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A woman stands a distance \(d\) from a loud motor that emits sound uniformly in all directions. The sound intensity at her position is an uncomfortable \(3.2 \times 10^{-3} \mathrm{W} / \mathrm{m}^{2}\) . There are no reflections. At a position twice as far from the motor, what are (a) the sound intensity and (b) the sound intensity level relative to the threshold of hearing?

Short Answer

Expert verified
The new intensity is \(0.8 \times 10^{-3} \ \mathrm{W/m}^2\) and the intensity level is 89.03 dB.

Step by step solution

01

Understanding the Given Intensity

We are given that the sound intensity at the woman's position is \(3.2 \times 10^{-3} \ \mathrm{W/m}^2\). This will be used to find the intensity at double the distance.
02

Applying the Inverse Square Law

The sound intensity \(I\) at a distance \(d\) from a source in open space decreases with the square of the distance. If the initial intensity is \(I_1\) at a distance \(d\), then the intensity \(I_2\) at a new distance \(2d\) is given by:\[I_2 = \frac{I_1}{(2)^2} = \frac{3.2 \times 10^{-3}}{4} \ \mathrm{W/m}^2\]Calculate to find \(I_2\).
03

Calculating the Intensity

Using the formula from Step 2:\[I_2 = \frac{3.2 \times 10^{-3}}{4} = 0.8 \times 10^{-3} \ \mathrm{W/m}^2\]Thus, the intensity at twice the distance will be \(0.8 \times 10^{-3} \ \mathrm{W/m}^2\).
04

Calculating Sound Intensity Level

The sound intensity level \(L\) in decibels (dB) relative to the threshold of hearing \(I_0\), which is \(1 \times 10^{-12} \ \mathrm{W/m}^2\), is calculated using the formula:\[L = 10 \log_{10} \left( \frac{I}{I_0} \right)\]Replace \(I\) with \(0.8 \times 10^{-3} \ \mathrm{W/m}^2\) to find \(L\).
05

Calculating Sound Intensity Level

Substitute \(I_2 = 0.8 \times 10^{-3} \ \mathrm{W/m}^2\) into the formula:\[L = 10 \log_{10} \left( \frac{0.8 \times 10^{-3}}{1 \times 10^{-12}} \right)\]Calculate the value inside the logarithm:\[L = 10 \log_{10} (0.8 \times 10^9)\]Now calculate the dB level.
06

Final Calculation of Sound Intensity Level

Solving the logarithm:\[L = 10 \log_{10} (8 \times 10^8) = 10 (\log_{10} 8 + 8) = 10 (0.903 + 8)\]\[L = 10 \times 8.903 = 89.03 \ \mathrm{dB}\]

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Key Concepts

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

Inverse Square Law
Sound intensity refers to the power per unit area and it diminishes as distance from the source increases. This phenomenon is explained by the Inverse Square Law. According to this principle, the intensity of sound is inversely proportional to the square of the distance from its source.

In simpler terms, if you double the distance from the sound source, the intensity becomes one-fourth. This is because the sound energy spreads out over a larger area. For example, if the initial sound intensity at distance \(d\) is \(I_1\), then at a distance of \(2d\), the intensity \(I_2\) would be calculated as:

\[I_2 = \frac{I_1}{(2)^2} = \frac{I_1}{4}\]
This concept is crucial for understanding how sound dissipates over distance and why sounds appear quieter the farther away you are from the source.
Decibels
The decibel (dB) is a logarithmic unit used to express sound intensity levels. It is relative to a reference point called the threshold of hearing. The human ear has a vast dynamic range. Therefore, the logarithmic scale compresses this range into manageable numbers.

The decibel formula for sound intensity level is given by:

\[L = 10 \log_{10} \left( \frac{I}{I_0} \right)\]
where \(I\) is the intensity of sound. \(I_0\) is the standard reference intensity of \(1 \times 10^{-12} \ \mathrm{W/m}^2\). This formula helps convert absolute sound intensity into a more comprehensible dB scale to easily convey how sound varies in loudness.

To put into context, an increase of 10 dB represents a tenfold increase in intensity, but it is perceived by the human ear as approximately twice as loud.
Physics Problem Solving
Physics problem solving involves translating real-world situations into mathematical models. This requires breaking down problems into manageable steps and applying relevant laws and formulas.

In our example, understanding how sound behaves with distance using the Inverse Square Law was the stepping stone. Calculating where the sound intensity diminishes allowed us to solve the problem step-by-step.
  • Identify what is given: sound intensity at a certain distance.
  • Apply the Inverse Square Law to find intensity at a new distance.
  • Use sound intensity to compute decibel levels using the dB formula.
  • Each step involves logical reasoning and application of physics laws to derive a solution. Consistent practice develops better problem-solving skills.

    Approach problem-solving with patience and clarity. This way, even complex problems become easier to handle.
    Threshold of Hearing
    The threshold of hearing is the quietest sound that the average human ear can detect. It is usually represented by an intensity level of \(1 \times 10^{-12} \ \mathrm{W/m}^2\).

    This baseline is critical for measuring sound levels in decibels. It serves as the reference point in our earlier dB formula, \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\).

    Understanding this concept is essential to relating everyday sounds to their decibel levels. Louder sounds correspond to higher values above the threshold, often perceived as an exponential increase in loudness.

    By knowing this, we appreciate how different environments impact our auditory experience. For instance, quiet rooms typically hover around 30 dB, while busy traffic can reach 70-85 dB. Each sound level comparison refers back to the baseline threshold, emphasizing our hearing system's sensitivity.

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    Most popular questions from this chapter

    A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. The speakers are 30.0 m apart and the total power of the sound coming from each speaker is 0.500 W. Find the total sound intensity that the man hears (a) when he is at his initial position halfway between the speakers, and (b) after he has walked 4.0 m directly toward one of the speakers.

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