Problem 49 - Physics

Question

At a distance of \(3.8 \mathrm{~m}\) from a siren, the sound intensity is \(3.6 \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2}\). Assuming that the siren radiates sound uniformly in all directions, find the total power radiated.

Step-by-Step Solution

Answer

Total power radiated is approximately 65.2 W.

Step 1: Understanding the Problem

We are given the sound intensity at a certain distance from a siren and need to find the total power radiated by the siren. The intensity is given as a function of distance from a point source.

Step 2: Understanding Intensity Formula

The intensity of sound at a distance from a point source is given by the formula \( I = \frac{P}{A} \), where \( I \) is the intensity, \( P \) is the power, and \( A \) is the area over which the power is spread. For a point source radiating uniformly, the area is the surface area of a sphere \( 4\pi r^2 \).

Step 3: Expressing Area in Terms of Distance

Given that the distance from the siren is 3.8 m, the formula for the surface area at this distance is \( A = 4\pi (3.8)^2 \). Calculate this to find the area over which the sound spreads.

Step 4: Solve for Power

Rewrite the intensity formula to solve for power: \( P = I \times A \). Substitute the intensity \( 3.6 \times 10^{-2} \, \mathrm{W/m^2} \) and the calculated area to find the total power \( P \).

Step 5: Calculate Total Power

Calculate \( A = 4\pi (3.8)^2 \). Then use \( P = I \times A = 3.6 \times 10^{-2} \times 4\pi (3.8)^2 \) to find \( P \). This will give the total power radiated by the siren.

Key Concepts

Sound Power

Sound power is the total energy that a sound source, like our siren, radiates over time. It is measured in watts (W) and represents the overall output, independent of the surrounding environment. For instance, irrespective of whether the siren is placed in a small room or in an open field, the sound power remains the same as long as the siren operates uniformly.

Understanding sound power helps us to quantify how much energy is being spread through space. This concept is essential for computations that allow us to determine the impact of sound in different settings.

In our exercise, knowing the total sound power helps us understand how energy is being transmitted from the siren to further points, such as 3.8 meters away. By calculating it from known values of intensity and distance, we find how energetically the siren is operating.

Intensity Formula

The intensity formula is crucial in connecting the concept of sound power with how we perceive sound at certain distances. It is given by: \[ I = \frac{P}{A} \]where:

• \( I \) is the sound intensity (watts per square meter, \( \mathrm{W/m^2} \)).
• \( P \) is the sound power.
• \( A \) is the area over which the sound power disperses.

This formula allows us to calculate the intensity based on a known power and area, or vice versa. For point sources emitting sound in all directions, as in the problem, the spreading area will generally be spherical. Understanding this gives the key to deciphering how intense the sound will be at a certain distance. Use this relationship every time you need to connect the environment and source characteristics.

Spherical Wavefront

When sound radiates from a point source like a siren, it does so by creating a spherical wavefront. Think of it like the ripples in a pond when you throw a stone. They spread out evenly in all directions.

As you measure distance from the source, sound covers larger and larger areas, forming an expanding sphere. The further you move from the siren, the more area the sound must cover, which spreads the power over the larger sphere's area.

For any given distance \( r \), the surface area, \( A \), of the sphere is calculated using: \[ A = 4\pi r^2 \]This formula helps us find the area which is essential for determining sound intensity. For example, at 3.8 meters from the source, it assists in computing the total sound area and thus the sound intensity we hear.

Uniform Radiation

Uniform radiation is a fundamental assumption in our problem. It implies that the sound energy is distributed evenly in all directions from the source.

This means that, ideally, the sound intensity would be the same at any point on the surface of an imaginary sphere surrounding the sound source.

Uniform radiation simplifies calculations and helps correctly apply the intensity formula with a spherical wavefront. It avoids complicating factors such as obstacles or directional sound sources which could affect how sound spreads in real-world settings.

By assuming the siren radiates sound uniformly, it ensures our calculations only depend on distance and sound power, allowing a straightforward understanding of sound propagation in this scenario.