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Chapter 12: Problem 62

A very noisy chain saw operated by a tree surgeon emits a total acoustic power of \(20.0 \mathrm{~W}\) uniformly in all directions. At what distance from the source is the sound level equal to (a) \(100 \mathrm{~dB}\), (b) \(60 \mathrm{~dB} ?\)

Short Answer

Expert verified
(a) 3.99 m, (b) 63.2 m.

Step by step solution

01

Understand the Problem

We need to find the distance from the sound source where the sound level is at specific decibels (dB). We will use the formula for sound intensity and the relationship between sound intensity and sound level in decibels.
02

Formula for Sound Intensity

Sound intensity is given by \( I = \frac{P}{A} \), where \( P \) is the acoustic power and \( A \) is the area over which the sound spreads. For a point source emitting uniformly in all directions, \( A = 4\pi r^2 \), where \( r \) is the distance from the source. Thus, \( I = \frac{P}{4\pi r^2} \).
03

Convert Sound Level to Intensity

Sound level in decibels is related to intensity by the formula \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I_0 = 1 \times 10^{-12} \mathrm{~W/m^2} \) is the reference intensity. To find the intensity \( I \) corresponding to a given sound level \( L \), rearrange to \( I = I_0 \cdot 10^{L/10} \).
04

Calculate Sound Intensity for 100 dB

For \( L = 100 \mathrm{~dB} \), \( I = 1 \times 10^{-12} \times 10^{100/10} = 1 \times 10^{-12} \times 10^{10} = 0.1 \mathrm{~W/m^2} \).
05

Solve for Distance for 100 dB

Using \( I = \frac{P}{4\pi r^2} \) where \( I = 0.1 \mathrm{~W/m^2} \) and \( P = 20 \mathrm{~W} \), solve for \( r \): \( r = \sqrt{\frac{P}{4\pi I}} = \sqrt{\frac{20}{4\pi \times 0.1}} \). Calculating this gives \( r \approx 3.99 \mathrm{~m} \).
06

Calculate Sound Intensity for 60 dB

For \( L = 60 \mathrm{~dB} \), \( I = 1 \times 10^{-12} \times 10^{60/10} = 1 \times 10^{-12} \times 10^6 = 1 \times 10^{-6} \mathrm{~W/m^2} \).
07

Solve for Distance for 60 dB

Using \( I = \frac{P}{4\pi r^2} \) where \( I = 1 \times 10^{-6} \mathrm{~W/m^2} \) and \( P = 20 \mathrm{~W} \), solve for \( r \): \( r = \sqrt{\frac{P}{4\pi I}} = \sqrt{\frac{20}{4\pi \times 1 \times 10^{-6}}} \). Calculating this gives \( r \approx 63.2 \mathrm{~m} \).

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Key Concepts

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

Acoustic Power
Acoustic power refers to the total energy emitted by a sound source per unit of time. In other words, it's the sound energy radiating from a source like a chain saw or speaker system.
  • The unit of acoustic power is Watts (W).
  • A powerful source emits a higher amount of acoustical power.
  • In this exercise, the chain saw emits an acoustic power of 20 Watts.
Acoustic power spreads out uniformly in all directions if there are no obstacles. Understanding how this energy spreads helps in calculating how far we can be from the source before the sound level changes.
Decibels
Decibels (dB) are a unit for measuring sound level, which is essential for determining how loud a sound seems to be. The decibel scale is logarithmic, which means that each increase of 10 dB represents a tenfold increase in sound intensity.
  • The equation used for converting intensity to decibels is: \[ L = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \]where \( L \) is the sound level in dB, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity (\( 1 \times 10^{-12} \) W/m虏).
  • A sound at 100 dB is significantly more intense than one at 60 dB.
  • This calculation helps us understand the relationship between intensity and perceived loudness.
Distance Calculation
To find the distance at which a specific sound level is heard, we use the formula for sound intensity distribution.
  • The formula is: \[ I = \frac{P}{4\pi r^2} \]where \( I \) is the intensity, \( P \) is the power (20 W in our exercise), and \( r \) is the distance from the source.
  • Solving for \( r \) gives us the distance: \[ r = \sqrt{\frac{P}{4\pi I}} \]
  • For a sound level of 100 dB, the distance \( r \) is approximately 3.99 m, and for 60 dB, it is about 63.2 m.
  • This shows that sound dissipates as it travels, and the farther you are from the source, the lower the sound level will be.
Sound Level
Sound level is a measure of how intense a sound is, expressed in decibels, which relate to human perception of loudness.
  • It helps us to quantify how loud or soft a sound appears.
  • Higher sound levels signify louder sounds, which could be harmful over prolonged exposure; for instance, consistently high sound levels (over 85 dB) could lead to hearing damage.
  • This understanding is vital in areas like occupational safety, urban planning, and audio engineering to ensure environments remain safe and comfortable.
Grasping sound levels effectively allows for better planning in both personal and professional settings to avoid excessive noise exposure.

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