/*! 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 20 If the intensity of sound is dou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the intensity of sound is doubled, by how many decibels does the sound level increase?

Short Answer

Expert verified
The sound level increases by approximately 3.01 decibels.

Step by step solution

01

Understanding the Decibel Scale

The decibel (dB) scale is a logarithmic scale used to measure sound intensity. The formula relating intensity \(I\) to decibels \(L\) is \(L = 10 \log_{10}\left(\frac{I}{I_0}\right)\), where \(I_0\) is the reference intensity.
02

Calculating Initial Decibel Level

Let's assume the initial intensity of sound is \(I_1\). Then the initial sound level is \(L_1 = 10 \log_{10}\left(\frac{I_1}{I_0}\right)\).
03

New Intensity after Doubling

When the intensity is doubled, the new intensity \(I_2 = 2I_1\).
04

Calculating New Decibel Level

The new sound level with intensity \(I_2\) is \(L_2 = 10 \log_{10}\left(\frac{2I_1}{I_0}\right)\).
05

Finding the Change in Decibels

The change in decibels \(\Delta L = L_2 - L_1 = 10 \log_{10}\left(\frac{2I_1}{I_0}\right) - 10 \log_{10}\left(\frac{I_1}{I_0}\right)\).
06

Using Logarithmic Properties

Applying the property of logarithms that \(\log_{10}(a) - \log_{10}(b) = \log_{10}\left(\frac{a}{b}\right)\), we simplify \(\Delta L = 10 \log_{10}\left(\frac{2I_1}{I_0} \times \frac{I_0}{I_1}\right) = 10 \log_{10}(2)\).
07

Calculating the Numerical Value

Using a calculator, \(\log_{10}(2) \approx 0.301\). So, \(\Delta L = 10 \times 0.301 = 3.01\). Thus, the sound level increases by approximately 3.01 decibels.

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 sound intensity level is measured using the decibel (dB) scale. This scale helps us express sound levels in a manageable way due to the vast range of audible sound intensities.
The formula used in the decibel scale is:
  • \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \)
where:
  • \( L \) represents the sound level in decibels (dB),
  • \( I \) refers to the intensity of the sound,
  • \( I_0 \) is the reference intensity, typically \( 10^{-12} \) watts per square meter, the quietest sound the average human ear can detect.
Using the decibel scale allows us to deal with these wide-ranging values in a more practical numerical range. As the sound intensity changes, so does the decibel level, making it crucial to understand this relationship for accurate sound measurement.
The decibel scale's logarithmic nature is key to understanding how sound intensity translates to human perception. Typically, an increase of about 10 dB corresponds to a tenfold increase in perceived loudness.
Logarithmic Scale
The decibel scale is not linear; it is a logarithmic scale. But what does this mean exactly? A logarithmic scale is one where each step of the scale represents a multiplication of the previous step, rather than an addition as in linear scales.
In terms of sound intensity, every 10 dB increase represents a tenfold increase in intensity.
This implies:
  • Going from 30 dB to 40 dB is not just an increase of "10 dB," but effectively the sound has gained ten times more intensity.
This property of the logarithmic scale helps simplify the mathematics of sound and reflects the human ear's response to changes in sound intensity. Our ears are more sensitive to changes in sound intensity rather than actual sound intensity, making the logarithmic approach in decibel calculations valuable for practical purposes.
Intensity Doubling
When we mention intensity doubling, it means the sound's power per unit area has doubled. This is a key concept in sound measurement.
If the original intensity is \( I_1 \), after doubling, it becomes \( I_2 = 2I_1 \).
  • This means the sound wave carries twice the amount of energy per unit of area.
In the logarithmic decibel scale, intensity doubling does not double the decibel level. Instead, it results in a lesser increase measured on the dB scale due to its logarithmic nature.
This factor is unique to logarithmic calculations, and illustrates why the decibel scale is suitable for measuring sound. It helps us understand how even significant changes in intensity translate to relatively small changes in perception.
Change in Decibels
When the intensity of a sound is doubled, determining how many decibels it increases is a common calculation in acoustics. The change in decibels, \( \Delta L \), can be calculated using logarithmic properties:
First, calculate the new decibel level with the increased intensity:
  • \( L_2 = 10 \log_{10}\left(\frac{2I_1}{I_0}\right) \)
Next, find the difference between the new and original decibel levels:
  • \( \Delta L = L_2 - L_1 \)
  • \( \Delta L = 10 \log_{10}\left(\frac{2I_1}{I_0}\right) - 10 \log_{10}\left(\frac{I_1}{I_0}\right) \)
Utilizing the properties of logarithms, simplify this to:
  • \( \Delta L = 10 \log_{10}(2) \)
Finally, the numerical solution reveals that \( \Delta L \approx 3.01 \) dB. Thus, doubling the sound's intensity results in an increase of approximately 3.01 decibels. This subtle increase shows the efficacy of the dB scale in expressing real-world sound variations.

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 car moving at \(108 \mathrm{~km} \mathrm{~h}^{-1}\) finds another car in front of it going in the same direction at \(72 \mathrm{~km} \mathrm{~h}^{-1}\). The first car sounds a horn that has a dominant frequency of \(800 \mathrm{~Hz}\). What will be the apparent frequency heard by the driver in the front car ? Speed of sound in air \(=330 \mathrm{~m} \mathrm{~s}^{-1}\).

A bullet passes past a person at a speed of \(220 \mathrm{~m} / \mathrm{s}\). Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air \(=330 \mathrm{~m} / \mathrm{s}\).

A tuning fork of frequency \(256 \mathrm{~Hz}\) produces 4 beats per second with a wire of length \(25 \mathrm{~cm}\) vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?

A copper rod of length \(1 \cdot 0 \mathrm{~m}\) is clamped at its middle point. Find the frequencies between \(20 \mathrm{~Hz}\) and \(20,000 \mathrm{~Hz}\) at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is \(38 \mathrm{~km} \mathrm{~s}^{-1}\).

An open organ pipe has a length of \(5 \mathrm{~cm}\). (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range ? Speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\) and the audible range is \(20-20,000 \mathrm{~Hz}\).

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.