/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A jet plane at takeoff can produ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

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

Step by step solution

01

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.
02

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 \( r \).
03

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 \( r \).
04

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.
05

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 \( I_f \).
06

Calculate Friend's Intensity Result

\( I_f = \frac{36000}{4 \times 6000^2} = 1.5 \times 10^{-7} \) W/m").
07

Find the Power of the Sound

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 Hz to about 20.0 kHz. (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short- wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 m/s. How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

A 0.800-m-long string with linear mass density \(\mu = 7.50\) g/m is stretched between two supports. The string has tension \(F\) and a standing-wave pattern (not the fundamental) of frequency 624 Hz. With the same tension, the next higher standing-wave frequency is 780 Hz. (a) What are the frequency and wavelength of the fundamental standing wave for this string? (b) What is the value of \(F\)?

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string? (b) How long does it take a point on the string to travel a distance of 8.00 m, once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?

A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 406 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W?

A certain transverse wave is described by $$y(x, t) = (6.50 \, \mathrm{mm}) \mathrm{cos} \, 2\pi \Big( \frac{x}{28.0 \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \Big)$$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

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