Kyrisch Posted June 9, 2008 Share Posted June 9, 2008 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? Link to comment Share on other sites More sharing options...

ecoli Posted June 9, 2008 Share Posted June 9, 2008 Isn't it given by trigonometry? http://en.wikipedia.org/wiki/Sine#Periodic_functions Link to comment Share on other sites More sharing options...

Kyrisch Posted June 9, 2008 Author Share Posted June 9, 2008 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]). Link to comment Share on other sites More sharing options...

kevinalm Posted June 9, 2008 Share Posted June 9, 2008 Here's a link: http://farside.ph.utexas.edu/teaching/336k/lectures/node38.html Fitzpatrick explains it pretty well. Link to comment Share on other sites More sharing options...

thedarkshade Posted June 9, 2008 Share Posted June 9, 2008 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, Shade! Link to comment Share on other sites More sharing options...

swansont Posted June 9, 2008 Share Posted June 9, 2008 Note also the dependence on the small-angle approximation Link to comment Share on other sites More sharing options...

kevinalm Posted June 9, 2008 Share Posted June 9, 2008 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. Link to comment Share on other sites More sharing options...

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