/*! 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 35 Loudness of Sound The loudness \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Loudness of Sound The loudness \(L\) of a sound (measured in decibels, dB) is inversely proportional to the square of the distance \(d\) from the source of the sound. A person who is 10 \(\mathrm{ft}\) from a lawn mower experiences a sound level of 70 \(\mathrm{dB}\) . How loud is the lawn mower when the person is 100 \(\mathrm{ft}\) away?

Short Answer

Expert verified
The loudness of the lawn mower at 100 ft is 49 dB.

Step by step solution

01

Understand the Relationship

The loudness \(L\) of sound is inversely proportional to the square of the distance \(d\). This can be expressed as \(L \propto \frac{1}{d^2}\), or \(L = \frac{k}{d^2}\), where \(k\) is the constant of proportionality.
02

Determine the Constant of Proportionality

To find the constant \(k\), use the information given: when \(d = 10\, \mathrm{ft}\), \(L = 70\, \mathrm{dB}\). Substitute these values into the equation \(L = \frac{k}{d^2}\) to get:\[ 70 = \frac{k}{10^2} \]\[ 70 \times 100 = k \]\[ k = 7000 \]
03

Calculate the Loudness at a Different Distance

Now that we have \(k = 7000\), we can find the loudness when the distance \(d = 100\, \mathrm{ft}\). Substitute \(d = 100\) and \(k = 7000\) into the equation \(L = \frac{k}{d^2}\):\[ L = \frac{7000}{100^2} \]\[ L = \frac{7000}{10000} \]\[ L = 0.7 \]
04

Convert the Loudness Back to Decibels

The calculation in the last step yields a relative scale value, but in context, the loudness in terms of decibels is proportional, so state that the loudness at 100 ft is \(0.7\) times the loudness at 10 ft, i.e., \(0.7 \times 70\, \mathrm{dB}\). Thus:\[ L = 0.7 \times 70 = 49 \text{ dB} \]

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.

Decibels
Sound is measured in units called decibels (dB), which might seem a bit puzzling at first. Imagine decibels as the yardstick for sound. They help us know how loud or soft a sound is. The decibel scale is logarithmic, not linear, which means each increase of 10 dB isn’t just a bit louder; it’s ten times louder. That’s why sounds measured in decibels can range from quite low (like a whisper at around 30 dB) to extremely high (a rock concert might hit 120 dB).
This helps explain why a small change in the distance from the sound source can make a big difference in how loud it sounds. So, understanding decibels isn't just about numbers; it's about grasping how sound works in our world. Whether it's the buzz of a lawn mower or your favorite song blasting through speakers, the decibel scale helps us quantify and compare those sounds.
Distance and Sound
When it comes to sound, distance plays a big role in how loud we perceive it to be. This is based on the principle that loudness is inversely proportional to the square of the distance from the source. Simply put, the further you move away from a sound, the quieter it becomes; and doubling the distance means reducing the loudness by one-fourth.
  • The lawn mower problem is a great example. At 10 feet, it’s measured at 70 dB.
  • But when you’re 100 feet away, the sound isn’t just a little quieter; it drops significantly to 49 dB.
This drop in loudness with distance is a crucial concept in physics because it explains why sounds fade as we walk away from their source. This understanding is not only useful for calculating sound levels at varying distances but also for practical applications like setting up speakers at an event.
Proportionality Constant
In our exercise, the constant of proportionality, denoted as \( k \), is crucial for calculating how sound levels alter with distance. Once you establish this constant, you can use it to find the loudness at any distance from the sound source. Here's how it works:
  • You start with the equation \( L = \frac{k}{d^2} \), where \( L \) is loudness, \( k \) is the constant we're solving for, and \( d \) is the distance from the sound source.
  • Using the known values — in this case, 70 dB at 10 feet — plug them into the equation to find \( k \).
  • After calculating, you find \( k = 7000 \). This value remains constant irrespective of distance — it simply scales with the changing distance \( d \).
By understanding and using this constant, we can confidently predict how loud a sound will be at different distances without having to measure it directly each time. It’s a powerful tool in acoustics and helps simplify complex sound calculations.

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

Enter Equations Carefully A student wishes to graph the equations $$ y=x^{1 / 3} \quad \text { and } \quad y=\frac{x}{x+4} $$ on the same screen, so he enters the following information into his calculator: $$ Y_{1}=x^{\wedge} 1 / 3 \quad Y_{2}=x / x+4 $$ The calculator graphs two lines instead of the equations he wanted. What went wrong?

Find the area of the triangle formed by the coordinate axes and the line $$ 2 y+3 x-6=0 $$

Skidding in a Curve A car is traveling on a curve that forms a circular arc. The force \(F\) needed to keep the car from skidding is jointly proportional to the weight \(w\) of the car and the square of its speed \(s,\) and is inversely proportional to the radius \(r\) of the curve. (a) Write an equation that expresses this variation. (b) A car weighing 1600 lb travels around a curve at 60 \(\mathrm{mi} / \mathrm{h}\) . The next car to round this curve weighs 2500 \(\mathrm{lb}\) and requires the same force as the first car to keep from skidding. How fast is the second car traveling?

cost of Driving The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May her driving cost was \(\$ 380\) for 480 \(\mathrm{mi}\) and in June her cost was \(\$ 460\) for 800 \(\mathrm{mi}\) . Assume that there is a linear relationship between the monthly cost \(C\) of driving a car and the distance driven \(d\) (a) Find a linear equation that relates \(C\) and \(d\) (b) Use part (a) to predict the cost of driving 1500 \(\mathrm{mi}\) per month. (c) Draw the graph of the linear equation. What does the slope of the line represent? (d) What does the \(y\) -intercept of the graph represent? (d) Why is a linear relationship a suitable model for this situation?

Kepler's Third Law Kepler's Third Law of planetary motion states that the square of the period \(T\) of a planet (the time it takes for the planet to make a complete revolution about the sun) is directly proportional to the cube of its average distance \(d\) from the sun. (a) Express Kepler's Third Law as an equation. (b) Find the constant of proportionality by using the fact that for our planet the period is about 365 days and the average distance is about 93 million miles. (c) The planet Neptune is about \(2.79 \times 10^{9}\) mi from the sun. Find the period of Neptune.

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.