Problem 58 - Pearson Physics

Question

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

Step-by-Step Solution

Answer

The intensity is approximately \(0.0765\, \mathrm{W/m^2}\).

Step 1: Understand the Point Source Model

In a point source model, sound emanates uniformly in all directions forming a sphere. The intensity of sound at a distance from the source is dependent on how the sound spreads over the surface area of this sphere.

Step 2: Determine the Surface Area of the Sphere

The surface area of a sphere with radius \( r \) is given by the formula \( 4\pi r^2 \). Here, \( r = 5.1 \mathrm{~m} \), thus the surface area is \( 4\pi (5.1)^2 \).

Step 3: Find the Intensity of the Sound

Sound intensity is the power per unit area where the area is the sphere's surface area surrounding the source. Use the formula for intensity: \( I = \frac{P}{A} \), where \( P = 25 \mathrm{~W} \) is the sound power, and \( A \) is the surface area calculated in the previous step.

Step 4: Calculate the Intensity

Calculate \( A = 4\pi (5.1)^2 \approx 326.9 \mathrm{~m^2} \). Then, compute the intensity: \( I = \frac{25}{326.9} \approx 0.0765\, \mathrm{W/m^2} \).

Key Concepts

Point Source Model

In acoustics, the point source model is a fundamental concept for understanding how sound behaves. Imagine a tiny, single point from which sound waves spread out evenly in all directions. This point source sends waves just like ripples from a stone thrown into a pond, resulting in a spherical spread of sound. It's a useful approximation for many real-world problems.

• Point Source models assume isotropic emission, meaning sound is uniform in all directions.
• It helps in determining how sound intensity changes with distance.

When using this model, the more you move away from the sound source, the larger the spherical surface over which the sound spreads. Hence, the intensity of sound diminishes with increasing distance.

Surface Area of a Sphere

The surface area of a sphere is crucial when evaluating how sound spreads in the point source model. As sound waves expand from their point of origin, they cover the surface of an imaginary sphere.The formula to find the surface area is:\[A = 4\pi r^2\]where \( r \) is the radius of the sphere. The radius corresponds to your distance from the point source.

• This formula helps us understand how much area the sound energy covers as it moves outward.
• With this, we figure out how 'stretched' the sound waves become over a given distance.

For example, if you are 5.1 meters away from the sound source, the area is computed to be \(326.9\ \mathrm{m^2}\), representing the vast space the sound occupies.

Sound Power

Sound power, denoted by \( P \), is an important element in the analysis of sound intensity. It is the total amount of sound energy emitted by the source per second and is measured in watts (W).

• Sound power is different from sound intensity. While power concerns the source, intensity refers to energy at a particular distance.
• Intensity relates to area by the formula \( I = \frac{P}{A} \).

In this scenario, a 25 W point source lets us calculate the sound intensity at any distance from it. By understanding both the concepts of power and surface area, you can determine how loud the sound will seem as you move further away from the source. In our example, at 5.1 meters away, the sound intensity ends up as approximately 0.0765 \( \mathrm{W/m^2} \), which illustrates how sound grows quiet over distance.