Problem 57 - Pearson Physics

Question

A sound wave passes through an area of \(1.8 \mathrm{~m}^{2}\) with an intensity of \(4.4 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). What is the power of this sound?

Step-by-Step Solution

Answer

The power of the sound wave is \(7.92 \times 10^{-4} \mathrm{~W}\).

Step 1: Understand the Problem

We need to find the power of a sound wave given its intensity and the area through which it passes. The problem provides the values of intensity \(I = 4.4 \times 10^{-4} \mathrm{~W/m}^2\) and area \(A = 1.8 \mathrm{~m}^2\).

Step 2: Apply the Formula for Power

The formula to find the power \(P\) of a sound wave is: \[ P = I \times A \] where \(I\) is the intensity and \(A\) is the area.

Step 3: Substitute the Given Values

Substitute \(I = 4.4 \times 10^{-4} \mathrm{~W/m}^2\) and \(A = 1.8 \mathrm{~m}^2\) into the formula: \[ P = (4.4 \times 10^{-4}) \times 1.8 \]

Step 4: Calculate the Power

Multiply the values: \[ P = 7.92 \times 10^{-4} \mathrm{~W} \] This calculation gives us the power of the sound wave.

Key Concepts

Sound Intensity

Sound intensity is a measure of the energy carried by a sound wave per unit area, and is usually expressed in watts per square meter (W/m²). Understanding sound intensity helps us determine how powerful a sound wave is when it passes through a particular area. A higher intensity indicates a louder or more powerful sound, while a lower intensity reflects a quieter sound.

Whenever sound travels, it spreads out from the source, typically spherically, unless guided by a specific pathway. This spreading means the intensity decreases as the distance from the source increases. Consider it like ripples in a pond: the farther you are from the splash, the softer the ripple.

• Intensity (I): Defined as the power (P) passing through a unit area (A), given by the formula: \( I = \frac{P}{A} \).

By rearranging this formula, we can calculate the power if the intensity and area are known, using \( P = I \times A \).

Physics Problem Solving

Physics problem solving is about breaking down the problem into understandable components.The power of physics lies in simplifying complex situations by using fundamental equations.In the case of sound power calculation, it's crucial to follow a structured approach.

Here are easy steps to solve similar problems:

• Identify known values: Start by identifying what you know. In this exercise, the known values are sound intensity and area.
• Choose the right formula: Once you know the variables you have and those you need, find the appropriate formula. Here, use \( P = I \times A \).
• Substitute and solve: Insert the known values into the formula and perform the necessary calculations.
• Verify results: Double-check your calculations to make sure there are no mistakes.

Through these steps, students will develop a systematic approach that can help them tackle numerous physics problems with confidence.

Area and Power Relation

Understanding the relationship between area and power is essential in various fields, especially when dealing with waves like sound. The power transmitted by a wave is how much energy it carries over a period and this can be spread over different areas depending on how the wave propagates.

The key concept is the direct proportionality between power (P) and area (A) when intensity (I) is constant. This means that for the same sound wave:

• Doubling the area: When you double the area the wave covers, the power of the sound also doubles, assuming the same intensity.
• Halving the area: If the area is reduced by half, the power of the sound also halves.

This relation is useful in explaining how sound behaves in a variety of environments, like open fields versus enclosed rooms. Sound engineers and acousticians manipulate these variables to achieve desired results in settings like concert halls to ensure sound reaches the audience evenly.