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Suppose that sound is emitted uniformly in all directions by a public address system. The intensity at a location \(22 \mathrm{~m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). What is the intensity at a spot that is \(78 \mathrm{~m}\) away?

Short Answer

Expert verified
The intensity at 78 m is approximately \(2.37 \times 10^{-5} \, \text{W/m}^2\).

Step by step solution

01

Understand the relationship between intensity and distance

Intensity is related to the distance from the sound source by the inverse square law. This states that intensity is inversely proportional to the square of the distance from the source. Mathematically, this can be written as \( I \propto \frac{1}{r^2} \), where \( I \) is the intensity and \( r \) is the distance from the source.
02

Write down the inverse square law equation

Using the inverse square law, the equation relating intensity at two different distances is: \[ \frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 \]Where \( I_1 \) is the intensity at the first distance \( r_1 = 22 \, \text{m} \), and \( I_2 \) is the intensity at the second distance \( r_2 = 78 \, \text{m} \).
03

Plug values into the equation

The known intensity \( I_1 = 3.0 \times 10^{-4} \, \text{W/m}^2 \). Plug in the known distances and the known intensity into the equation:\[ \frac{3.0 \times 10^{-4}}{I_2} = \left( \frac{78}{22} \right)^2 \]
04

Solve for the unknown intensity \(I_2\)

First calculate \( \left( \frac{78}{22} \right)^2 \):\[ \left( \frac{78}{22} \right)^2 \approx 12.64 \] Now solve for \( I_2 \):\[ I_2 = \frac{3.0 \times 10^{-4}}{12.64} \approx 2.37 \times 10^{-5} \, \text{W/m}^2 \]
05

Conclusion

The intensity at a location 78 m away from the sound source is approximately \( 2.37 \times 10^{-5} \, \text{W/m}^2 \).

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Key Concepts

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

Sound Intensity
Sound intensity refers to the power of sound waves passing through a unit area perpendicular to the direction of the wave. It's measured in watts per square meter (W/m²). Understanding sound intensity is important as it gives us insight into how strong a sound is perceived in different environments. Think of it this way: when you stand closer to a speaker at a concert, the sound feels more intense compared to standing at the back of the venue. This is because sound intensity decreases as you move further from the source. Key points about sound intensity:
  • It helps us measure and understand how loud a sound is at any point in space.
  • A higher intensity means a louder sound is perceived.
  • The unit of measurement is watts per square meter.
By understanding sound intensity, you can appreciate how sound engineers adjust audio systems for optimal listening experiences, whether in a small room or a large auditorium.
Distance and Intensity Relationship
The relationship between distance and intensity is governed by the inverse square law. This fundamental principle in physics explains how the intensity of sound decreases as you move away from the source. The inverse square law states that intensity is inversely proportional to the square of the distance. Mathematically, it is expressed as \( I \propto \frac{1}{r^2} \), where \( I \) is the intensity and \( r \) is the distance. In simpler terms, if you double the distance from the sound source, the intensity becomes quartered. The inverse square law is crucial in:
  • Designing audio systems and public address systems to ensure even sound distribution.
  • Understanding the impact of sound pollution and how far sound travels in different environments.
  • Using it to solve physics problems involving sound intensity and distance changes.
This principle not only helps in academic exercises but also has practical applications in engineering and environmental sciences.
Physics Problem Solving
Solving physics problems involving sound intensity and distance usually follows a systematic approach. It's important to understand how mathematical formulas apply to real-world scenarios. The solution described in the exercise employs the inverse square law to find the intensity of sound at a different distance.Here's how you can approach similar problems:
  • Start by identifying the known and unknown quantities, like distances and intensities.
  • Apply the inverse square law in the form of the equation \( \frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 \).
  • Plug in the known values and solve for the unknown intensity.
  • Ensure your final answer makes logical sense by verifying the units and calculations.
Key to problem-solving in physics is breaking down the problem into manageable pieces and understanding the physical meaning behind mathematical expressions. By practicing regularly, you'll develop a keen sense for interpreting physics problems.

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Most popular questions from this chapter

A rocket in a fireworks display explodes high in the air. The sound spreads out uniformly in all directions. The intensity of the sound is \(2.0 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(120 \mathrm{~m}\) from the explosion. Find the distance from the source at which the intensity is \(0.80 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{2}\)

Two submarines are underwater and approaching each other head-on. Sub A has a speed of \(12 \mathrm{~m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{~m} / \mathrm{s}\). Sub A sends out a 1550 -Hz sonar wave that travels at a speed of \(1522 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency detected by sub \(\mathrm{B}\) ? (b) Part of the sonar wave is reflected from \(\mathrm{B}\) and returns to \(\mathrm{A}\). What frequency does A detect for this reflected wave?

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{~m}^{2},\) has a linear density of \(7.0 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of \(46 \mathrm{~m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). What will be the speed of the wave when the temperature is lowered by \(14 \mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

A loudspeaker has a circular opening with a radius of \(0.0950 \mathrm{~m} .\) The electrical power needed to operate the speaker is \(25.0 \mathrm{~W}\). The average sound intensity at the opening is \(17.5 \mathrm{~W} / \mathrm{m}^{2}\). What percentage of the electrical power is converted by the speaker into sound power?

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticeable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of \(750 \mathrm{~km}\) and traveled a distance of \(3700 \mathrm{~km}\) in \(5.3 \mathrm{~h}\). (a) What was the speed (in \(\mathrm{m} / \mathrm{s}\) ) of the wave? For reference, the speed of a 747 jetliner is about \(250 \mathrm{~m} / \mathrm{s}\). Find the wave's (b) frequency and (c) period.

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