Problem 58 - Pearson Physics

Question

What is the intensity of a sound from a 25 -W point source at a distance of \(5.1 \mathrm{~m}\) ?

Step-by-Step Solution

Answer

The intensity of the sound is approximately 0.0765 W/m².

Step 1: Understand the problem

We need to find the intensity of sound from a point source with a given power and at a certain distance. The formula for intensity is related to power and area.

Step 2: Recall the formula for intensity

The intensity \( I \) of a sound is given by the formula \( I = \frac{P}{A} \), where \( P \) is the power of the source and \( A \) is the surface area over which the power is distributed.

Step 3: Find the area of a sphere at given distance

Since sound propagates spherically from the point source, the area \( A \) is the surface area of a sphere. \( A = 4\pi r^2 \), where \( r \) is the distance from the source. Substituting \( r = 5.1 \) m, we get \( A = 4\pi (5.1)^2 \).

Step 4: Calculate the surface area

Compute \( A = 4\pi (5.1)^2 = 4 \pi \times 26.01 \approx 326.725 \) m².

Step 5: Calculate the intensity

Using the formula \( I = \frac{P}{A} \), with \( P = 25 \) W and \( A \approx 326.725 \) m², the intensity \( I \) is \( I = \frac{25}{326.725} \approx 0.0765 \) W/m².

Key Concepts

Point Source

In acoustics, a point source is a theoretical source of sound that's so small it can be considered as a single point from which sound radiates uniformly in all directions. This simplification helps us analyze and calculate how sound intensity diminishes with distance. Real-world examples might include a small, concentrated loudspeaker or a person speaking in an open environment. From a practical perspective:

• The point source assumption is valid when the distance from the sound source is much larger than the size of the source itself.
• Sound travels outward in a spherical pattern, spreading the energy over an increasing area as the distance grows.
• This model helps link the power output of the source to the intensity at a given distance from the source.

Surface Area of a Sphere

When sound emanates from a point source, it distributes itself on the surface of an expanding sphere as it travels. The surface area of this sphere is crucial in determining how sound intensity is calculated at various distances. The equation to determine the surface area of a sphere is:\[ A = 4\pi r^2 \]where:

• \( A \) is the surface area,
• \( r \) is the radius or distance from the point source,
• \( \pi \) is the mathematical constant approximately equal to 3.14159.

This equation illustrates that as the distance from the sound source doubles, the surface area increases fourfold, significantly impacting the intensity at that distance.

Power and Intensity Relationship

The relationship between power and intensity in acoustics is a fundamental principle that helps us understand how sound energy spreads over distance. Sound power (\( P \)) refers to the total energy emitted by the source per unit of time, measured in watts (W).
Sound intensity (\( I \)), on the other hand, quantifies how much power passes through a given area, and is measured in watts per square meter (W/m²). The mathematical relationship is given by:
\[ I = \frac{P}{A} \]where:

• \( I \) is the intensity.
• \( P \) is the power of the sound source.
• \( A \) is the area over which the sound is spread.

This ratio becomes central when dealing with point sources, as the power remains constant, but the area increases with distance, leading to a decrease in intensity.

Acoustics

Acoustics is the science of sound, focusing on how it is produced, controlled, transmitted, and received. Understanding acoustics is vital for designing audio systems, buildings, and even musical instruments to achieve desired sound properties.

• It deals with the study of mechanical waves in gases, liquids, and solids.
• A critical part involves analyzing how sound waves travel and interact with different environments.
• In practical applications, such as concert halls or recording studios, acoustics informs decisions to enhance sound quality and minimize unwanted echoes or noise.

Sound Propagation

Sound propagation refers to how sound waves travel through different media. When a sound is generated by a point source, it propagates in a spherical pattern, travelling through the air or other materials.

• In general, sound needs a medium like air, water, or solid materials to travel, as it cannot propagate through a vacuum.
• Factors such as temperature, medium, and obstacles can affect how sound propagates.
• As sound travels away from its source, its intensity decreases because the energy is spread over a larger area, which is why distant sounds seem fainter.

Understanding sound propagation is essential for optimizing its delivery and comprehension in diverse environments, ensuring clarity and fidelity in communication and entertainment systems.