/*! 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 63 Solve the given applied problems... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the given applied problems involving variation. The acoustical intensity \(I\) of a sound wave is proportional to the square of the pressure amplitude \(P\) and inversely proportional to the velocity \(v\) of the wave. If \(I=0.474 \mathrm{W} / \mathrm{m}^{2}\) for \(P=20.0 \mathrm{Pa}\) and \(v=346 \mathrm{m} / \mathrm{s},\) find \(I\) if \(P=15.0 \mathrm{Pa}\) and \(v=320 \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
The new acoustical intensity \( I \) is approximately \( 0.288 \mathrm{W}/\mathrm{m}^2 \).

Step by step solution

01

Understand the Variation Formula

The problem states that the acoustical intensity, \( I \), is proportional to the square of the pressure amplitude, \( P^2 \), and inversely proportional to the velocity, \( v \). This relationship can be expressed as: \( I = k \frac{P^2}{v} \), where \( k \) is the constant of proportionality.
02

Determine the Constant of Proportionality

Use the given values to find \( k \). We have \( I = 0.474 \), \( P = 20.0 \), and \( v = 346 \). Substitute into the formula: \( 0.474 = k \frac{(20.0)^2}{346} \). Simplify this to find \( k \).
03

Solve for the Constant

Calculate \( (20.0)^2 = 400 \). Then, substitute into the equation: \( 0.474 = k \frac{400}{346} \). Multiply both sides by \( 346 \) to isolate \( k \): \( 0.474 \times 346 = 400k \). Thus, \( k = \frac{0.474 \times 346}{400} \).
04

Calculate the New Intensity

Substitute the new values of \( P = 15.0 \) and \( v = 320 \) into the variation formula: \( I = k \frac{(15.0)^2}{320} \). First, calculate \( k \) from the previous step, then find \( I \) using these new values.
05

Final Calculation

Using \( k = \frac{0.474 \times 346}{400} \), calculate \( I = k \frac{(15.0)^2}{320} \). Simplify \( (15.0)^2 = 225 \), then find \( I \). Substitute \( k \) into the equation: \( I = \frac{0.474 \times 346}{400} \times \frac{225}{320} \). Calculate this to find the final intensity \( I \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Acoustical Intensity
Acoustical intensity refers to the power per unit area carried by a sound wave. It's a measure of how much sound energy is passing through a specific area, and it's usually measured in watts per square meter (W/m²). In simple terms, it's the concentration of sound energy at a point.
Understanding acoustical intensity is crucial because it's how we perceive the loudness of sounds. The greater the intensity, the louder the sound appears to us.
  • It relies on factors such as pressure amplitude and the velocity of the sound wave.
  • Its formula is derived from the relationship involving these factors to determine how different scenarios alter sound perception.
  • In practical applications, it helps in assessing noise pollution levels and designing acoustics in architectural spaces.
Whenever you encounter a change in either the pressure amplitude or the velocity of a sound wave, the acoustical intensity will adjust accordingly, thanks to the mathematical relationship described in acoustical variation problems.
Pressure Amplitude
Pressure amplitude is a vital concept in acoustics. It represents the maximum change in pressure at any point in a sound wave and is measured in pascals (Pa). Essentially, pressure amplitude indicates the strength or power of a sound wave at its peak.
When you think about how loud a sound appears, pressure amplitude plays a big role. A greater pressure amplitude means a more powerful and potentially louder sound.
  • A larger amplitude leads to a louder sound.
  • It's a crucial factor in equations for acoustical intensity, specifically affecting the square term in the variation formula.
In acoustical intensity problems, the square of the pressure amplitude directly influences the intensity. Therefore, when calculating or predicting changes in intensity, variations in amplitude are key. Even a small change in pressure amplitude can cause significant changes in the acoustical intensity due to its squared effect in calculations.
Constant of Proportionality
The constant of proportionality, denoted as "k", is a factor in the equation for acoustical intensity. It binds together the pressure amplitude and velocity with acoustical intensity in a consistent formula across different conditions.
This constant ensures that the proportionality between these elements is maintained.
  • It's derived from given conditions, where all other variables are known.
  • Its value helps determine how variations in pressure amplitude and velocity impact acoustical intensity in a predictable manner.
  • By solving for \( k \), we establish a foundation for predicting future changes or conditions.
Essentially, once you have the value of \( k \), you can adjust the other variables and always trust that the formula will accurately reflect how those changes affect acoustical intensity. This makes "k" a crucial element in solving variation problems.
Mathematical Modeling
Mathematical modeling in variation problems like acoustical intensity is crucial for making sense of how different factors interact in the real world.
By developing equations that describe these interactions, mathematical models allow us to predict and understand various phenomena with accuracy.
  • In the context of acoustics, it helps explain how sound behaves under different conditions.
  • Models provide a systematic way to explore what happens when either pressure amplitude or velocity changes.
  • They transform abstract concepts into actionable equations, ensuring real-world applications like sound engineering and acoustics design are grounded in solid mathematics.
With mathematical modeling, you can break down complex real-world acoustical situations into manageable problems, solve them, and apply the solutions practically. This makes understanding concepts like acoustical intensity not just theoretical but also practical and applicable in various scientific and engineering fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the given applied problems involving variation. The escape velocity \(v\) a spacecraft needs to leave the gravitational field of a planet varies directly as the square root of the product of the planet's radius \(R\) and its acceleration due to gravity \(g .\) For Mars and earth, \(R_{M}=0.533 R_{e}\) and \(g_{M}=0.400 g_{e^{*}}\) Find \(v_{M}\) for Mars if \(v_{e}=11.2 \mathrm{km} / \mathrm{s}\)

A motorist traveled $$5.0 \mathrm{km}$$ (accurately measured) in 4 min 54 s, and the speedometer showed \(58 \mathrm{km} / \mathrm{h}\) for the same interval. What is the percent error in the speedometer reading?

Find the required ratios. Power is defined as the ratio of work done to the time required to do the work. If an engine performs \(3.65 \mathrm{kJ}\) of work in \(15.0 \mathrm{s}\), find the power developed by the engine. (See Appendix B.)

Solve the given applied problems involving variation. The power \(P\) required to propel a ship varies directly as the cube of the speed \(s\) of the ship. If 5200 hp will propel a ship at \(12.0 \mathrm{mi} / \mathrm{h},\) what power is required to propel it at \(15.0 \mathrm{mi} / \mathrm{h} ?\)

Solve the given applied problems involving variation. The intensity \(I\) of sound varies directly as the power \(P\) of the source and inversely as the square of the distance \(r\) from the source. Two sound sources are separated by a distance \(d\), and one has twice the power output of the other. Where should an observer be located on a line between them such that the intensity of each sound is the same?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.