/*! 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 89 A recording engineer works in a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A recording engineer works in a soundproofed room that is \(44.0 \mathrm{dB}\) quieter than the outside. If the sound intensity that leaks into the room is \(1.20 \times 10^{-10} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity outside?

Short Answer

Expert verified
The intensity outside is approximately \(3.02 \times 10^{-6} \mathrm{W}/\mathrm{m}^2\).

Step by step solution

01

Understand the Decibel Scale

The decibel (dB) scale is a logarithmic scale used to measure sound intensity. The intensity level in decibels (dB) is given by the formula: \ \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] \ Here, \( I \) is the intensity of the sound in watts per square meter (W/m²) and \( I_0 \) is the reference intensity, usually \( 1.0 \times 10^{-12} \mathrm{W}/\mathrm{m}^2 \). This means if the room is 44.0 dB quieter, the outside intensity is 44.0 dB greater than the inside intensity.
02

Calculate Inside Intensity Level

Given the inside intensity \( I_{in} = 1.20 \times 10^{-10} \mathrm{W}/\mathrm{m}^2 \), we calculate its dB level using: \ \[ L_{in} = 10 \log_{10} \left( \frac{1.20 \times 10^{-10}}{1.0 \times 10^{-12}} \right) \] \ Calculate \( L_{in} \): \ \[ L_{in} = 10 \log_{10} (120) \] \ \( \log_{10} (120) \approx 2.08 \), hence \ \[ L_{in} \approx 20.8 \mathrm{dB} \]
03

Determine Outside Intensity Level

Since the soundproofing makes the room 44.0 dB quieter, the outside intensity level \( L_{out} \) is: \ \[ L_{out} = L_{in} + 44.0 = 20.8 + 44.0 = 64.8 \mathrm{dB} \]
04

Calculate Outside Intensity

Use the outside intensity level \( L_{out} \) to find the intensity \( I_{out} \) using the formula: \ \[ 64.8 = 10 \log_{10} \left( \frac{I_{out}}{1.0 \times 10^{-12}} \right) \] \ Solving for \( I_{out} \): \ \[ \log_{10} \left( \frac{I_{out}}{1.0 \times 10^{-12}} \right) = 6.48 \] \ \[ \frac{I_{out}}{1.0 \times 10^{-12}} = 10^{6.48} \] \ \[ I_{out} = 10^{6.48} \times 1.0 \times 10^{-12} \] \ \[ I_{out} \approx 3.02 \times 10^{-6} \mathrm{W}/\mathrm{m}^2 \]
05

Conclusion

The intensity of the sound outside the soundproofed room is approximately \( 3.02 \times 10^{-6} \mathrm{W}/\mathrm{m}^2 \).

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.

decibel scale
The decibel scale is widely used in measuring sound intensity and is essential for understanding how loud a sound is. Unlike linear scales, the decibel scale is logarithmic, making it particularly suited for sound measurements because human hearing is also logarithmic. This means if a sound is twice as intense, it doesn’t seem twice as loud to us. Instead, every 10-decibel increase represents a sound that is perceived to be about twice as loud.
  • The formula for the decibel level, given a sound intensity \( I \), is \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \).
  • \( I_0 \) is the reference intensity, set at \( 1.0 \times 10^{-12} \mathrm{W}/\mathrm{m}^2 \), which is the quietest sound a human ear can typically hear.

In practical terms, when a soundproof room is described as being 44 dB quieter than outside, it means the sound levels have been reduced dramatically to create a quieter environment.
soundproofing
Soundproofing is the process of making an environment resistant to the transmission of sound. It is crucial in various settings like recording studios, where minimizing external noise is important. Good soundproofing can significantly reduce unwanted noise levels, and its effectiveness is often expressed as a number of decibels.
  • In our exercise, the room is said to be 44 dB quieter, highlighting substantial soundproofing effectiveness.
  • Soundproofing uses materials and construction techniques to stop sound waves from entering or leaving a space.

The goal is to achieve a peaceful indoor environment, despite potentially noisy surroundings. Understanding and calculating decibel reduction can help in assessing how well a space is soundproofed.
intensity calculation
Calculating sound intensity involves understanding how much energy is transmitted through an area by sound waves. Given that sound intensity is measured in watts per square meter (\( \mathrm{W}/\mathrm{m}^2 \)), it directly reflects the energy involved in producing a sound.
To calculate sound intensity from a known decibel level, the formula used is derived from the decibel scale with:
  • \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \)
  • Rearranging gives \( I = I_0 \times 10^{\frac{L}{10}} \).

In our scenario, we computed the outside intensity by knowing the inside intensity and the decibel difference due to soundproofing, resulting in \( I_{out} \approx 3.02 \times 10^{-6} \mathrm{W}/\mathrm{m}^2 \). This calculation shows how mathematical relationships in acoustics apply to real-world problems.
logarithmic scale
The logarithmic scale is a way to represent data that varies over a large range of values in a more manageable format. Sounds cover a vast range of intensities, and their perception by the human ear changes logarithmically rather than linearly.
  • A logarithmic scale compresses these large ranges into more perceptible increments, which is why it’s so useful for measuring sound intensity.
  • Each increment on this scale corresponds to a multiplying factor of change in intensity, allowing us to manage and calculate sound levels easily.

In the context of our exercise, the logarithmic scale helps us determine the sound intensity levels both inside and outside a soundproofed room. Understanding this scale is fundamental to making sense of decibel-based measurements, as it bridges the gap between raw data and our perception of sound loudness.

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

A 3.49-rad/s ( \(33 \frac{1}{3}\) rpm) record has a \(5.00-\mathrm{kHz}\) tone cut in the groove. If the groove is located \(0.100 \mathrm{m}\) from the center of the record (see drawing), what is the wavelength in the groove?

As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{m} .\) A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The sound reaches microphone 1 first, and then, \(1.46 \mathrm{ms}\) later, it reaches microphone \(2 .\) Find the distances \(L_{1}\) and \(L_{2}\).

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

A member of an aircraft maintenance crew wears protective earplugs that reduce the sound intensity by a factor of \(350 .\) When a jet aircraft is taking off, the sound intensity level experienced by the crew member is \(88 \mathrm{dB}\). What sound intensity level would the crew member experience if he removed the protective earplugs?

To navigate, a porpoise emits a sound wave that has a wavelength of \(1.5 \mathrm{cm} .\) The speed at which the wave travels in seawater is \(1522 \mathrm{m} / \mathrm{s}\). Find the period of the wave.

See all solutions

Recommended explanations on Physics 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.