Simple Harmonic Motion Differential Equation

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Homework Help Overview

The problem involves a particle of mass m moving under a force proportional to its displacement, leading to a differential equation related to simple harmonic motion. The original poster attempts to derive the equation and verify the solution form, while also seeking to express the position at later times given initial conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the differential equation and the verification of the proposed solution. There are attempts to differentiate the solution and compare it with the derived equation, alongside questions about the implications of the results and the substitution of known values.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the verification of the solution. There is recognition of the need to substitute known values into the equations, and some participants express confusion about the implications of their calculations.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to verify the solution and the initial conditions provided. There is an emphasis on understanding the relationship between acceleration, displacement, and the constants involved.

Matuku
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Homework Statement


A particle of mass m moves in one dimension under the action of a force given by -kx where x is the displacement of the body at time t, and k is a positive constant. Using F=ma write down a differential equation for x, and verify that its solution is x=Acos([tex]\omega[/tex]t+[tex]\phi[/tex]), where [tex]\omega[/tex]2=k/m (omega squared, that is). If the body starts from rest at the point x=A at time t=0, find an expression for x at later times.

Homework Equations





The Attempt at a Solution


I think the differential equation they're looking for is,
a=-kx/m

As a=d2x/dt2

But from here I can't see where to go; integration of course leads to the wrong formula.
 
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You have the differential equation so just substitute the solution in and show that both sides are equal.
 
Kurdt said:
You have the differential equation so just substitute the solution in and show that both sides are equal.

So differentiate the solution given to us to get it in terms of acceleration and then just compare that with the a=-kx/m?
 
Thats correct.
 
I'm sorry I'm still sightly confused; I now have:

[tex]a=-A\omega^{2}cos(\omega t+\phi)=\frac{-kx}{m}[/tex]

Which implies that [tex]x=Acos(\omega t+\phi)[/tex] but doesn't really show why? Is this what you intended or am I missing something?
 
You're also told what omega is in the question, and you haven't substituted for x on the right hand side.
 
Oh of course it is! We're told what x is!
 

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