Simple Harmonic Oscillator: From Hooke's Law to Harmonious Motion

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Homework Help Overview

The discussion revolves around demonstrating simple harmonic motion (SHM) starting from Hooke's Law, specifically through the formulation of the differential equation derived from the force equation F = -kx. Participants are exploring the relationship between the mathematical representation of SHM and the solutions to the corresponding differential equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various approaches to solving the differential equation, including the use of F = ma and the characteristic equation. There are attempts to derive the solution using integration and logarithmic functions, with some questioning where the derivation may have gone awry.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on the derivation process. Some have pointed out potential errors in the integration steps, while others are considering the implications of complex roots in the characteristic equation. There is a sense of collaborative exploration, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of direct guidance provided. There is also an acknowledgment of confusion regarding the transition from the derived forms to the expected solutions, particularly concerning the appearance of complex numbers.

Bill Foster
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Homework Statement



Show simple harmonic motion starting from Hooke's Law.


The Attempt at a Solution



[tex]F=-kx[/tex]

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

[tex]\frac{1}{x}\frac{d^2x}{dt^2}=-\frac{k}{m}[/tex]

[tex]=\frac{1}{x}\frac{d}{dt}\frac{dx}{dt}=-\frac{k}{m}[/tex]

[tex]\int\int\frac{1}{x}d\left(dx\right)=-\frac{k}{m}\int dt\int dt[/tex]

[tex]\int\frac{1}{x}dx=-\frac{k}{2m}t^2[/tex]

[tex]\ln\left(x\right)=-\frac{k}{2m}t^2[/tex]

[tex]x\left(t\right)=e^{-\frac{k}{2m}t^2}[/tex]

But it should be:

[tex]x\left(t\right)=e^{-i\sqrt{\frac{k}{2m}}t}[/tex]

I thought I knew how to do this. :redface:
 
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Well normally you could just put F=ma and then put it the form [itex]a= - \omega^2 x[/itex] and that would be sufficient to show SHM.


Else to solve the DE, you would need to know that the equation y''+ky=0 has solutions y1=cos(kt) and y2=sin(kt)
 
I want to solve the differential equation. But as you can see, somewhere I need to take the square root of the exponential argument.

Where, in my steps, did I miss that?
 
Nobody knows?
 
I don't see how to use this to get to the right answer, but:

[tex]\int\frac{1}{x}dx \neq \ln (x)[/tex]

[tex]\int\frac{1}{x}dx = \ln |x| + C[/tex]

So the derivation would continue:

[tex]\ln |x| + C =-\frac{k}{2m}t^2[/tex]

[tex]|x| = e^{-\frac{k}{2m}t^2 + C}[/tex]

Like I said, not sure how to use this, or if it helps. I could see squaring x, then taking the square root, in lieu of the absolute value signs, but, then, I don't see where the i comes out. Most of the derivations I've seen 'guess' at the solution of the 2nd order DE to be of the form [itex]a cos(\omega t)[/itex], and go from there.

I'm going to look at this more tomorrow, I can't believe I don't know either.
 
dotman said:
I'm going to look at this more tomorrow, I can't believe I don't know either.


[tex]m\frac{d^2x}{dt^2}=-kx \Rightarrow \frac{d^2x}{dt^2}+\frac{k}{m}x=0[/tex]


the auxiliary equation is r2+(k/m)=0 so r=±√(k/m)i

when you have roots in the form r=λ±μi what is the general solution x(t) equal to?
 

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