/*! 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 10 The pressure in a traveling soun... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 10

The pressure in a traveling sound wave is given by the equation $$\Delta P(x, t)=(1.50 \mathrm{~Pa}) \sin \pi[(0.900 \mathrm{rad} / \mathrm{m}) x-(315 \mathrm{rad} / \mathrm{s}) t]$$ Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the wave.

Short Answer

Expert verified
Pressure amplitude: 1.50 Pa. Frequency: 50.1 Hz. Wavelength: 6.98 m. Speed of wave: 350 m/s.

Step by step solution

01

Identify Pressure Amplitude

The pressure amplitude is the coefficient in front of the sine function in the wave equation. Here, it is given directly as 1.50 Pa.
02

Determine the Wave Function Components

The general wave function for pressure is given by ewline ewline\( \Delta P(x, t) = P_0 \sin(kx - \omega t) \) ewline ewline Compare this with the given equation ewline ewline\( \Delta P(x, t)=(1.50 \mathrm{~Pa})\sin \pi[(0.900 \mathrm{rad} / \mathrm{m}) x-(315 \mathrm{rad} / \mathrm{s}) t]\) ewline ewline We can identify \(\pi k = 0.900 \mathrm{rad}/\mathrm{m}\) and \(\pi \omega = 315 \mathrm{rad}/\mathrm{s}\).
03

Extract Wave Number and Angular Frequency

Move \(\pi\) to simplify: ewline ewline \(k = 0.900 \frac{\mathrm{rad}}{\mathrm{m}} \) ewline ewline \(\omega = 315 \frac{\mathrm{rad}}{\mathrm{s}} \)
04

Calculate Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by ewline ewline \(\omega = 2\pi f \) ewline ewline Therefore, ewline ewline \( f = \frac{\omega}{2\pi} = \frac{315}{2\pi} \approx 50.1 \mathrm{Hz} \)
05

Calculate Wavelength

The wave number \( k \) is related to the wavelength \( \lambda \) by ewline ewline \( k = \frac{2\pi}{\lambda}\) ewline ewline Therefore, ewline ewline \(\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.900} \approx 6.98 \mathrm{m} \)
06

Calculate Speed of the Wave

The speed \( v \) of the wave is given by the relation ewline ewline \( v = \lambda f \) ewline ewline Substituting the values, ewline ewline \( v = 6.98 \mathrm{m} \times 50.1 \mathrm{Hz} \approx 350\mathrm{m}/\mathrm{s} \)

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Amplitude
The pressure amplitude is an essential property of any sound wave. It represents the maximum pressure variation from the wave's mean value. If we look at the given wave equation: ewline ewline \(螖P(x, t) = (1.50 \mathrm{\ Pa}) \sin \pi [(0.900 \mathrm{\ rad} / \mathrm{\ m}) x - (315 \mathrm{\ rad} / \mathrm{\ s}) t] \)ewline ewline The pressure amplitude \( \Delta P \) is the coefficient in front of the sine function, which in this case is 1.50 Pa. This value indicates the maximum deviation of pressure from its average value caused by the wave.
Frequency Calculation
Frequency is a measure of how often the wave oscillates per second. From the wave equation, the angular frequency \( \omega \) is given. Remember the relationship between angular frequency and frequency: ewline ewline \(\omega = 2\pi f \)ewline ewline By rearranging, we can find the frequency \( f \):ewline \( f = \dfrac{\omega}{2\pi} \)ewline ewline Given \( \omega = 315 \ \,\mathrm{\ rad/s} \), we have: ewline \( f = \dfrac{315}{2 \pi} \approx 50.1 \,\mathrm{\ Hz} \)ewline ewline Meaning, the sound wave oscillates around 50 times per second.
Wavelength Determination
The wavelength \( \lambda \) refers to the distance over which the wave's shape repeats. It connects to the wave number \( k \) by the equation: ewline ewline \( k = \dfrac{2\pi}{\lambda} \)ewline ewline Rearrange it to solve for \( \lambda \):ewline \( \lambda = \dfrac{2\pi}{k} \)ewline ewline From the problem, \( k = 0.900 \ \, \mathrm{\ rad/m} \), thus: ewline \( \lambda = \dfrac{2\pi}{0.900} \approx 6.98 \mathrm{\ m} \)ewline ewline This means the sound wave's repeating pattern spans nearly 7 meters.
Wave Speed
The wave speed \( v \) is the speed at which the wave propagates through the medium. It can be found using the relationship: ewline ewline \( v = \lambda f \)ewline ewline By substituting the values we calculated: ewline \( \lambda = 6.98 \mathrm{\ m} \) and \( f = 50.1 \mathrm{\ Hz} \), we get: ewline \( v = 6.98 \, \mathrm{\ m} \times 50.1 \mathrm{\ Hz} \approx 350 \mathrm{\ m/s} \)ewline ewline This indicates that the sound wave travels at a speed of 350 meters per second in the medium.

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

Trooper \(B\) is chasing speeder \(A\) along a straight stretch of road. Both are moving at a speed of \(160 \mathrm{~km} / \mathrm{h}\). Trooper \(B\), failing to catch up, sounds his siren again. Take the speed of sound in air to be \(343 \mathrm{~m} / \mathrm{s}\) and the frequency of the source to be \(500 \mathrm{~Hz}\). What is the Doppler shift in the frequency heard by speeder \(A\) ?

(a) Find the speed of waves on a violin string of mass \(800 \mathrm{mg}\) and length \(22.0 \mathrm{~cm}\) if the fundamental frequency is \(920 \mathrm{~Hz}\). (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

Fig. \(18-32, S\) is a small loudspeaker driven by an audio oscillator and amplifier, adjustable in frequency from 1000 to \(2000 \mathrm{~Hz}\) only. Tube \(D\) is a piece of cylindrical sheet-metal pipe \(45.7 \mathrm{~cm}\) long and open at both ends. (a) If the speed of sound in air is \(344 \mathrm{~m} / \mathrm{s}\) at the existing temperature, at what frequencies will resonance occur in the pipe when the frequency emitted by the speaker is varied from \(1000 \mathrm{~Hz}\) to \(2000 \mathrm{~Hz}\) ? (b) Sketch the standing wave (using the style of Fig. \(18-16 b\) ) for each resonant frequency.

A \(1.0 \mathrm{~W}\) point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a) \(10 \mathrm{~m}\) from the source and (b) \(2.5 \mathrm{~m}\) from the source.

Two point sources of sound waves of identical wavelength \(\lambda\) and amplitude are separated by distance \(D=\) \(2.0 \lambda\). The sources are in phase. (a) How many points of maximum signal (that is, maximum constructive interference) lie along a large circle around the sources? (b) How many points of minimum signal (destructive interference) lie around the circle?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Famous Physicists

Read Explanation

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electric Charge Field and Potential

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.