Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 27

The sound level in decibels is typically expressed as \(\beta=10 \log \left(I / I_{0}\right),\) but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on the amplitude squared, so the expression is \(\beta=20 \log \left(P / P_{0}\right),\) where \(P_{0}\) is the smallest pressure difference noticeable by the ear: \(P_{0}=2.00 \cdot 10^{-5} \mathrm{~Pa}\). A loud rock concert has a sound level of \(110 . \mathrm{dB}\), find the amplitude of the pressure wave generated by this concert.

Short Answer

Expert verified
Answer: The amplitude of the pressure wave generated by the rock concert is approximately \(1.58 \mathrm{Pa}\).

Step by step solution

01

Write down the given information

We have the following information: - Sound level at the rock concert, \(\beta = 110\mathrm{dB}\) - Smallest pressure difference noticeable by the ear, \(P_{0} = 2.00 \cdot 10^{-5} \mathrm{Pa}\)
02

Write down the expression for sound level in decibels

The expression for sound level in decibels (\(\beta\)) is given by: $$ \beta=20 \log \left(\frac{P} {P_{0}}\right) $$
03

Plug in the given information into the expression for \(\beta\)

We will plug in the given values of \(\beta\) and \(P_{0}\) into the expression for \(\beta\): $$ 110 = 20 \log \left(\frac{P} {2.00 \cdot 10^{-5}}\right) $$
04

Solve for \(P\)

Now, we need to solve this equation for \(P\). Follow these steps: 1. Divide both sides by 20: $$ 5.5 = \log \left(\frac{P} {2.00 \cdot 10^{-5}}\right) $$ 2. Apply anti-logarithm to eliminate the logarithm: $$ 10^{5.5} = \frac{P} {2.00 \cdot 10^{-5}} $$ 3. Multiply both sides by \(2.00 \cdot 10^{-5}\): $$ P = 10^{5.5} \times 2.00 \cdot 10^{-5} \mathrm{Pa} $$ 4. Calculate the value of \(P\): $$ P \approx 1.58 \mathrm{Pa} $$ Thus, the amplitude of the pressure wave generated by the rock concert is approximately \(1.58 \mathrm{Pa}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 train whistle emits a sound at a frequency \(f=3000 .\) Hz when stationary. You are standing near the tracks when the train goes by at a speed of \(v=30.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in the frequency \((|\Delta f|)\) of the whistle as the train passes? (Assume that the speed of sound is \(v=343 \mathrm{~m} / \mathrm{s}\).)

Find the resonance frequency of the ear canal. Treat it as a half-open pipe of diameter \(8.0 \mathrm{~mm}\) and length \(25 \mathrm{~mm}\). Assume that the temperature inside the ear canal is body temperature \(\left(37^{\circ} \mathrm{C}\right)\).

Electromagnetic radiation (light) consists of waves. More than a century ago, scientists thought that light, like other waves, required a medium (called the ether) to support its transmission. Glass, having a typical mass density of \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3},\) also supports the transmission of light. What would the elastic modulus of glass have to be to support the transmission of light waves at a speed of \(v=2.0 \cdot 10^{8} \mathrm{~m} / \mathrm{s} ?\) Compare this to the actual elastic modulus of window glass, which is \(5 \cdot 10^{10} \mathrm{~N} / \mathrm{m}^{2}\).

A soprano sings the note \(C 6(1046 \mathrm{~Hz})\) across the mouth of a soda bottle. To excite a fundamental frequency in the soda bottle equal to this note, describe how far the top of the liquid must be below the top of the bottle.

On a windy day, a child standing outside a school hears the school bell. If the wind is blowing toward the child from the direction of the bell, will it alter the frequency, the wavelength, or the velocity of the sound heard by the child?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Engineering Physics

Read Explanation

Electromagnetism

Read Explanation

Magnetism

Read Explanation

Mechanics and Materials

Read Explanation

Medical Physics

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.

Sign-up for free