Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 41

Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time \(t\) the displacement is given by \(x=A \cos \omega t\) and \(y=A \cos \left(\omega t+\phi_{y}\right)\). Describe the path of the electrons and determine its equation when (a) \(\phi_{y}=0^{\circ}\), (b) \(\phi_{\mathrm{v}}=30^{\circ}\), and \((c) \phi_{y}=90^{\circ}\).

Short Answer

Expert verified
(a) For \(\phi_{y}=0^{\circ}\), path is a straight line, \(y=x\). (b) For \(\phi_{y}=30^{\circ}\), path is an ellipse, equation is not easily obtainable. (c) For \(\phi_{y}=90^{\circ}\), path is a circle, \((x/A)^2 + (y/A)^2 = 1\).

Step by step solution

01

Define displacement equations

The given displacement equations are \(x=A \cos \omega t\) for x-direction and \(y=A \cos \left(\omega t+ \phi_{y}\right)\) for y-direction where \(\phi_{y}\) is the phase difference for y.
02

Situation (a): In case of \(\phi_{y}=0^{\circ}\)

Here, phase \(\phi_{y}\) is \(0^{\circ}\) or 0 radians. So the displacement in y-direction is \(y=A \cos \omega t\). We notice that displacement equations for x and y are identical. Therefore, the path of electron forms a straight line with slope 1, and the equation of the path is \(y = x\).
03

Situation (b): In case of \(\phi_{y}=30^{\circ}\)

Here, phase \(\phi_{y}\) is \(30^{\circ}\) or \(\pi /6\) radians. In this case, the displacement equations for x and y are not identical. Therefore, the path of the electron forms an ellipse. However, obtaining its equation requires a difficult process that goes beyond high school level physics.
04

Situation (c): In case of \(\phi_{y}=90^{\circ}\)

In this case, phase \(\phi_{y}\) is \(90^{\circ}\) or \(\pi /2\) radians. Therefore, the displacement in y-direction becomes \(y=A \sin \omega t\). With x and y both ranging from -A to +A, the path of the electron is a circle. The equation of the path will be: \((x/A)^2 + (y/A)^2 = 1\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider the forced oscillations of a damped block-spring system. Show that at resonance \((a)\) the amplitude of oscillation is \(x_{\mathrm{m}}=F_{\mathrm{m}} / b \omega\), and \((b)\) the maximum speed of the oscillating block is \(v_{\max }=F_{\mathrm{m}} / b\)

A damped harmonic oscillator involves a block (m = \(1.91 \mathrm{~kg}\) ), a spring \((k=12.6 \mathrm{~N} / \mathrm{m})\), and a damping force \(F=\) \(-b v_{x} .\) Initially, it oscillates with an amplitude of \(26.2 \mathrm{~cm}\); because of the damping, the amplitude falls to three-fourths of this initial value after four complete cycles. (a) What is the value of \(b ?(b)\) How much energy has been "lost" during these four cycles?

A pendulum whose upper end is attached so as to allow the pendulum to swing freely in any direction can be used to repeat an experiment first shown publicly by Foucault in Paris in 1851 . If the pendulum is set oscillating, the plane of oscillation slowly rotates with respect to a line drawn on the floor, even though the tension in the wire supporting the bob and the gravitational pull of the Earth on the bob lie in a vertical plane. (a) Show that this is a result of the fact that the Earth is not an inertial reference frame. ( \(b\) ) Show that for a Foucault pendulum at a latitude \(\theta\), the period of rotation of the plane, in hours, is \(24 \sin \theta .(c)\) Explain in simple terms the result at \(\theta=90^{\circ}\) (the poles) and \(\theta=0^{\circ}\) (the equator).

An automobile can be considered to be mounted on four springs as far as vertical oscillations are concerned. The springs of a certain car of mass \(1460 \mathrm{~kg}\) are adjusted so that the vibrations have a frequency of \(2.95 \mathrm{~Hz}\). (a) Find the force constant of each of the four springs (assumed identical). (b) What will be the vibration frequency if five persons, averaging \(73.2\) kg each, ride in the car?

An oscillating block-spring system has a mechanical energy of \(1.18 \mathrm{~J}\), an amplitude of \(9.84 \mathrm{~cm}\), and a maximum speed of \(1.22 \mathrm{~m} / \mathrm{s} .\) Find \((a)\) the force constant of the spring, \((b)\) the mass of the block, and \((c)\) the frequency of oscillation.
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Engineering Physics

Read Explanation

Waves Physics

Read Explanation

Wave Optics

Read Explanation

Measurements

Read Explanation

Thermodynamics

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.

Sign-up for free