/*! 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 17 (a) Show that if \(I_{2}=10.01_{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 17

(a) Show that if \(I_{2}=10.01_{1},\) then $\beta_{2}=\beta_{1}+10.0 \mathrm{dB} .$ (A factor of 10 increase in intensity corresponds to a 10.0 dB increase in intensity level.) (b) Show that if \(I_{2}=2.0 I_{1}\) then \(\beta_{2}=\beta_{1}+3.0 \mathrm{dB} .\) (A factor of 2 increase in intensity corresponds to a 3.0 -dB increase in intensity level. tutorial: decibels)

Short Answer

Expert verified
Question: Show the following relationships between intensity and intensity level in decibels: a) If \(I_{2} = 10 I_{1}\), then \(\beta_{2} = \beta_{1} + 10.0 dB\). b) If \(I_2 = 2 I_1\) then \(\beta_2 = \beta_1 + 3.0 dB\). Answer: a) We have proved that if \(I_{2} = 10 I_{1}\), then \(\beta_{2} = \beta_{1} + 10.0\,\mathrm{dB}\). b) We have proved that if \(I_2 = 2 I_1\), then \(\beta_2 = \beta_1 + 3.0\,\mathrm{dB}\).

Step by step solution

01

Write the intensity level formula for \(I_1\) and \(I_2\)

First, we write the decibel intensity formula for \(I_1\) and \(I_2\): \(\beta_1 = 10\log_{10}\left(\frac{I_1}{I_0}\right)\) \(\beta_2 = 10\log_{10}\left(\frac{I_2}{I_0}\right)\)
02

Substitute the given value \(I_2 = 10 I_1\)

Now, let's substitute the value \(I_2 = 10 I_1\) into the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(\frac{10 I_1}{I_0}\right)\)
03

Simplify the expression inside the logarithm

We can simplify the expression inside the logarithm in the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(10\frac{I_1}{I_0}\right)\)
04

Use the logarithm rules

According to logarithm rules, for any positive numbers x and any positive integer n, \(\log (x^n) = n\log(x)\). So we can simplify the expression further: \(\beta_2 = 10\log_{10}(10) + 10\log_{10}\left(\frac{I_1}{I_0}\right)\)
05

Calculate the logarithm and compare with \(\beta_1\)

Since \(\log_{10}(10) = 1\), we can find the value of \(\beta_2\) and compare it with \(\beta_1\): \(\beta_2 = 10 \cdot 1 + 10\log_{10}\left(\frac{I_1}{I_0}\right)\) So we have proved that: \(\beta_2 = \beta_1 + 10.0\,\mathrm{dB}\) #b) Show that if \(I_2 = 2 I_1\) then \(\beta_2 = \beta_1 + 3.0 dB\)#
06

Write the intensity level formula for \(I_1\) and \(I_2\)

First, we write the decibel intensity formula for \(I_1\) and \(I_2\): \(\beta_1 = 10\log_{10}\left(\frac{I_1}{I_0}\right)\) \(\beta_2 = 10\log_{10}\left(\frac{I_2}{I_0}\right)\)
07

Substitute the given value \(I_2 = 2 I_1\)

Now, let's substitute the value \(I_2 = 2 I_1\) into the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(\frac{2 I_1}{I_0}\right)\)
08

Simplify the expression inside the logarithm

We can simplify the expression inside the logarithm in the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(2\frac{I_1}{I_0}\right)\)
09

Use the logarithm rules

According to logarithm rules, for any positive numbers x and z, \(\log (x z) = \log(x) + \log(z)\). So we can simplify the expression further: \(\beta_2 = 10\log_{10}(2) + 10\log_{10}\left(\frac{I_1}{I_0}\right)\)
10

Calculate the logarithm and compare with \(\beta_1\)

Since \(\log_{10}(2) \approx 0.3010\), we can find the value of \(\beta_2\) and compare it with \(\beta_1\): \(\beta_2 = 10 \cdot 0.3010 + 10\log_{10}\left(\frac{I_1}{I_0}\right)\) So we have proved that: \(\beta_2 = \beta_1 + 3.0\,\mathrm{dB}\)

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 91影视!

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 bat emits chirping sounds of frequency \(82.0 \mathrm{kHz}\) while hunting for moths to eat. If the bat is flying toward the moth at a speed of $4.40 \mathrm{m} / \mathrm{s}\( and the moth is flying away from the bat at \)1.20 \mathrm{m} / \mathrm{s},$ what is the frequency of the sound wave reflected from the moth as observed by the bat? Assume it is a cool night with a temperature of \(10.0^{\circ} \mathrm{C} .\) [Hint: There are two Doppler shifts. Think of the moth as a receiver, which then becomes a source as it "retransmits" the reflected wave. \(]\)
In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.
A cello string has a fundamental frequency of \(65.40 \mathrm{Hz}\) What beat frequency is heard when this cello string is bowed at the same time as a violin string with frequency of \(196.0 \mathrm{Hz} ?\) [Hint: The beats occur between the third harmonic of the cello string and the fundamental of the violin. \(]\)
What are the four lowest standing wave frequencies for an organ pipe that is \(4.80 \mathrm{m}\) long and closed at one end?
Your friend needs advice on her newest "acoustic sculpture." She attaches one end of a steel wire, of diameter \(4.00 \mathrm{mm}\) and density $7860 \mathrm{kg} / \mathrm{m}^{3},$ to a wall. After passing over a pulley, located \(1.00 \mathrm{m}\) from the wall, the other end of the wire is attached to a hanging weight. Below the horizontal length of wire she places a 1.50 -m-long hollow tube, open at one end and closed at the other. Once the sculpture is in place, air will blow through the tube, creating a sound. Your friend wants this sound to cause the steel wire to oscillate at the same resonant frequency as the tube. What weight (in newtons) should she hang from the wire if the temperature is \(18.0^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Work Energy and Power

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation
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.