Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 5

Find the equation of motion of a particle moving along the \(x\) axis if the potential energy is \(V=\frac{1}{2} k x^{2} .\) (This is a simple harmonic oscillator.)

Short Answer

Expert verified
The equation of motion is \( x(t) = A\cos \left( \sqrt{\frac{k}{m}} t + \phi \right) \).

Step by step solution

01

Identify the Potential Energy and Force

The potential energy given is \[V=\frac{1}{2} k x^{2}\]. Identify the force acting on the particle using the relation \[ F = -\frac{dV}{dx} \].
02

Compute the Force

Differentiate the potential energy with respect to x:\[ F = -\frac{d}{dx} \left( \frac{1}{2} k x^{2} \right) = -k x \].
03

Apply Newton's Second Law

Newton's second law states that \[ F = ma \] where \(a\) is the acceleration. Substitute the expression for force to get \[ -kx = m\frac{d^{2}x}{dt^{2}} \].
04

Derive the Equation of Motion

Rearrange the equation to obtain the standard form of the equation of motion for simple harmonic motion:\[ \frac{d^{2}x}{dt^{2}} + \frac{k}{m}x = 0 \].
05

Write the Solution for the Equation of Motion

The general solution for the equation of motion \( \frac{d^{2}x}{dt^{2}} + \omega^{2}x = 0 \) where \( \omega = \sqrt{\frac{k}{m}} \), is \[ x(t) = A\cos(\omega t + \phi) \].

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!

Key Concepts

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

Simple Harmonic Oscillator
A simple harmonic oscillator is a system that experiences a restoring force proportional to its displacement from equilibrium.
Imagine you've attached a mass to a spring. When you pull it and then release, the mass vibrates back and forth. This is simple harmonic motion.
Based on the exercise, we know the potential energy is given by \[ V = \frac{1}{2} k x^{2} \]. Here, \(k\) is the force constant and \(x\) is the displacement.
Newton's Second Law
Newton's second law is all about understanding how forces influence motion.
It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \[ F = ma \].
In our problem, by combining this with the force derived from the potential energy, we find \[-kx = m\frac{d^{2}x}{dt^{2}} \]. This is the foundational principle that allows us to describe the motion of the oscillator.
Potential Energy
Potential energy represents the stored energy of an object due to its position or state.
In our scenario, the given potential energy is \[ V = \frac{1}{2} k x^{2} \]. This form indicates we are dealing with a spring-like system.
When we differentiate this potential energy with respect to \(x\), we get the force: \[ F = - \frac{dV}{dx} = -kx \], which is the classic restoring force for a simple harmonic oscillator.

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

A uniform flexible chain of given length is suspended at given points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right) .\) Find the curve in which it hangs. Hint: It will hang so that its center of gravity is as low as possible.

Use Fermat's principle to find the path of a light ray through a medium of index of refraction proportional to the given function. \((x+y)^{1 / 2} \quad\) Hint: Make the change of variables \((45^{\circ}\) rotation). $$X=\frac{1}{\sqrt{2}}(x+y), \quad Y=\frac{1}{\sqrt{2}}(x-y) ; \quad \text { what is } \quad d X^{2}+d Y^{2} ?$$

(a) Consider the case of two dependent variables. Show that if \(F=F\left(x, y, z, y^{\prime}, z^{\prime}\right)\) and we want to find \(y(x)\) and \(z(x)\) to make \(I=\int_{x_{1}}^{x_{2}} F d x\) stationary, then \(y\) and \(z\) should each satisfy an Euler equation as in (5.1). Hint: Construct a formula for a varied path \(Y\) for \(y\) as in Section \(2[Y=y+\epsilon \eta(x) \text { with } \eta(x) \text { arbitrary }]\) and construct a similar formula for \(z\) llet \(Z=z+\epsilon \zeta(x),\) where \(\zeta(x)\) is another arbitrary function]. Carry through the details of differentiating with respect to \(\epsilon,\) putting \(\epsilon=0,\) and integrating by parts as in Section \(2 ;\) then use the fact that both \(\eta(x)\) and \(\zeta(x)\) are arbitrary to get (5.1). (b) Consider the case of two independent variables. You want to find the function \(u(x, y)\) which makes stationary the double integral $$\int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} F\left(u, x, y, u_{x}, u_{y}\right) d x d y$$. Hint: Let the varied \(U(x, y)=u(x, y)+\epsilon \eta(x, y)\) where \(\eta(x, y)=0\) at \(x=x_{1}\) \(x=x_{2}, y=y_{1}, y=y_{2},\) but is otherwise arbitrary. As in Section \(2,\) differentiate \(x=y=y=y\), with respect to \(\epsilon,\) set \(\epsilon=0,\) integrate by parts, and use the fact that \(\eta\) is arbitrary. Show that the Euler equation is then $$\frac{\partial}{\partial x} \frac{\partial F}{\partial u_{x}}+\frac{\partial}{\partial y} \frac{\partial F}{\partial u_{y}}-\frac{\partial F}{\partial u}=0$$ (c) Consider the case in which \(F\) depends on \(x, y, y^{\prime},\) and \(y^{\prime \prime} .\) Assuming zero values of the variation \(\eta(x)\) and its derivative at the endpoints \(x_{1}\) and \(x_{2},\) show that then the Euler equation becomes $$\frac{d^{2}}{d x^{2}} \frac{\partial F}{\partial y^{\prime \prime}}-\frac{d}{d x} \frac{\partial F}{\partial y^{\prime}}+\frac{\partial F}{\partial y}=0$$

Find the geodesics on the cone \(x^{2}+y^{2}=z^{2} .\) Hint: Use cylindrical coordinates.

Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler. \(\int_{x_{1}}^{x_{2}} \sqrt{1+y^{2} y^{\prime 2}} d x\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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