Problem 48 - Physics

Question

A typical adult ear has a surface area of \(2.1 \times 10^{-3} \mathrm{~m}^{2} .\) The sound intensity during a normal conversation is about \(3.2 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{2}\) at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?

Step-by-Step Solution

Answer

The power intercepted by the ear is \(6.72 \times 10^{-9}\) Watts.

Step 1: Understand the Problem

We are asked to find out how much power is captured by the ear from sound waves during a normal conversation. We have the area of the ear and the sound intensity.

Step 2: Recall the Power Formula

The power intercepted by a surface is given by the formula: \( P = I \times A \), where \( I \) is the intensity of the sound, and \( A \) is the area of the surface affected.

Step 3: Identify Values

From the problem, we know that the area \( A = 2.1 \times 10^{-3} \) m\(^2\) and the sound intensity \( I = 3.2 \times 10^{-6} \) W/m\(^2\).

Step 4: Substitute and Calculate

Substitute the values into the power formula: \[ P = (3.2 \times 10^{-6} \text{ W/m}^{2}) \times (2.1 \times 10^{-3} \text{ m}^{2}) \].

Step 5: Perform the Multiplication

Calculate \( 3.2 \times 2.1 = 6.72 \). Since we are dealing with powers of 10, use the property \( 10^{-6} \times 10^{-3} = 10^{-9} \). Thus, \( P = 6.72 \times 10^{-9} \).

Step 6: Result Interpretation

The power intercepted by the ear is \( 6.72 \times 10^{-9} \) Watts.

Key Concepts

Power Formula

When calculating the power intercepted by a surface from sound waves, we rely on the power formula. This is a fundamental equation in physics:
\[ P = I \times A \]Here, \( P \) stands for power, \( I \) is the intensity of the sound in watts per square meter (W/m\(^2\)), and \( A \) represents the area in square meters (m\(^2\)) over which the sound is distributed.
The power formula is essentially saying that the power captured depends on how intense the sound is and how large the surface area is. For example, if the sound intensity increases, more power is captured. Similarly, if the area expands, more power is intercepted. This formula is a straightforward yet powerful way to understand the relationship between sound characteristics and surface interaction.
Remember, it is imperative to have consistent units when applying this formula. If the intensity is in W/m\(^2\) and the area in m\(^2\), the resulting power will naturally be in watts (W). This makes it easy to apply and interpret in various physics problems.

Surface Area Calculation

The concept of surface area calculation is crucial when dealing with problems involving sound interception. In the given exercise, we have a specific area of the ear that sound waves strike. Knowing the surface area helps determine how much of the sound's energy is captured.
In the problem, we have a surface area \( A = 2.1 \times 10^{-3} \) m\(^2\). This tells us how much of over which the conversation sound is distributed.

• Step 1: Identify the surface area involved.
• Step 2: Ensure the area is measured in square meters for uniformity in calculations.

This attention to measurement is central in physics, ensuring all units work together seamlessly in formulas to provide meaningful and accurate results.Surface areas can vary dramatically based on the object in question. A larger surface, such as a wall, will intercept more sound energy compared to a smaller surface, like an ear. Thus, accurate measurement and understanding of the surface area are pivotal in such calculations.

Physics Problem Solving

Solving physics problems effectively requires a structured approach. Let's dive into this with the problem of calculating power from sound intensity:

• **Understanding the Problem:** Begin by clearly identifying what you need to find. In our case, it's about calculating power intercepted by the ear.
• **Using the Right Formulas:** Know the relevant formulas. Here, we employ the formula for power, \( P = I \times A \), to relate sound intensity and surface area to power.
• **Substitution and Calculation:** Carefully substitute the known values into the equation, ensuring consistency in units. Then compute the values, keeping track of number relationships, such as powers of ten in scientific notation.
• **Interpretation of Results:** Once calculated, interpret the outcome. What does \( 6.72 \times 10^{-9} \) Watts mean in practical terms? It indicates the amount of power energy from the conversation reaching the ear surface.

Following these steps helps break down complex queries into manageable parts, making physics more approachable and less daunting. While the concepts might appear challenging at first, adhering to a problem-solving strategy demystifies even the most intricate physics challenges.