Differential mechanics equation

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Homework Help Overview

The problem involves a differential equation derived from a force acting on a particle of mass m, expressed as F=kvx, where k is a positive constant. The original poster seeks to find the position x as a function of time t, given the initial conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up a differential equation but expresses uncertainty about its correctness and how to solve it. Another participant suggests rewriting the equation and discusses the implications of the force on the particle's motion, raising questions about the initial conditions and the context of the force.

Discussion Status

Participants are actively engaging with the problem, with one providing a reformulation of the differential equation and discussing the behavior of the particle under different conditions. The original poster seeks further clarification on the next steps after the reformulation, indicating an ongoing exploration of the topic.

Contextual Notes

There is a mention of the initial conditions and the behavior of the particle based on its velocity and position, which may influence the interpretation of the problem. The original poster also notes a lack of confidence in their understanding of differential equations.

thenewbosco
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here is the question and my solution until i have become stumped:

The force acting on a particle of mass m is given by : F=kvx where k is a positive constant. The particle passes through the origin with speed vo at t=0. Find x as a function of t.

what i have done is set up the following differential equation:

[tex]k\frac{dx}{dt}x=m\frac{d^2x}{dt^2}[/tex]

is this correct? and if so, how do i solve this type of differential equation?
i am not so strong at de's
 
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Rewrite your equation as

[tex] \frac{d^2 x}{d t^2} - 2\mu x \frac{dx}{dt} = 0[/tex]

(just relabelling constants, [itex]\mu = k/2m[/itex]), Then you notice that

[tex] 2\mu x \frac{dx}{dt} = \frac{d}{dt}\left(\mu x^2\right)[/tex]

So the entire LHS of your equation can be written as... [something]

By the way, are you sure that this is the force you're given?

The solutions you get from this should be pretty non-trivially dependent on your initial conditions (as is the case with most non-linear ODEs), for example if you analyse your force, when x is positive, and v is positive, your force is positive, and so your particle will accelerate off to infinity. If x is positive, and v is negative, then the particle is accelerated towards the origin. If x is negative, and v is negative, the particle is accelerated toward the origin (slowing the particle), until it reaches a steady state when v=0, (since the force acting on the particle is zero). If x is negative, and v is positive, the particle is accelerated away from the origin along -x (deccelerating the particle). If the particles initial momentum is not enough to reach the origin, (or just enough to reach it) it will then come to a halt (again because v=0, and also F=0). If it has enough momentum to go beyond the origin, then again it will escape to infinity.

In what context did you come across this force?
 
Last edited:
thanks for the help. however i am still not sure what to do after rewriting like this, is there anymore help someone can provide?
 
You might find this link useful:

http://www.math.uvic.ca/other/bohun/m201/m201sup/m201bern.html

By the way, post what you have got so far...
 
Last edited by a moderator:

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