/*! 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 43 A Generator A generator at one e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Generator A generator at one end of a very long string creates a wave given by $$ y(x, t)=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}[(2.0 \mathrm{rad} / \mathrm{m}) x+(8.0 \mathrm{rad} / \mathrm{s}) t] $$ and one at the other end creates the wave $$ y(x, t)=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}[(2.0 \mathrm{rad} / \mathrm{m}) x-(8.0 \mathrm{rad} / \mathrm{s}) t] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. At what \(x\) values are the (d) nodes and (e) antinodes?

Short Answer

Expert verified
Frequency: 1.27 Hz, Wavelength: 3.14 m, Speed: 4.0 m/s, Nodes: \(n \cdot 1.57 \) m, Antinodes: \((n + 0.5) \cdot 1.57 \) m.

Step by step solution

01

Identify the given wave equations

The given wave equations are:1. \( y_1(x, t) = (6.0 \, \mathrm{cm}) \, \cos \left( \frac{\pi}{2} \left[ (2.0 \, \mathrm{rad/m}) \, x + (8.0 \, \mathrm{rad/s}) \, t \right] \right) \)2. \( y_2(x, t) = (6.0 \, \mathrm{cm}) \, \cos \left( \frac{\pi}{2} \left[ (2.0 \, \mathrm{rad/m}) \, x - (8.0 \, \mathrm{rad/s}) \, t \right] \)
02

Calculate the frequency

The general wave equation is given by:\( y(x, t) = A \, \cos (kx \pm \omega t) \)Comparing with \( y_1(x, t) \), we can identify \( \omega = 8.0 \, \mathrm{rad/s} \). Frequency \( f \) can be found using the relation \( \omega = 2 \pi f \):\[ f = \frac{\omega}{2 \pi} = \frac{8.0 \, \mathrm{rad/s}}{2 \pi} = \frac{8.0}{2\pi} \, \mathrm{Hz} \approx 1.27 \, \mathrm{Hz} \]
03

Calculate the wavelength

Comparing the wave equation with the general form, we identify \( k = 2.0 \, \mathrm{rad/m} \). The wavelength \( \lambda \) is given by:\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2.0 \, \mathrm{rad/m}} = \pi \, \mathrm{m} \approx 3.14 \, \mathrm{m} \]
04

Calculate the speed of the wave

The speed of the wave \( v \) can be found using the relation \( v = f \lambda \):\[ v = (1.27 \, \mathrm{Hz}) \cdot (3.14 \, \mathrm{m}) \approx 4.0 \, \mathrm{m/s} \]
05

Determine the positions of nodes

Nodes occur where the amplitude of the resultant wave is zero. This happens for standing waves at positions:\[ kx = n\pi, \quad n = 0, \pm1, \pm2, \ldots \]Given \( k = 2.0 \, \mathrm{rad/m} \), the positions of nodes are:\[ x = \frac{n\pi}{2.0} = n \cdot 1.57 \, \mathrm{m}, \quad n = 0, \pm1, \pm2, \ldots \]
06

Determine the positions of antinodes

Antinodes occur where the amplitude is maximum. This happens at positions:\[ kx = (n + 0.5)\pi, \quad n = 0, \pm1, \pm2, \ldots \]Given \( k = 2.0 \, \mathrm{rad/m} \), the positions of antinodes are:\[ x = \frac{(n + 0.5)\pi}{2.0} = (n + 0.5) \cdot 1.57 \, \mathrm{m}, \quad n = 0, \pm1, \pm2, \ldots \]

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 91Ó°ÊÓ!

