 \]

Step 3: Calculate Total Intensity

For the total intensity \( I_t \) corresponding to \( \beta_2 = 109 \text{ dB} \), we have:\[ 109 = 10 \log_{10} \left( \frac{I_t}{I_0} \right) \]Rewriting gives:\[ \frac{I_t}{I_0} = 10^{10.9} \]Thus,\[ I_t = 10^{10.9} \cdot I_0 \]

Step 4: Determine Number of People

Since each person produces the same intensity \( I_1 \), the total intensity is the sum of the intensities of all people shouting. Therefore,\[ I_t = N \cdot I_1 \]Substitute \( I_t = 10^{10.9} \cdot I_0 \) and \( I_1 = 10^{6} \cdot I_0 \) into the equation:\[ 10^{10.9} \cdot I_0 = N \cdot 10^{6} \cdot I_0 \]Cancel \( I_0 \) and solve for \( N \):\[ N = 10^{10.9 - 6} = 10^{4.9} \approx 7943 \]

Step 5: Conclusion

Therefore, approximately 7943 people are present at the game shouting together to achieve an intensity level of 109 dB.

Key Concepts

Decibel Scale

The decibel scale is a logarithmic unit used to measure sound intensity level. It's a representation that allows us to express very large or very small power quantities in an easily understandable form.
With the decibel scale, a tenfold increase in sound intensity corresponds to an increase of 10 dB. For example:

• A sound intensity of 60 dB is 10 times more intense than 50 dB.
• Similarly, a sound of 70 dB is 100 times more intense than 50 dB, as each 10 dB increment represents a tenfold increase.

This scale is crucial because our ears perceive sound intensity logarithmically. We actually perceive changes on this scale as linear changes in loudness, which is why it's so effective in sound measurement.

Sound Intensity Level

Sound Intensity Level, measured in decibels (dB), quantifies the intensity of sound over a given area. Intensity refers to the power per unit area carried by a wave.

• Sound intensity is mathematically expressed with the formula: \( \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \)
• Here, \(I\) is the sound intensity, and \(I_0\) is a reference intensity, typically \( 10^{-12} \text{ W/m}^2 \).

This threshold of hearing represents the quietest sound that can be heard by the average human ear. Sound intensity level helps us understand not only the loudness but also the impact sound can have on surroundings.

Logarithmic Scale

The logarithmic scale is crucial in plotting sound intensity levels because it transforms multiplicative processes into additive ones. This is particularly useful in sound measurements where intensity levels can vary exponentially.

• The key advantage is ease of use when dealing with a wide range of values, allowing compression of exponentially varying data.
• In sound, using a logarithmic scale like decibels makes it easier to compare different levels, as multiplication becomes simple additions.

Without the logarithmic scale, working with sound levels spanning so many orders of magnitude would be overwhelmingly complex and unintuitive for most users.

Physics Problem Solving

Physics problem-solving often requires clear steps and an organized approach. Dealing with concepts such as sound intensity involves understanding the equations and what they represent.

• Understanding the formula: grasp what each variable signifies.
• Solve systematically: break down the problem into digestible parts, as was done in the original exercise through several steps.
• Check assumptions: e.g., assuming all people produce identical sound intensity.

Careful calculation and simplification turn complex physical equations into manageable solutions, helping verify results and assumptions in the given scenario.