/*! 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 5 In a traveling electromagnetic w... [FREE SOLUTION] | 91Ó°ÊÓ

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In a traveling electromagnetic wave, the electric field is represented mathematically as $$ E=E_{0} \sin \left[\left(1.5 \times 10^{10} \mathrm{s}^{-1}\right) t-\left(5.0 \times 10^{1} \mathrm{m}^{-1}\right) x\right] $$ where \(E_{0}\) is the maximum field strength. This equation is an adaptation of Equation \(16.3 .\) (a) What is the frequency of the wave? (b) This wave and the wave that results from its reflection can form a standing wave, in a way similar to that in which standing waves can arise on a string (see Section 17.5 ). What is the separation between adjacent nodes in the standing wave?

Short Answer

Expert verified
(a) \(2.39 \times 10^{9} \mathrm{Hz}\). (b) \(0.0628 \mathrm{m}\).

Step by step solution

01

Identify the angular frequency

The given equation for the electric field of the wave is \( E = E_{0} \sin\left[(1.5 \times 10^{10} \mathrm{s}^{-1}) t - (5.0 \times 10^{1} \mathrm{m}^{-1}) x\right] \). Here, the term \((1.5 \times 10^{10} \mathrm{s}^{-1})\) is the angular frequency, denoted by \(\omega\), which is related to the frequency \(f\) by the formula \(\omega = 2\pi f\).
02

Calculate the frequency

Given \(\omega = 1.5 \times 10^{10} \mathrm{s}^{-1}\), use the formula \(\omega = 2\pi f\) to find \(f\). Rearranging gives \(f = \frac{\omega}{2\pi}\). Substitute the angular frequency \(\omega\) into this formula: \[ f = \frac{1.5 \times 10^{10}}{2\pi} \approx 2.39 \times 10^{9} \mathrm{Hz} \].
03

Identify the wave number

The wave number, denoted by \(k\), is the spatial analogue to angular frequency and is given by the coefficient of \(x\) in the equation for \(E\). Here, \(k = 5.0 \times 10^1 \mathrm{m}^{-1}\). This corresponds to the relation \(k = \frac{2\pi}{\lambda}\), where \(\lambda\) is the wavelength.
04

Calculate the wavelength

Using \(k = \frac{2\pi}{\lambda}\), solve for \(\lambda\). Rearrange to find \(\lambda = \frac{2\pi}{k}\). Substitute the wave number \(k\): \[ \lambda = \frac{2\pi}{5.0 \times 10^1} \approx 0.1256 \mathrm{m} \].
05

Find the separation between nodes

In a standing wave, the separation between adjacent nodes is half of the wavelength. Since \(\lambda = 0.1256 \mathrm{m}\), the node separation is \(\frac{\lambda}{2} = \frac{0.1256}{2} \mathrm{m} \approx 0.0628 \mathrm{m}\).

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Key Concepts

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

Frequency of Waves
Electromagnetic waves, like light and radio waves, oscillate as they propagate. A critical property of these waves is their frequency. The frequency of a wave refers to how many cycles of the wave pass a given point in a second. It's measured in Hertz (Hz).

In the case of a wave described by the equation \(E=E_{0} \sin \left[\left(1.5 \times 10^{10} \mathrm{s}^{-1}\right) t-\left(5.0 \times 10^{1} \mathrm{m}^{-1}\right) x\right]\), the term \(1.5 \times 10^{10} \mathrm{s}^{-1}\) is known as the angular frequency \(\omega\).

This angular frequency is connected to the frequency \(f\) by the relationship \(\omega = 2\pi f\). By rearranging this formula to \(f = \frac{\omega}{2\pi}\) and substituting the given \(\omega\), we can calculate the frequency. For our example, \(f \approx 2.39 \times 10^{9} \mathrm{Hz}\).

This value tells us the number of complete wave cycles that occur every second.
Standing Wave Nodes
Standing waves occur when two waves of the same type and frequency meet while traveling in opposite directions. This interaction forms a wave that appears to "stand" still, revealing specific points where there is no movement at all. These points are called nodes.

In the context of standing waves formed by electromagnetic waves, such as light reflecting in a cavity or box, the node separation is directly related to the wavelength of the wave. The nodes occur at distances \(\lambda/2\) apart, where \(\lambda\) is the wavelength.

In our given problem, once we've calculated the wavelength, which is approximately \(0.1256 \mathrm{m}\), we know the distance between each node in the standing wave is half that wavelength, which is approximately \(0.0628 \mathrm{m}\). This separation determines how the wave patterns overlap and create nodes.
Wave Number
The wave number is an essential factor that describes the spatial characteristics of waves. It is denoted by the symbol \(k\) and is defined as the number of wavelengths per unit distance, usually meters.

In the equation \( E = E_{0} \sin\left[(1.5 \times 10^{10} \mathrm{s}^{-1}) t - (5.0 \times 10^{1} \mathrm{m}^{-1}) x\right] \), the term \(5.0 \times 10^{1} \mathrm{m}^{-1}\) represents the wave number. Mathematically, \(k = \frac{2\pi}{\lambda}\), establishing a direct inverse relationship with the wavelength.

A higher wave number indicates a wave with shorter wavelengths. Understanding \(k\) helps visualize how compactly the wave is "packed" in space. This property is vital in fields such as physics and engineering, where understanding wave mechanics is crucial.
Wavelength Calculation
Wavelength is a fundamental property of waves—it signifies the length of one complete wave cycle from crest to crest or trough to trough. Calculating the wavelength provides insight into how the wave behaves in space.

From the relation \(k = \frac{2\pi}{\lambda}\), the wavelength \(\lambda\) can be found by rearranging to \(\lambda = \frac{2\pi}{k}\). Substituting \(k = 5.0 \times 10^1 \mathrm{m}^{-1}\), we determine \(\lambda \approx 0.1256 \mathrm{m}\).

This calculation shows us how "long" the wave is from one full cycle to the next. A critical application is in radio communications and optics, where wavelength informs signal transmission and light properties. By understanding how to calculate wavelength, we gain insight into the interaction and propagation of electromagnetic waves.

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Most popular questions from this chapter

A laser emits a narrow beam of light. The radius of the beam is \(1.0 \times 10^{-3} \mathrm{m},\) and the power is \(1.2 \times 10^{-3} \mathrm{W} .\) What is the intensity of the laser beam?

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m}\). Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is \(5.6 \times 10^{10} \mathrm{m}\) from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.

Unpolarized light whose intensity is \(1.10 \mathrm{W} / \mathrm{m}^{2}\) is incident on the polarizer in Figure \(24.20 .\) (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of \(\theta=75^{\circ}\) with respect to the polarizer, what is the intensity of the light that reaches the photocell?

A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated \(2.3 \mathrm{m}\) from his television set. A reporter at the press conference is located \(4.1 \mathrm{m}\) from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of \(343 \mathrm{m} / \mathrm{s}\) for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.

A lidar (laser radar) gun is an alternative to the standard radar gun that uses the Doppler effect to catch speeders. A lidar gun uses an infrared laser and emits a precisely timed series of pulses of infrared electromagnetic waves. The time for each pulse to travel to the speeding vehicle and return to the gun is measured. In one situation a lidar gun in a stationary police car observes a difference of \(1.27 \times 10^{-7} \mathrm{s}\) in round-trip travel times for two pulses that are emitted 0.450 s apart. Assuming that the speeding vehicle is approaching the police car essentially head-on, determine the speed of the vehicle.

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