Problem 27 - University Physics with Modern Physics

Question

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 \(\mathrm{W} / \mathrm{m}^{2}\) at a distance of 4.3 \(\mathrm{m}\) from the source. (a) What is the intensity at a distance of 3.1 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

Step-by-Step Solution

Answer

(a) Intensity at 3.1 m is approximately 0.068 W/m². (b) The source emits about 29,023.2 J in one hour.

Step 1: Understanding Sound Intensity Formula

Sound intensity (\( I \)) is defined as the power (\( P \)) per unit area, where the area of a sphere is calculated as \( 4 \pi r^2 \) for a point source radiating equally in all directions. Thus, the formula for intensity is given by \( I = \frac{P}{4 \pi r^2} \).

Step 2: Solving for Source Power

Using the given intensity at 4.3 m, \( I = 0.026 \text{ W/m}^2 \), we rearrange the formula to solve for power. Therefore, \( P = I \times 4 \pi r^2 = 0.026 \times 4 \pi \times (4.3)^2 \approx 8.062 \text{ W} \).

Step 3: Finding Intensity at 3.1 m

Now, using the power found above, calculate the intensity at 3.1 m using \( I = \frac{P}{4 \pi (3.1)^2} \). Thus, \( I \approx \frac{8.062}{4 \pi \times (3.1)^2} \approx 0.068 \text{ W/m}^2 \).

Step 4: Calculating Energy Emission

To find the total energy emitted per hour, first determine the total power output per second, which is \( 8.062 \text{ W} \). As 1 W is 1 J/s, in an hour (\( 3600 \text{ s} \)), the energy emitted will be \( 8.062 \times 3600 \approx 29023.2 \text{ J} \).

Key Concepts

Point Source

A point source is an idealized concept used to describe the origin of a sound wave. This means the sound originates from a single point in space, much like how a pebble causes ripples in a pond. From a point source, sound emissions spread out equally in all directions, forming spheres around the source. This characteristic is crucial when modeling sound waves in physics.

For sound waves spreading from a point source, the distance from the source is important. The greater the distance from the point source, the more the sound disperses and the area over which it spreads grows larger. This is essential when calculating sound intensity, as the intensity decreases when measured further from the source.

When examining the behavior of sound from a point source:

• Sound spreads out spherically, which means any calculations involving sound intensity assume it radiates evenly.
• This assumption simplifies real-world problems where a sound may not emanate uniformly due to obstacles or absorption.
• The model works well for basic exercises and theoretical studies, enabling us to derive other properties like intensity and energy emission efficiently.

Energy Emission

Energy emission describes the total energy that a sound source releases over time. This is directly related to the power output of the sound source and the duration of time it is active.

The principle revolves around converting the sound's power output over a specific period into total energy. For instance, when a source has a constant power (such as 8.062 W, from the prior solution), it's key to multiply this power by the time in seconds to find the total energy emitted.

In the given solution, the sound energy emitted per hour was calculated using the formula:

• Calculating total seconds in an hour: 3600 seconds.
• Using the power of 8.062 W, total energy becomes: \( 8.062 \times 3600 = 29023.2 \text{ J} \).

Understanding energy emission is essential in various fields:

• In acoustics, it helps determine how much energy is needed to produce a certain sound level over a distance.
• In engineering, knowing energy outputs guides the design of audio equipment and speakers.
• It is crucial in environmental studies to assess noise pollution and its effects on nearby habitats.

Power Output

Power output is a fundamental indicator of a source's strength, describing how much energy is converted into sound per unit time. It is expressed in watts (W), representing joules per second, similar to electrical power.

To calculate power output, we rearrange the intensity formula, utilizing the measured intensity and distance from the sound source. The formula used is:

• \(I = \frac{P}{4 \pi r^2}\) where \(I\) is the intensity, \(P\) is the power, and \(r\) is the radius or distance from the point source.
• Solving for power: \(P = I \times 4 \pi r^2\).
• For example, given \(I = 0.026\text{ W/m}^2\) at a distance of 4.3 m, the power output P is approximately 8.062 W.

Power output helps in various applications:

• Designing audio systems to achieve desired sound levels.
• Assessing the viability of sound systems in large spaces, like auditoriums.
• Comparing efficiencies among different sound-producing devices.

Understanding the concept of power output ensures sound devices are both effective and tailored to their intended settings.