Problem 50 - Pearson Physics

Question

What is the power of a point source of a sound that has an intensity of \(3.2 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(48 \mathrm{~m}\) ?

Step-by-Step Solution

Answer

The power of the sound source is approximately 0.0927 W.

Step 1: Understanding the Relationship

The problem involves calculating the power of a sound source, and to do that we can use the formula relating intensity (I), power (P), and distance (r):\[ I = \frac{P}{4\pi r^2} \]where \(I\) is the intensity, \(P\) is the power of the source, and \(r\) is the distance from the source.

Step 2: Rearranging the Formula

We need to find the power \(P\), so we rearrange the formula to solve for \(P\):\[ P = I \times 4\pi r^2 \]This formula allows us to calculate the power given intensity and distance.

Step 3: Substituting Given Values

We are given that the intensity \(I = 3.2 \times 10^{-6}\, \mathrm{W/m^2}\) and the distance \(r = 48\, \mathrm{m}\). Substituting these values into the power formula gives us:\[ P = 3.2 \times 10^{-6} \times 4 \pi \times (48)^2\]

Step 4: Calculating the Power

First, calculate \(r^2\):\[ 48^2 = 2304 \]. Then substitute and calculate \( P \):\[ P = 3.2 \times 10^{-6} \times 4 \times 3.1416 \times 2304 \]. Performing these calculations gives:\[ P \approx 0.0927 \text{ W} \].

Key Concepts

Point Source in Acoustics

In the field of acoustics, a point source is an idealized model used to represent a sound source that emits sound uniformly in all directions. Imagine a small, spherical source from which sound radiates outward equally. This model helps simplify the calculations involved in determining sound properties like intensity and power at various distances away from the source. For instance, when you stand near a speaker (assuming it behaves like a point source), the sound you hear is similar in all directions around the speaker. This concept is vital because it establishes the foundation for using simple equations to predict how sound behaves as it travels through space. Understanding a point source makes it easier for students to grasp more complex sound behavior by generalizing the source's impact uniformly across space.

Understanding the Intensity Formula

The intensity formula plays a critical role in defining sound behavior. Intensity (\( I \)) is defined as the power per unit area and is measured in watts per square meter (\( ext{W/m}^2 \)). The intensity formula, \( I = \frac{P}{4\pi r^2} \), connects power (\( P \)), intensity (\( I \)), and distance (\( r \)). It describes how sound intensity diminishes as you move away from the point source.

• Power, \( P \): Total energy emitted by the source per second.
• Distance, \( r \): The separation between the point source and the point where intensity is measured.

This formula shows that intensity decreases with the square of the distance from the source. As the distance doubles, the area over which the sound spreads quadruples, reducing the intensity perceived by any single point in the area. Understanding this principle is essential for calculations involving acoustic intensity and its change over distance.

Calculating Sound Power

Power calculation is fundamental in acoustics to understand how much energy a sound source emits. When given intensity and distance, you can calculate the power (\( P \)) of a sound source by rearranging the intensity formula: \[ P = I \times 4\pi r^2 \].
This expression tells us that power is the product of intensity and the surface area of a sphere (which is \( 4\pi r^2 \)). By using this relationship, you can solve for the power if you know how intense the sound is at a given distance.

• Example: Given intensity \( I = 3.2 \times 10^{-6} \, \mathrm{W/m^2} \) and distance \( r = 48 \, \mathrm{m} \), substitute into the formula to find power \( P \).

Understanding power calculation helps in designing and evaluating acoustic devices, ensuring that they provide the correct level of sound energy for their intended use.

Distance in Acoustics

Distance is a crucial factor in acoustics as it affects sound intensity and perceived loudness. As sound travels through space from a point source, it spreads out over larger areas, thus reducing its intensity. This concept is particularly relevant in environments like theaters, concert halls, or even outdoor settings.

• Inverse Square Law: Describes how intensity decreases proportionally to the square of the distance.
• Practical Implications: This can determine optimal speaker placement to achieve uniform sound coverage in a room.

When you calculate the distance from a source and plug it into the intensity formula, you can predict how loud the sound will be at that distance. This knowledge is essential for sound engineers and designers who need to control how sound is delivered across different environments.
Understanding distance in acoustics helps in effectively managing sound distribution, ensuring the best possible audio experience in various contexts.