Classical Mechanics

  • Thread starter matt222
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  • #1
matt222
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Homework Statement


Repelling force F=kx. What is the motion equation

Homework Equations



F=ma

The Attempt at a Solution


what I did so far, I found the equation of motion which is equal to x= sqrt(2E/K)Sin(wt+b), I am not sure about my answer.

For the second part it is kind of tricky for me, I know a point where the potential energy has a minimum is called point of stable equilibrium, and a point where the the potential has a maximum is called a point of unstable eqiulibrium. I couldn't really got what I should do for the second part
 
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Answers and Replies

  • #2
JesseC
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You've got the the wrong solution :) I think you have mistaken it for a simple harmonic oscillator! A repulsive force of the form you wrote may have solutions in terms of exponentials (most probably simplest) or hyperbolic functions like sinh, cosh etc (not necessary in this case i think although they are basically equivalent).
 
  • #3
matt222
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it is indeed SHM, since we have mdv/dt=F and we have been givin the repelling force we only need to integerat it and found the answer that's what I got,
 
  • #4
JesseC
251
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Homework Statement



Find the solution for the motion of a body subject to a linear repelling force F=kx. Show that this motion to be expected in the neighborhood of a point of unstable equilibrium.

It is most definitely not SHM :D ! The difference between it and SHM is a minus sign... which makes all the difference. SHM relies on an **attractive** force about a stable equilibrium, not a repulsive one around an unstable equilibrium as you yourself wrote in the question.

For it to be SHM:

[tex] F = -k x [/tex]

However in your case

[tex] F = k x [/tex]

Thus the equation of motion you are trying to solve is:

[tex] m {\frac{d^2x}{dt^2}} = kx [/tex] (1)

Which has solutions of the form:

[tex] x = A e ^ {\omega t}+B e ^ {-\omega t} [/tex]

where

[tex]\omega^2=\frac{k}{m}[/tex]

Where A and B are constants. You can easily check this solution is correct by plugging it back into (1) and showing that it works.

You should always check your solution by putting it back into the differential equation if you're not sure.
 
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