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Period of a Simple Harmonic Oscillator


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I was trying to explain to a friend of mine why the time it takes for one swing of a pendulum is independent of both the mass on the end and the distance it is drawn back (for small angles, of course) but could not for the life of me remember the derivation of the equation for period of a simple harmonic oscillator [math] T = 2\pi\sqrt{\frac{-x}{a}}[/math]. Could someone help me out with this?

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From trigonometry it can be found that [math] x = -l[/math] and [math] a = g [/math] which makes the period for a pendulum [math] T = 2\pi\sqrt{\frac{l}{g}}[/math] but that doesn't help with that actual derivation of the general formula for period ([math]T = 2\pi\sqrt{\frac{-x}{a}}[/math]).

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Yes, the period is indeed independent from the mass. It only varies from the length (directly) and gravitational attraction (indirectly), as Kyrisch's formula points out.


You have have a clear picture of the pendulum, then I'll just move on to derivation"


[math]F=mg\frac{x}{l}[/math] where x=elongation


we know that F=ma, so:


[math]ma=mg\frac{x}{l}[/math] so the masses cancel and then we're left with


[math]a=mg\frac{x}{l}[/math] and as [math]a=\omega^2 x [/math] then:


[math]\omega^2 x = g\frac{x}{l}[/math] again we cancel x on both sides and now you clearly see that we finally get


[math]T=2\pi \sqrt{\frac{l}{g}}[/math]


I hope this helpz,


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Generally speaking, the resonant frequency of a harmonic oscillator is only amplitude independant if the restoring force is linear, ie of the form F=-kx where k is a constant. Which of course is why a pendulum is amplitude dependant, the restoring force only satisfies this condition approximately, to the extent that the small angle aprx. sin x = x holds.

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