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I would like to know how one could show and prove that a given motion is simple harmonic motion.

Once given an answer, I'll apply that technique to an example I am trying to figure out.

Thank you in advance!

I believe a motion can be proved simple harmonic, if the relation between its is as such: $$ a_x = - \omega^2\cdot x $$

And as such the period time is: $$ T =\frac{2\pi}{\omega} $$

Question-so-far: How do you prove such for a given force $F = \frac{G\cdot m_e \cdot M}{R_E} \cdot r$ ? Or any force that has non-trivial constants?

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You should specify what information you are given. For example, the Brachistochrone motion is simple harmonic, but showing this is nontrivial, and requires solving the Brachistochrone. –  Ron Maimon Jan 17 '12 at 5:51
    
Related: physics.stackexchange.com/q/1018/2451 –  Qmechanic Jan 20 '13 at 17:35
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2 Answers 2

If the total force $F$ on a mass $m$ follows Hooke's law,

$$F~=~-kx,$$

then one can use Newton's 2nd law

$$F=ma,$$

to infer that the motion is a simple harmonic motion

$$ a =-\omega^2x, \qquad\qquad \frac{2\pi}{T}~=~\omega~=~ \sqrt{\frac{k}{m}}~,$$

cf. OP's correct belief. Now it only remains to solve the ODE

$$ \frac{d^2x(t)}{dt^2}~=~-\omega^2x(t), $$

which is a pure math exercise.

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$$F = ma$$ $$F = -mg \sin \theta$$ $$ma = -\frac{mgx}{L}$$ $$a = -\frac{gx}{L}$$ Since $-g/L$ is a constant, $a \propto -x$. Hence proved.

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Almost nothing this terse is ever helpful. We expect better quality than that. -1 –  Brandon Enright Feb 24 at 16:53
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