Problem 25 - University Physics with Modern Physics

Question

A jet plane at takeoff can produce sound of intensity 10.0 W/m\(^2\) at 30.0 m away. But you prefer the tranquil sound of normal conversation, which is 1.0 \(\mu\)W/m\(^2\). Assume that the plane behaves like a point source of sound. (a) What is the closest distance you should live from the airport runway to preserve your peace of mind? (b) What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? (c) What power of sound does the jet produce at takeoff?

Step-by-Step Solution

Answer

(a) 3000 m, (b) 1.5 \( \times \) 10\(^{-7}\) W/m"), (c) 36000\( \pi \) W.

Step 1: Understanding Sound Intensity and Point Source

We need to find the distances from the sound source (the jet), where the sound intensity reduces to the level of a normal conversation (1.0 \( \mu \)W/m"). Use the concept that intensity \( I \) from a point source is inversely proportional to the square of distance \( r \): \( I = \frac{P}{4\pi r^2} \), where \( P \) is the power of the sound source.

Step 2: Find Closest Distance for Peaceful Sound

We need to solve for \( r \) where \( I = 1.0 \times 10^{-6} \) W/m"). Using the inverse-square law: initially \( 10.0 = \frac{P}{4\pi (30)^2} \). Solving gives \( P = 10.0 \times 4\pi (30)^2 \). Set \( \frac{P}{4\pi r^2} = 1.0 \times 10^{-6} \), solve for \( r \).

Step 3: Calculate Distance Equation

\( P = 10.0 \times 4\pi \times 900 = 36000\pi \) W. Now \( 1.0 \times 10^{-6} = \frac{36000\pi}{4\pi r^2} \), thus \( r^2 = \frac{36000\pi}{4\pi \times 1.0 \times 10^{-6}} \), solve to find the distance \( r \).

Step 4: Simplify Distance Calculation

\( r^2 = \frac{36000}{4 \times 1.0 \times 10^{-6}} \), \( r = \sqrt{9 \times 10^9} \), \( r = 3000 \) m is the closest distance.

Step 5: Calculate Intensity at Friend's Location

Friend lives twice the distance, \( 2r = 6000 \) m, use \( I_f = \frac{36000\pi}{4\pi (6000)^2} \). Calculate \( I_f \).

Step 6: Calculate Friend's Intensity Result

\( I_f = \frac{36000}{4 \times 6000^2} = 1.5 \times 10^{-7} \) W/m").

Step 7: Find the Power of the Sound

From Step 3, the power \( P \) is \( 36000\pi \) W.

Key Concepts

Inverse Square Law

The inverse square law is a principle that states that the intensity of a physical phenomenon, like sound, decreases with the square of the distance from the source. This concept is crucial when dealing with sound emitted from a point source, such as a jet engine.

To put it simply, if you double the distance from the sound source, the intensity of the sound you perceive becomes one-fourth of its original value. Mathematically, this can be expressed as:

• \( I = \frac{P}{4\pi r^2} \)

where \( I \) is the sound intensity, \( P \) is the power of the sound, and \( r \) is the distance from the source.

For instance, if a jet at takeoff generates a tremendous noise level of 10.0 W/m² at 30 meters, moving further away doesn't just reduce the intensity linearly. Instead, the intensity diminishes significantly as you increase your distance from the noise source.

Utilizing the inverse square law helps us calculate at what distance noise becomes bearable, such as moving to where the noise is akin to a normal conversation (1.0 µW/m²), by solving for \( r \) with our formula.

Point Source of Sound

A point source of sound is an idealized concept where the sound source emits sound uniformly in all directions. It's like imagining a sound-emitting ball where sound rays spread outwards evenly. This assumption simplifies calculations and helps us understand real-world scenarios better.

Jet planes, despite their complexity, are often treated as point sources when calculating sound intensity. This approximation works because the sound from the plane spreads out in all directions without being obstructed or focused in a single direction.

When dealing with a point source, the spreading of sound affects how we'd expect the intensity to change with distance. Because of the uniform spread, using the inverse square law, we can predict how loud or quiet a source will seem as we move further away or come closer. This becomes crucial in determining the closest distance one should live from sources of high sound intensity, such as an airport runway, to maintain a comfortable level of noise exposure.

Sound Power Calculation

Sound power is the total energy that a source, like a jet engine, emits in the form of sound. It's measured in watts and represents how sound spreads from the source.

To calculate sound power using the given problem, we start by using the known formula:

• \( P = I \times 4\pi r^2 \)

where \( I \) is the initial intensity measured at a specific distance, and \( r \) is that distance. In our scenario, being 30 meters away with an intensity of 10.0 W/m², we use this to find that the jet's sound power is substantial.

We derived a power of \( 36000\pi \) watts for the plane at takeoff. Knowing this, we can further use it to calculate intensities at various distances, such as where the sound level would equal that of conversational tones or the intensity your friend would experience at double the safe distance.

Calculating sound power is a vital step in assessing environmental noise impacts and designing effective noise reduction strategies. It helps predict how far sound travels and how it might affect different areas.