Problem 51 - Pearson Physics

Question

The power of a point source of sound and the distance at which the sound is heard are given below for four different cases. Rank cases A through D in order of increasing intensity. Indicate ties where appropriate. Case A: \(P=10 \mathrm{~W}, r=1 \mathrm{~m}\) Case B: \(P=20 \mathrm{~W}, r=2 \mathrm{~m}\) Case C: \(P=50 \mathrm{~W}, r=5 \mathrm{~m}\) Case D: \(P=100 \mathrm{~W}, r=5 \mathrm{~m}\)

Step-by-Step Solution

Answer

Order of increasing intensity: C, D, B, A.

Step 1: Understanding Intensity

The intensity of sound at a distance from a point source is defined by the formula \( I = \frac{P}{4\pi r^2} \), where \( I \) is the intensity, \( P \) is the power of the source, and \( r \) is the distance from the source. Our goal is to compute the intensity for each case.

Step 2: Calculating Intensity for Case A

For Case A, the power \( P = 10 \, \mathrm{W} \) and the distance \( r = 1 \, \mathrm{m} \). Applying the formula, we get: \[ I_A = \frac{10}{4\pi (1)^2} = \frac{10}{4\pi} \].

Step 3: Calculating Intensity for Case B

For Case B, the power \( P = 20 \, \mathrm{W} \) and the distance \( r = 2 \, \mathrm{m} \). Applying the formula, we get: \[ I_B = \frac{20}{4\pi (2)^2} = \frac{20}{16\pi} = \frac{5}{4\pi} \].

Step 4: Calculating Intensity for Case C

For Case C, the power \( P = 50 \, \mathrm{W} \) and the distance \( r = 5 \, \mathrm{m} \). Applying the formula, we get: \[ I_C = \frac{50}{4\pi (5)^2} = \frac{50}{100\pi} = \frac{1}{2\pi} \].

Step 5: Calculating Intensity for Case D

For Case D, the power \( P = 100 \, \mathrm{W} \) and the distance \( r = 5 \, \mathrm{m} \). Applying the formula, we get: \[ I_D = \frac{100}{4\pi (5)^2} = \frac{100}{100\pi} = \frac{1}{\pi} \].

Step 6: Ranking Intensities

We have calculated the intensities: \( I_A = \frac{10}{4\pi} \), \( I_B = \frac{5}{4\pi} \), \( I_C = \frac{1}{2\pi} \), \( I_D = \frac{1}{\pi} \). Convert these to comparable terms: \( I_A \approx 0.7958 \), \( I_B \approx 0.3979 \), \( I_C \approx 0.1592 \), \( I_D \approx 0.3183 \). Rank them in increasing order of intensity: Case C, Case D, Case B, Case A.

Key Concepts

Point Source

In acoustics, a **point source** refers to an idealized source of sound that emits waves equally in all directions. This source is considered to exist at a single point in space. Unlike real-world sources, which may have dimensions and shapes, the point source simplifies calculations by assuming the distribution of sound in a spherical pattern.

This simplification is useful because it allows us to use basic geometrical principles such as the inverse square law to calculate the intensity at varying distances from the point. Point sources are foundational in acoustics because they help in understanding how sound propagates through a medium. For practical purposes, anything from a small speaker to machinery can be modeled as a point source when considering distances much larger than the size of the source itself.

Inverse Square Law

The **Inverse Square Law** explains how the intensity of sound diminishes as it moves away from the source. According to this law, the intensity is inversely proportional to the square of the distance from the source. This means that if you double the distance from a sound source, the intensity becomes one-fourth.

Mathematically, the relationship can be expressed as:\[I = \frac{P}{4\pi r^2}\]where:

• \( I \) is the intensity of sound
• \( P \) is the power of the sound source
• \( r \) is the distance from the source

This principle is critical in acoustics as it governs not just how sound spreads in an open area, but also provides insights into designing audio systems and understanding sound pollution. By using this law, sound intensity can be effectively managed in practical applications, ensuring optimal sound levels are maintained.

Sound Power

**Sound Power** is the total energy emitted by a sound source per unit time. It is measured in watts (W) and is a crucial parameter because it remains constant regardless of the distance from the source. Unlike sound intensity, which can vary depending on location relative to the source, sound power is a fixed characteristic of the source itself.

Understanding sound power helps in evaluating how effective a sound source would be over large distances. In our exercise, each case presents a different power level, illustrating how sources with higher power levels can produce greater intensity of sound, assuming distance and other factors remain the same.

When measuring or characterizing sound, distinguishing between sound intensity and sound power is essential, as each provides different insights into sound behavior and control.

Distance and Intensity

**Distance and Intensity** are directly related in the study of sound propagation. As distance increases from the sound source, the intensity of the sound decreases following the inverse square law. This relationship explains why sounds become fainter as you move away from the source. It is crucial for designing sound systems and for understanding the impact of environmental noise.

The exercise showcases this concept by calculating and comparing the intensity for various configurations of power and distance. For instance, case A with higher power and closer proximity results in higher intensity compared to a lower power or greater distance.

Sound engineers and environmental scientists use the understanding of distance and intensity to address concerns ranging from venue acoustics to urban noise pollution. This knowledge is key to balancing sound quality with factors like audience positioning and environmental considerations.