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Chapter 16: Problem 27

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{m}\), a speed of \(5.2 \mathrm{m} / \mathrm{s},\) and a frequency of \(14 \mathrm{Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

Short Answer

Expert verified
The wave equation is \( y(x, t) = 0.35 \sin(16.91x - 28\pi t) \).

Step by step solution

01

Identify the Wave Equation

We are asked to write the wave equation. The general form of a traveling wave equation is \( y(x, t) = A \sin(kx - \omega t + \phi) \) for a wave moving in the \(+x\) direction, where \( A \) is the amplitude, \( k \) is the wave number, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.
02

Plug in the Amplititude

From the problem, the amplitude \( A = 0.35 \mathrm{m} \). Therefore, the wave equation starts as \( y(x, t) = 0.35 \sin(kx - \omega t + \phi) \).
03

Calculate the Angular Frequency \( \omega \)

The angular frequency \( \omega \) is calculated using \( \omega = 2 \pi f \), where \( f \) is the frequency. Here, \( f = 14 \ \mathrm{Hz} \). So, \( \omega = 2 \pi \times 14 = 28\pi \ \mathrm{rad/s} \).
04

Calculate the Wave Number \( k \)

The wave number \( k \) is given by \( k = \frac{2 \pi}{\lambda} \). The wave speed \( v \) is given by \( v = f \lambda \) or rearranged, \( \lambda = \frac{v}{f} \). With \( v = 5.2 \ \mathrm{m/s} \) and \( f = 14 \ \mathrm{Hz} \), we find \( \lambda = \frac{5.2}{14} \approx 0.3714 \ \mathrm{m} \). Thus, \( k = \frac{2\pi}{0.3714} \approx 16.91 \ \mathrm{m}^{-1} \).
05

Assume the Phase Constant \( \phi \)

The problem doesn't specify a phase constant, so we can assume \( \phi = 0 \) for simplicity in writing the wave equation.
06

Write the Full Wave Equation

Plug all the values into the wave equation form. We have \( A = 0.35 \), \( k = 16.91 \), \( \omega = 28\pi \), and \( \phi = 0 \). So the wave equation is \( y(x, t) = 0.35 \sin(16.91x - 28\pi t) \).

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Key Concepts

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

Wave Properties
Waves are fascinating phenomena that can be seen in many forms, from ocean waves to sound waves. Here, we focus on properties crucial to understanding a traveling wave, such as amplitude, speed, and frequency.
  • Amplitude: This is the maximum displacement of points on a wave, measured from the rest position. In our example, the amplitude is 0.35 meters, meaning each peak (or trough) reaches 0.35 meters from the midline.
  • Speed: This tells us how fast the wave is moving. For our wave, the speed is 5.2 meters per second in the positive x-direction.
  • Frequency: Frequency refers to how many waves pass a point in one second, measured in Hertz (Hz). Here, it is 14 Hz, meaning 14 wave cycles pass a point every second.
Understanding these properties helps us write the appropriate wave equation, which describes the wave's behavior.
Angular Frequency
Angular frequency is a critical concept when analyzing wave motion. It measures how quickly the wave oscillates in terms of radians per second. Unlike regular frequency, which is measured in cycles per second (Hz), angular frequency uses radians as its unit.
The formula to calculate angular frequency \( \omega \) is \( \omega = 2 \pi f \), where \( f \) is the frequency.
In our exercise, \( \omega = 2 \pi \times 14 = 28\pi \ \mathrm{rad/s} \).
This conversion from frequency to angular frequency helps connect wave properties to circular motion ideas, as waves exhibit repetitive oscillations similar to circles.
Wave Number
The wave number \( k \) is another fundamental property of waves. It relates to the wavelength, which is the distance between two consecutive wave peaks (or troughs).
The wave number is calculated using the formula \( k = \frac{2 \pi}{\lambda} \), where \( \lambda \) is the wavelength.
To find \( \lambda \), use the wave speed \( v \) and frequency \( f \) in the formula \( \lambda = \frac{v}{f} \). In this case, \( \lambda = \frac{5.2}{14} \approx 0.3714 \ \mathrm{m} \).
Thus, \( k = \frac{2\pi}{0.3714} \approx 16.91 \ \mathrm{m}^{-1} \).
Wave number helps us incorporate spatial characteristics of wave propagation into our equations.
Traveling Waves
A traveling wave is a wave that moves through space, transferring energy from one point to another. The movement is described by a wave equation, which shows how displacement changes over time at any point on the wave.
In our scenario, the wave is moving in the positive x-direction, and the equation given is \( y(x, t) = A \sin(kx - \omega t + \phi) \).
This equation tells us:
  • \( A \) is the amplitude, deciding the wave's height.
  • \( kx \) captures how the wave changes in space.
  • \( \omega t \) describes how it oscillates in time.
  • \( \phi \) is the phase constant, which can shift the wave horizontally.
Understanding traveling waves helps us model and predict how they carry energy through various mediums.

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Most popular questions from this chapter

A woman stands a distance \(d\) from a loud motor that emits sound uniformly in all directions. The sound intensity at her position is an uncomfortable \(3.2 \times 10^{-3} \mathrm{W} / \mathrm{m}^{2}\). There are no reflections. At a position twice as far from the motor, what are (a) the sound intensity and (b) the sound intensity level relative to the threshold of hearing?

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?

Two sources of sound are located on the \(x\) axis, and each emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at \(x=+123 \mathrm{m} .\) The source at the origin emits four times as much power as the other source. Where on the \(x\) axis are the two sounds equal in intensity? Note that there are two answers.

Two submarines are under water and approaching each other head-on. Sub A has a speed of \(12 \mathrm{m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{m} / \mathrm{s}\). Sub \(\mathrm{A}\) sends out a \(1550-\mathrm{Hz}\) sonar wave that travels at a speed of \(1522 \mathrm{m} / \mathrm{s}\). (a) What is the frequency detected by sub \(B ?\) (b) Part of the sonar wave is reflected from sub \(B\) and returns to sub \(A\). What frequency does sub A detect for this reflected wave?

A car is parked \(20.0 \mathrm{m}\) directly south of a railroad crossing. A train is approaching the crossing from the west, headed directly east at a speed of \(55.0 \mathrm{m} / \mathrm{s}\). The train sounds a short blast of its \(289-\mathrm{Hz}\). horn when it reaches a point \(20.0 \mathrm{m}\) west of the crossing. What frequency does the car's driver hear when the horn blast reaches the car? The speed of sound in air is \(343 \mathrm{m} / \mathrm{s} .\)
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