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

Short Answer

Expert verified
There are approximately 7943 people at the game.

Step by step solution

01

Understand the Decibel Formula

The sound intensity level in decibels (dB) is calculated using the formula \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( L \) is the intensity level in dB, \( I \) is the intensity in watts per square meter, and \( I_0 \) is the reference intensity, typically \( 10^{-12} \) W/m².
02

Determine Individual and Total Intensity

For one person shouting, the intensity level is 60.0 dB, so we set \( L_1 = 60.0 \). For the total shouting, \( L_N = 109 \) dB. Using the formula \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), we convert these levels back to intensities \( I_1 \) and \( I_N \).
03

Convert Decibel Levels to Intensities

Use \( I_1 = I_0 \times 10^{\frac{60}{10}} \) to find the intensity of one person shouting. \( I_1 = 10^{-12} \times 10^6 = 10^{-6} \) W/m². Similarly, calculate the total intensity \( I_N = I_0 \times 10^{\frac{109}{10}} = 10^{-12} \times 10^{10.9} \).
04

Calculate Total Shouting Intensity

Calculate \( I_N = 10^{-12} \times 10^{10.9} = 7.9433 \times 10^{-3} \) W/m². This is the combined intensity of all the people shouting.
05

Solve for the Number of People

Since each person contributes \( I_1 = 10^{-6} \) W/m², the number of people \( N \) can be found using \( N = \frac{I_N}{I_1} = \frac{7.9433 \times 10^{-3}}{10^{-6}} = 7943.3 \).
06

Round to the Nearest Whole Number

Since the number of people must be a whole number, round \( 7943.3 \) to \( 7943 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Decibel Formula
Decibels are often used to express sound intensity levels, simplifying comparisons of vastly different energy levels. The decibel formula is essential in acoustics, enabling easier understanding and manipulation of sound data. Intensities can be challenging to comprehend due to their potentially large or small values. For instance, sound intensities range dramatically from whisper-like soft sounds to the roaring engines of an airplane. The decibel formula resolves this by using a logarithmic scale, compressing large dynamic ranges into manageable numbers. The formula itself, \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), describes the relationship between the sound intensity \( I \) and a reference intensity \( I_0 \). Here, \( I_0 \) typically represents the faintest sound a human ear can hear, set at \( 10^{-12} \) W/m². By employing this formula, we can effectively calculate and communicate sound intensity levels with ease.
Logarithms in Physics
Logarithms, particularly in the realm of physics, simplify calculations involving exponential growth or decay. This simplification is crucial when dealing with physical phenomena like sound intensity, which can span several orders of magnitude.When using a logarithmic scale, immense or minuscule values are compressed into more manageable figures. Hence, logarithms transform multiplicative relationships into additive ones. This is why you see the sound intensity formula expressed logarithmically. Breaking down changes in sound intensity requires adding or subtracting numbers instead of multiplying, thanks to logarithms. For instance, when sound intensity increases from \(60.0\) to \(109\) dB, mixing decibels becomes a straightforward task, making logarithms invaluable in solving acoustics-related problems.
Acoustics
Acoustics, the science of sound, dives into how sound waves interact within different environments. It encompasses a broad spectrum of phenomena including sound travel, reflection, absorption, and intensity measures.Acoustic principles are vital in various fields such as architecture, engineering, and environmental science. For example, acoustics help design stadiums to ensure spectators hear announcements clearly, or in soundproofing rooms to block external noise.In our problem, acoustics helps us understand how the intensity level at the center of a football field changes when more people join in shouting. With acoustics, sound intensity levels, like our example of \(60\) dB increasing to \(109\) dB, can be examined and utilized to determine the combined impact of multiple sound sources.
Intensity Calculation
Understanding the process of intensity calculation is essential when dealing with sound measurement. In this scenario, we calculate both the intensity of a single person's shout and the collective intensity of everyone.First, to find the intensity from a single individual's shout, we consider the given \(60.0\) dB sound level. Using the decibel formula, we calculate the corresponding intensity as \(I_1 = 10^{-6}\) W/m².Next, for all the shouting, use the given \(109\) dB. This gives us the total intensity \(I_N = 7.9433 \times 10^{-3}\) W/m².To find the number of participants, divide the total intensity by the intensity of one individual's shouting. \(N = \frac{I_N}{I_1} = \frac{7.9433 \times 10^{-3}}{10^{-6}} \). This gives us approximately \(7943\) people. Rounding ensures a whole number, as fractional people do not make sense in practical scenarios. This method effectively calculates and illustrates how intensity sums up in a loud environment like a football game.

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Most popular questions from this chapter

A source of sound is located at the center of two concentric spheres, parts of which are shown in the drawing. The source emits sound uniformly in all directions. On the spheres are drawn three small patches that may or may not have equal areas. However, the same sound power passes through each patch. The source produces 2.3 W of sound power, and the radii of the concentric spheres are \(r_{\mathrm{A}}=0.60\) m and \(r_{\mathrm{B}}=0.80 \mathrm{m} .\) (a) Determine the sound intensity at each of the three patches. \((\mathrm{b})\) The sound power that passes through each of the patches is \(1.8 \times 10^{-3} \mathrm{W} .\) Find the area of each patch.

Using an intensity of \(1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\) as a reference, the threshold of hearing for an average young person is 0 \(\mathrm{dB}\) . Person 1 and person \(2,\) who are not average, have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{dB}\) and \(\beta_{2}=+12.0 \mathrm{dB}\) . What is the ratio \(I_{1} / I_{2}\) of the sound intensity \(I_{1}\) when person 1 hears the sound at his own threshold of hearing compared to the sound intensity \(I_{2}\) when person 2 hears the sound at his own threshold of hearing?

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1250 Hz. The bird-watcher, however, hears a frequency of 1290 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

Argon (molecular mass \(=39.9\) u) is a monatomic gas. Assuming that it behaves like an ideal gas at \(298 \mathrm{K}(\gamma=1.67),\) find (a) the rms speed of argon atoms and (b) the speed of sound in argon.

The tension in a string is 15 N, and its linear density is 0.85 kg/m. A wave on the string travels toward the x direction; it has an amplitude of 3.6 cm and a frequency of 12 Hz. What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\)

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