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Two sound waves have intensities of \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) and \(500 \mu \mathrm{W} / \mathrm{cm}^{2}\). What is the difference in their intensity levels? Call the \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) sound \(A\), and the other \(B\). Then $$ \begin{array}{l} \beta_{A}=10 \log _{10}\left(\frac{I_{A}}{I_{0}}\right)=10\left(\log _{10} I_{A}-\log _{10} I_{0}\right) \\ \beta_{B}=10 \log _{10}\left(\frac{I_{B}}{I_{0}}\right)=10\left(\log _{10} I_{B}-\log _{10} I_{0}\right) \end{array} $$ Subtracting \(\beta_{A}\) from \(\beta_{B}\), $$ \begin{aligned} \beta_{B}-\beta_{A} &=10\left(\log _{10} I_{B}-\log _{10} I_{A}\right)=10 \log _{10}\left(\frac{I_{B}}{I_{A}}\right) \\ &=10 \log _{10}\left(\frac{500}{10}\right)=10 \log _{10} 50=(10)(1.70) \\ &=17 \mathrm{~dB} \end{aligned} $$

Step by step solution

01

Understand the Intensity Level Formula

The intensity level of a sound in decibels (dB) is given by the formula: \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where \(I\) is the intensity of the sound and \(I_0\) is the reference intensity, typically \(1 \mu \text{W/cm}^2\).

02

Calculate Intensity Levels for Sound A and B

Using the formula, calculate the intensity levels \(\beta_A\) for sound A with intensity \(I_A = 10 \mu\text{W/cm}^2\) and \(\beta_B\) for sound B with intensity \(I_B = 500 \mu\text{W/cm}^2\). \[ \beta_A = 10 \log_{10} \left( \frac{10}{1} \right) = 10 \times 1 = 10 \text{ dB} \] \[ \beta_B = 10 \log_{10} \left( \frac{500}{1} \right) = 10 \times 2.7 = 27 \text{ dB} \] (The numbers 1 and 2.7 are results of \(\log_{10}(10)\) and \(\log_{10}(500)\) respectively.)

03

Determine the Difference in Intensity Levels

Subtract the intensity level of sound A from that of sound B: \[ \beta_B - \beta_A = 27 \text{ dB} - 10 \text{ dB} = 17 \text{ dB} \]

04

Confirm Calculation Using Ratio

As an alternative method, calculate the logarithmic difference directly: \[ \beta_B - \beta_A = 10 \log_{10} \left( \frac{I_B}{I_A} \right) = 10 \log_{10} \left( \frac{500}{10} \right) = 10 \log_{10} 50 \] Since \(\log_{10} 50 \approx 1.7\), the calculation is \[ 10 \times 1.7 = 17 \text{ dB} \] Confirming the difference is 17 dB.

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Key Concepts

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

Decibels

Decibels (dB) are a unit used to measure the intensity level of sounds. In essence, decibels quantify how loud a sound is perceived. The concept of decibels is fundamental in understanding sound measurement because it simplifies a broad range of sound intensities into manageable figures. Since sound intensity can vary vastly, decibels as a measurement allow us to express these variations in a more understandable way.

• Decibels are a relative unit, which means they compare the sound intensity to a reference level.
• This reference intensity level is typically set at the threshold of hearing, which is around \(1 \mu \text{W/cm}^2\).
• By using decibels, we convert complex sound intensity into an easier-to-interpret numerical format.

Understanding decibels not only aids in physics problems but also has practical applications in everyday life, such as in acoustics engineering and environmental noise standards.

Logarithmic Scale

The logarithmic scale is fundamental when dealing with sound intensities because it allows us to manage and compute large numerical ranges more easily. Unlike a linear scale, where numbers change by addition, a logarithmic scale changes by multiplication. This makes it especially suited for measuring sound that can vary drastically in its intensity.

• The formula \( \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \) uses the logarithmic scale to represent intensity levels in decibels.
• Each 10-unit jump on the decibel scale represents a tenfold increase in sound intensity.
• This feature of the logarithmic scale means that what might seem like a small change in the decibel measurement actually corresponds to a large change in sound intensity.

This representation is much more manageable and allows us to perceive changes and differences in sound levels effectively.

Intensity Ratio

Intensity ratio is a critical aspect when calculating the difference in intensity levels of two sounds using decibels. It refers to the comparison of their actual energy levels. In the given problem, identifying the intensity ratio was crucial to determining the difference in decibels between the two sound sources.

• The formula \( \beta_B - \beta_A = 10 \log_{10} \left( \frac{I_B}{I_A} \right) \) reveals that the difference in intensity levels is directly related to the intensity ratio of the two sounds.
• This formula helps understand that the intensity ratio tells us how many times louder one sound is compared to another.
• In this context, the intensity ratio of sound B to sound A being 50 (i.e., \(\frac{500}{10}\) ) made the decibel difference 17 dB.

Understanding intensity ratios is not just about solving problems; it’s about grasping the proportional changes in sound intensity observed every day.

Physics Problem Solving

Physics problem solving often involves applying mathematical relationships to physical situations to find unknown quantities. In the case of decibels and sound intensity, problem-solving becomes a matter of applying the decibel formula to calculate sound levels and differences.

• An essential step in physics problem-solving is understanding the relevant formulas and knowing how to manipulate them.
• Carefully substituting the given values into these formulas is key to correctly calculating the desired outcomes, as shown in the problem where sound A and sound B's levels were computed.
• Another crucial aspect is double-checking your calculations using alternative methods, like using both direct subtraction and intensity ratio approaches to confirm results.

Problem-solving in physics is not just about performing calculations but also about understanding the concepts and reasoning through the given data and relationships.