Problem 27 - University Physics with Modern Physics

Question

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 W/m\(^2\) at a distance of 4.3 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) The intensity at 3.1 m is approximately 0.047 W/m². (b) The sound energy emitted in one hour is about 21766 J.

Step 1: Understand the problem

We are given measurements related to sound intensity at two different distances from a point source. The sound spreads spherically. We need to first find the intensity at a different distance and then determine the energy emitted over time.

Step 2: Recall the formula for intensity

The intensity of a sound wave at a distance from a point source is given by \( I = \frac{P}{4\pi r^2} \), where \( I \) is the intensity, \( P \) is the power of the source, and \( r \) is the distance from the source.

Step 3: Use the inverse square law for part (a)

According to the inverse square law, intensity is inversely proportional to the square of the distance: \( I_1 \times r_1^2 = I_2 \times r_2^2 \). We use this law to find the intensity at 3.1 m. With \( I_1 = 0.026 \text{ W/m}^2 \), \( r_1 = 4.3 \text{ m} \), and \( r_2 = 3.1 \text{ m} \): \[ I_2 = \frac{I_1 \times r_1^2}{r_2^2} = \frac{0.026 \times (4.3)^2}{(3.1)^2} \approx 0.047 \text{ W/m}^2 \]

Step 4: Calculate power output of the source

To find the power, rearrange the intensity formula: \( P = I \times 4\pi r^2 \). Using \( I_1 \) and \( r_1 \), we get: \[ P = 0.026 \times 4\pi \times (4.3)^2 \]

Step 5: Calculate power numerically

Substituting in the values for \( r_1 \), we calculate: \[ P = 0.026 \times 4\pi \times 18.49 = 6.046 \text{ W} \]

Step 6: Calculate energy output for part (b)

Energy is power multiplied by time. Convert the time to seconds for one hour (3600 seconds): \[ E = P \times t = 6.046 \times 3600 \approx 21766 \text{ J} \]

Step 7: Verify final calculations

Cross-check all calculations to ensure each step is correct and ensure that units of measure are consistent. Each value should logically follow from the other.

Key Concepts

Inverse Square Law

The inverse square law is a principle that describes how sound intensity behaves in relation to distance from a point source. When a sound wave emanates from a point source, it spreads out uniformly in all directions. This spreading means that the intensity decreases as you move away from the source.

The law states that intensity (I) is inversely proportional to the square of the distance (r) from the source. Mathematically, this relationship can be expressed as:

• \( I \propto \frac{1}{r^2} \)
• Or, for two points at distances \( r_1 \) and \( r_2 \), we have: \( I_1 \times r_1^2 = I_2 \times r_2^2 \)

This principle means that if you double the distance from the sound source, the intensity of the sound wave becomes one-fourth of its original intensity.

In practical scenarios, like the one in the exercise, you can use this law to calculate the intensity of a sound at a given distance if you know the intensity at a different distance. This helps in understanding how sound diminishes as it travels away from its source.

Point Source

A point source in acoustics refers to an idealized source of sound that emits waves uniformly in all directions. It's theoretical but helps in simplifying complex problems involving sound propagation.

Visualize a hypothetical small, spherical object emitting sound waves. The waves radiate outwards, forming concentric spheres. Each sphere represents a wavefront where the sound is of equal intensity. This setup assumes no obstacles or other interference in the surrounding medium.

While real-world sound sources are rarely perfect point sources, the approximation allows calculations to focus on how sound behaves over distance without worrying about complex environmental factors. Engineers or scientists may assume items like loudspeakers or small alarms as point sources for practical calculations. Their compact design and approximately even distribution of sound in all directions allow them to model sound propagation easily.

Sound Wave Energy

Sound wave energy is the total energy carried by sound waves, usually measured in Joules. For a point source, the energy emitted over a particular time depends on its power output, which is the amount of energy produced per unit of time, often in Watts.

The sound power (P) is a constant flow of sound energy that a source emits. Power is related to intensity (I) through the formula: \( P = I \times 4\pi r^2 \), where \( r \) is the distance from the source. This equation helps convert local intensity measurements into total power emitted by the source.

To find out how much energy a source emits over time, multiply the power by the duration (in seconds). For example, a source with a power output of 6 Watts emits 6 Joules of energy every second. Over one hour (3600 seconds), the total energy expended is quite substantial. This concept is key to determining an acoustic source's efficiency and influence within a space.