Problem 21 - University Physics with Modern Physics

Question

CP A baby's mouth is 30 \(\mathrm{cm}\) from her father's ear and 1.50 \(\mathrm{m}\) from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Step-by-Step Solution

Answer

The difference in sound intensity levels is approximately 14.0 dB.

Step 1: Define the Formula for Sound Intensity Level

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

Step 2: Use the Inverse Square Law for Intensity Ratio

According to the inverse square law, the intensity \( I \) at a distance \( r \) is inversely proportional to \( r^2 \). So the ratio of intensities at two distances \( r_1 = 0.30 \) m (father) and \( r_2 = 1.50 \) m (mother) is \( \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \).

Step 3: Calculate the Intensity Ratio

Substitute the distances into the intensity ratio formula: \( \frac{I_1}{I_2} = \left( \frac{1.50}{0.30} \right)^2 = 25 \).

Step 4: Calculate the Difference in Sound Intensity Levels

The difference in sound intensity levels \( \Delta L \) is given by \( \Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right) = 10 \log_{10} (25) \).

Step 5: Compute the Logarithm and the Difference

Calculate \( \log_{10}(25) \approx 1.39794 \). Therefore, \( \Delta L = 10 \times 1.39794 = 13.9794 \). Round it to one decimal place to get \( \Delta L \approx 14.0 \) dB.

Key Concepts

Inverse Square Law

The inverse square law is a key principle in physics, particularly when discussing sound and light. It helps us understand how intensity changes with distance. According to this law, the intensity of a sound wave is inversely proportional to the square of the distance from the source. This means as you get farther from the sound source, the intensity dramatically decreases. For example, if you were twice as far from the sound source, the intensity would not just halve—it would decrease to a quarter.

• The formula: \( I \propto \frac{1}{r^2} \), where \( I \) is intensity and \( r \) is distance.
• In applications: vital in calculating how sound diminishes as it travels.

By doubling the distance, the intensity falls to one-quarter of its original value. This is crucial to know when you need to calculate sound levels at different distances, as shown in our solution.

Decibels

Decibels (dB) are a unit to measure the sound intensity level. They help us gauge how loud a sound is, based on its intensity. The decibel scale is logarithmic, meaning every 10 dB increase represents a tenfold increase in sound intensity. For instance, a sound at 20 dB is ten times more intense than one at 10 dB.

• The formula: \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \).
• \( I_0 \) is the reference intensity, set at the threshold of hearing \(10^{-12} \) W/m².

Because this scale is logarithmic, it efficiently compresses large ranges into manageable numbers and reflects how our ears perceive changes in loudness, making it a more user-friendly scale for sound engineers and physics calculations alike.

Intensity Ratio

The intensity ratio is a component that plays a crucial role in determining sound intensity differences over distances. It's a comparison between intensities at two different points. To find this ratio, you utilize the inverse square law.

• Formula: \( \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \).
• Used to compute differences in sound levels at varying distances.

In the problem’s example, finding the intensity ratio was essential to determine how much louder the baby sounds to the father compared to the mother. By understanding this ratio, we can easily compute the necessary adjustments or expectations in intensity levels at various listening points.

Problem Solving in Physics

Physics problems often require a step-by-step approach to find a solution accurately. They demand not only knowledge of formulas but also an understanding of problem context and how principles like the inverse square law apply. Here’s a basic strategy:

• Identify the problem: Understand what is being asked and recognize the principles involved, like the use of decibels and inverse square law in sound problems.
• Know the formulas: Be familiar with formulas and what each part signifies. For sound, it includes intensity level, intensity ratio, and applying logarithms.
• Apply the concepts: Use laws and calculations to propose a solution, adjusting values as needed and calculating differences when required.

Developing problem-solving skills in physics is about practice. Breaking down each variable and focusing on the steps can make even complex problems approachable and solvable, just as detailed in the original exercise solution.