Problem 25 - University Physics with Modern Physics

Question

\mathrm{A} jet plane at takeoff can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2} at 30.0 \)\mathrm{m} away. But you prefer the tranquil sound of normal conversation, which is 1.0$\mu \mathrm{W} / \mathrm{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) 30,000 m, (b) 0.25 µW/m², (c) 113,097 W.

Step 1: Understand the Intensity Formula

Sound intensity can be expressed as \( I = \frac{P}{A} \), where \( I \) is the intensity, \( P \) is the power of the sound source, and \( A \) is the surface area of a sphere (since the sound spreads out spherically). The area \( A \) of a sphere is \( 4 \pi r^2 \), with \( r \) being the radius, or the distance from the source.

Step 2: Calculate Required Distance for Desired Intensity

We want to find the distance where the intensity is equal to 1.0 \( \mu \mathrm{W} / \mathrm{m^2} = 1.0 \times 10^{-6} \mathrm{W} / \mathrm{m^2}\). Set the known intensity is \( 10.0 \mathrm{W} / \mathrm{m^2} \) at \( 30.0 \mathrm{m} \). Use the relation \( I_1 \times r_1^2 = I_2 \times r_2^2 \) to solve for new distance \( r_2 \):\[10.0 \times 30^2 = 1.0 \times 10^{-6} \times r_2^2\]Solving for \( r_2 \), we get:\[r_2^2 = \frac{10.0 \times 30^2}{1.0 \times 10^{-6}}\]This simplifies to:\[r_2 = \sqrt{\frac{10.0 \times 900}{1.0 \times 10^{-6}}} = \sqrt{9 \times 10^9} = 3 \times 10^4 \mathrm{m}\]Thus, you should live at least \( 30,000 \mathrm{m} \) away.

Step 3: Calculate Intensity at Double the Distance

Your friend lives twice as far, hence at a distance of \( 2 \times 3 \times 10^4 \mathrm{m} = 6 \times 10^4 \mathrm{m} \). Using the inverse square law and re-arranging for intensity:\[I_{friend} = \frac{I}{4}\]Since distance is doubled:\[I_{friend} = \frac{1.0 \times 10^{-6}}{4} = 0.25 \times 10^{-6} \mathrm{W} / \mathrm{m^2}\]

Step 4: Determine Jet Power Output in Watts

The power \( P \) can be found using the initial known values with the formula \( I = \frac{P}{4 \pi r^2} \).At \( r = 30 \mathrm{m} \), \( I = 10 \mathrm{W}/\mathrm{m}^2 \):\[P = I \times 4 \pi r^2 = 10 \mathrm{W}/\mathrm{m}^2 \times 4 \pi \times 30^2\]This simplifies to:\[P = 10 \times 4 \pi \times 900 = 36000 \pi \mathrm{W} \approx 113097 \mathrm{W}\]

Key Concepts

Inverse Square Law

The Inverse Square Law is a fundamental principle in physics that describes how sound intensity diminishes as it moves away from its source. When sound spreads out from a point source, like a jet engine, it propagates spherically. This means the intensity is inversely proportional to the square of the distance from the source. Since sound intensity (\( I \)) is tied to the distance (\( r \)), we use the formula:\[ I_1 \times r_1^2 = I_2 \times r_2^2 \]Where \( I_1 \) and \( r_1 \) are the initial intensity and distance, and \( I_2 \) and \( r_2 \) are the new intensity and distance. This relationship helps solve problems where desired intensity levels are needed at certain distances, providing a straightforward way to calculate how far someone must be from a noisy area like an airport to experience a specific sound level.

Sound Power

Sound power refers to the total energy the source emits in the form of sound waves per unit time. Unlike intensity, which describes how much power spreads over an area, sound power remains constant regardless of distance.For a jet engine, for example, the power of sound can be calculated using the intensity formula:\[ I = \frac{P}{4 \pi r^2} \]where \( P \) is the sound power and \( r \) is the distance from the source. The example shows how you can use the known intensity and distance to calculate the jet's sound power efficiently.For the jet, using the formula where \( r = 30 \text{ m} \) and \( I = 10 \text{ W/m}^2 \), we calculated that the jet emits around \( 113097 \text{ W} \). This power output reveals how much energy the jet engine releases in terms of sound, providing context for environmental noise considerations.

Jet Noise

Jet noise is one of the cities' more common sources of noise pollution, especially near busy airports. It generally originates from both the engine and its exhaust and can be extraordinarily loud.Due to the sheer power of jets, sound intensities like \( 10 \text{ W/m}^2 \) are feasible at takeoff. Thus, managing its disruption is crucial, particularly since it spreads widely and diminishes the quality of life if left unchecked.Jet noise also serves as an example of how significant and intense noise sources are whenever calculating distances for sound levels suitable for ordinary life activities. Understanding how much sound power jets emit helps in making necessary architectural choices, like sound-insulating windows or placing buildings strategically.

Decibel Scale

The decibel (dB) scale is a logarithmic way to measure sound intensity, making it easier to express sound levels over a vast range. Since the human ear perceives sound intensity logarithmically, the decibel scale aligns with our experience, providing a handy reference for comparing sound sources.For sound power levels, the decibel scale uses a defined reference point, often in micro-watts per square meter (\( \mu \text{W/m}^2 \)), to convey intensity levels more relatable to our perception. For instance, a normal conversation might measure around 60 dB, while a jet takeoff can exceed 120 dB.This scale aids in quantifying and managing exposure to various types of noises in our daily lives. Balancing the sound intensities through means such as industrial adjustments, personal protective equipment, and city planning ensures a more pleasant auditory environment overall.