/*! 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 22 (I) A transverse wave on a wire ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 22

(I) A transverse wave on a wire is given by \(D(x, t)=\) \(0.015 \sin (25 x-1200 t)\) where \(D\) and \(x\) are in meters and \(t\) is in seconds. (a) Write an expression for a wave with the same amplitude, wavelength, and frequency but traveling in the opposite direction. ( \(b\) ) What is the speed of either wave?

Short Answer

Expert verified
(a) Opposite wave: \( D(x, t) = 0.015 \sin(25x + 1200t) \). (b) Speed is 48 m/s.

Step by step solution

01

Identify the Wave Equation Parameters

The given transverse wave equation is \( D(x, t) = 0.015 \sin(25x - 1200t) \). The wave equation has the standard form \( D(x, t) = A \sin(kx - \omega t) \), where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.
02

Find the Amplitude

The amplitude \( A \) of the wave is the coefficient of the sine function. So, \( A = 0.015 \) meters.
03

Determine the Wave Number and Angular Frequency

From the equation \( D(x, t) = 0.015 \sin(25x - 1200t) \), we identify the wave number \( k = 25 \) rad/m and the angular frequency \( \omega = 1200 \) rad/s.
04

Calculate Wavelength

The wavelength \( \lambda \) is related to \( k \) by the equation \( \lambda = \frac{2\pi}{k} \). Thus, \( \lambda = \frac{2\pi}{25} \approx 0.251 \) meters.
05

Calculate Frequency

The frequency \( f \) is related to \( \omega \) by \( \omega = 2\pi f \), so \( f = \frac{1200}{2\pi} \approx 190.99 \) Hz.
06

Write the Opposite Direction Wave Equation

For the wave traveling in the opposite direction, reverse the sign of \( kt \) term. Therefore, the wave equation becomes \( D(x, t) = 0.015 \sin(25x + 1200t) \).
07

Calculate the Wave Speed

The speed \( v \) of the wave can be calculated using \( v = \frac{\omega}{k} \). Substitute in \( k = 25 \) rad/m and \( \omega = 1200 \) rad/s to get \( v = \frac{1200}{25} = 48 \) m/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.

Wave Equation
A wave equation is a mathematical representation of a wave and describes how the wave propagates through space and time. For a transverse wave, like one on a wire, it takes the form \( D(x, t) = A \sin(kx - \omega t) \), where:
  • \(D(x, t)\) is the wave displacement as a function of position \(x\) and time \(t\).
  • \(A\) is the amplitude.
  • \(k\) is the wave number, related to wavelength.
  • \(\omega\) is the angular frequency, related to the wave's frequency.
Understanding these parameters allows you to analyze and manipulate the wave. In our exercise, the equation is \( D(x, t) = 0.015 \sin(25x - 1200t) \), representing a specific wave behavior.
Amplitude
The amplitude of a wave is crucial because it indicates the maximum displacement from the rest position. In simple terms, it's the height of the wave peaks. For the given wave equation \( D(x, t) = 0.015 \sin(25x - 1200t) \), the amplitude is \(0.015\) meters. This means the wave reaches 0.015 meters above or below its equilibrium position. This measurement reflects the energy carried by the wave; a higher amplitude signifies more energy.
Wave Speed
Wave speed defines how fast the wave is traveling through the medium. It is critical in determining how wave-related phenomena unfold over time. The speed \(v\) of the wave is calculated using the formula \( v = \frac{\omega}{k} \), where \(\omega\) is the angular frequency and \(k\) is the wave number.
  • For our example, \(\omega = 1200 \) rad/s, and \(k = 25\) rad/m.
  • The calculated wave speed is \(v = \frac{1200}{25} = 48\) m/s.
This speed tells us how quickly the wave pulse or crest moves along the wire.
Wavelength
Wavelength \(\lambda\) represents the distance between two consecutive crests or troughs of the wave. Mathematically, it is calculated using \( \lambda = \frac{2\pi}{k} \), where \(k\) is the wave number. Understanding wavelength is important as it affects the wave's resilience and energy distribution.
  • For our wave, \(k = 25 \) rad/m, yielding a wavelength of \( \lambda = \frac{2\pi}{25} \approx 0.251 \) meters.
This means each wave cycle spans a length of approximately 0.251 meters in the medium.
Frequency
Frequency \(f\) describes how often the wave cycles or oscillates in one second, which is pivotal in understanding processes like sound perception and signal transmission. The angular frequency and frequency are connected through \( \omega = 2\pi f \).
  • Given \(\omega = 1200\) rad/s, the frequency is \(f = \frac{1200}{2\pi} \approx 190.99\) Hz.
This tells us the wave produces around 190.99 cycles per second, an essential parameter for applications requiring specific wave behavior settings.

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

(II) Two oppositely directed traveling waves given by \(D_{1}=(5.0 \mathrm{~mm}) \cos \left[\left(2.0 \mathrm{~m}^{-1}\right) x-(3.0 \mathrm{rad} / \mathrm{s}) t\right]\) and \(D_{2}=\) \((5.0 \mathrm{~mm}) \cos \left[\left(2.0 \mathrm{~m}^{-1}\right) x+(3.0 \mathrm{rad} / \mathrm{s}) t\right]\) form a standing wave. Determine the position of nodes along the \(x\) axis.

(II) A \(65-\mathrm{cm}\) guitar string is fixed at both ends. In the frequency range between 1.0 and \(2.0 \mathrm{kHz},\) the string is found to resonate only at frequencies \(1.2,1.5,\) and \(1.8 \mathrm{kHz} .\) What is the speed of traveling waves on this string?

(II) Suppose two linear waves of equal amplitude and frequency have a phase difference \(\phi\) as they travel in the same medium. They can be represented by $$ \begin{array}{l} D_{1}=A \sin (k x-\omega t) \\ D_{2}=A \sin (k x-\omega t+\phi) \end{array} $$ \(\begin{array}{llll}\text { (a) Use the trigonometric identity } & \sin \theta_{1}+\sin \theta_{2}=\end{array}\) \(2 \sin \frac{1}{2}\left(\theta_{1}+\theta_{2}\right) \cos \frac{1}{2}\left(\theta_{1}-\theta_{2}\right)\) to show that the resultant wave is given by $$ D=\left(2 A \cos \frac{\phi}{2}\right) \sin \left(k x-\omega t+\frac{\phi}{2}\right) $$ (b) What is the amplitude of this resultant wave? Is the wave purely sinusoidal, or not? (c) Show that constructive interference occurs if \(\phi=0,2 \pi, 4 \pi,\) and so on, and destructive interference occurs if \(\phi=\pi, 3 \pi, 5 \pi,\) etc. \((d)\) Describe the resultant wave, by equation and in words, if \(\phi=\pi / 2\).

A uniform cord of length \(\ell\) and mass \(m\) is hung vertically from a support. (a) Show that the speed of transverse waves in this cord is \(\sqrt{g h},\) where \(h\) is the height above the lower

Estimate the average power of a water wave when it hits the chest of an adult standing in the water at the seashore. Assume that the amplitude of the wave is \(0.50 \mathrm{~m}\), the wavelength is \(2.5 \mathrm{~m},\) and the period is \(4.0 \mathrm{~s}\).
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Classical Mechanics

Read Explanation

Measurements

Read Explanation

Scientific Method Physics

Read Explanation

Modern Physics

Read Explanation

Atoms and Radioactivity

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.