Key Concepts

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

Wave Frequency
The frequency of a wave, denoted as \( f \), is the number of wave cycles that pass a point per second. The unit of frequency is Hertz (Hz). For the given wave equation, you can compare it with the general form of a wave equation \( y(x, t) = A \cos(kx \pm \omega t) \). Here, \( \omega = 8.0 \, \mathrm{rad/s} \) is the angular frequency. Using the formula \( \omega = 2\pi f \):
\[ f = \frac{\omega}{2\pi} = \frac{8.0 \, \mathrm{rad/s}}{2\pi} = \frac{8.0}{2\pi} \approx 1.27 \, \mathrm{Hz} \] This means that each second, approximately 1.27 waves pass a fixed point.
Wave Wavelength
The wavelength \( \lambda \) of a wave is the distance between two consecutive points that are in phase, such as two crests or troughs. The given wave equation shows that the wave number, \( k \), is \( 2.0 \, \mathrm{rad/m} \). The relationship between the wave number and the wavelength is:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2.0 \, \mathrm{rad/m}} = \pi \, \mathrm{m} \approx 3.14 \, \mathrm{m} \] This indicates that each cycle of the wave stretches over a distance of about 3.14 meters.
Wave Speed
Wave speed \( v \) is the rate at which a wave propagates through a medium. It can be calculated using the formula:
\( v = f \lambda \). From the previous sections, we found that \( f \approx 1.27 \, \mathrm{Hz} \) and \( \lambda \approx 3.14 \, \mathrm{m} \). Plugging in these values, we get:
\[ v = (1.27 \, \mathrm{Hz}) \cdot (3.14 \, \mathrm{m}) \approx 4.0 \, \mathrm{m/s} \] Therefore, the wave moves at a speed of approximately 4.0 meters per second in the medium.
Nodes and Antinodes
In standing waves, nodes are points where the wave has zero amplitude due to destructive interference. Antinodes are points where the amplitude is maximum due to constructive interference.

**Nodes**: Nodes occur at positions where \( kx = n\pi \), where \( n = 0, \pm1, \pm2, \ldots \). Given \( k = 2.0 \, \mathrm{rad/m} \), the positions are:
\[ x = \frac{n\pi}{2.0} = n \cdot 1.57 \, \mathrm{m} \]

**Antinodes**: Antinodes occur at positions where \( kx = (n + 0.5)\pi \), where \( n = 0, \pm1, \pm2, \ldots \). Given \( k = 2.0 \, \mathrm{rad/m} \), the positions are:
\[ x = \frac{(n + 0.5)\pi}{2.0} = (n + 0.5) \cdot 1.57 \, \mathrm{m} \] These relationships describe where the maxima and minima of the wave are located.
Standing Waves
A standing wave is formed when two waves of the same frequency and amplitude travel in opposite directions, superimposing to create a wave that appears to be standing still. The provided wave equations represent such a situation. One wave travels in the positive direction:
\( y_1(x, t) = (6.0 \, \mathrm{cm}) \cos \left( \frac{\pi}{2} \left[ (2.0 \, \mathrm{rad/m}) \, x + (8.0 \, \mathrm{rad/s}) \, t \right] \right) \)
and the other in the negative direction:
\( y_2(x, t) = (6.0 \, \mathrm{cm}) \cos \left( \frac{\pi}{2} \left[ (2.0 \, \mathrm{rad/m}) \, x - (8.0 \, \mathrm{rad/s}) \, t \right] \). When these two waves overlap, they form a standing wave with nodes and antinodes as discussed. The nodes are points of zero displacement whereas antinodes are points of maximum displacement.

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

Electromagnetic Waves The speed of electromagnetic waves (which include visible light, radio, and \(x\) -rays) in vacuum is \(3.0 \times\) \(10^{8} \mathrm{~m} / \mathrm{s}\). (a) Wavelengths of visible light waves range from about \(400 \mathrm{~nm}\) in the violet to about \(700 \mathrm{~nm}\) in the red. What is the range of frequencies of these waves? (b) The range of frequencies for shortwave radio (for example, FM radio and VHF television) is \(1.5\) to \(300 \mathrm{MHz}\). What is the corresponding wavelength range? (c) X-ray wavelengths range from about \(5.0 \mathrm{~nm}\) to about \(1.0 \times\) \(10^{-2} \mathrm{~nm}\). What is the frequency range for \(\mathrm{x}\) -rays?

Amplitude of the Resultant Determine the amplitude of the resultant wave when two sinusoidal string waves having the same frequency and traveling in the same direction on the same string are combined, if their amplitudes are \(3.0 \mathrm{~cm}\) and \(4.0 \mathrm{~cm}\) and they have initial phases of \(0.0\) and \(\pi / 2 \mathrm{rad}\), respectively.

Three-Loop Standing Wave A string \(3.0 \mathrm{~m}\) long is oscillating as a three-loop standing wave with an amplitude of \(1.0 \mathrm{~cm}\). The wave speed is \(100 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency? (b) Write equations for two waves that, when combined, will result in this standing wave.

Resonant Frequencies A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Moving a Nonsymmetric Triangular Pulse A long taut spring is started at a time \(t=0\) with a pulse moving in the \(+x\) direction in the shape given by the function \(f(x)\) with $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{4} x+1 & -4

See all solutions

Recommended explanations on Physics Textbooks

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.