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December 11th, 2016, 03:43 PM  #11  
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December 11th, 2016, 05:33 PM  #12 
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  
December 11th, 2016, 05:34 PM  #13  
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December 11th, 2016, 05:50 PM  #14 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,443 Thanks: 531 Math Focus: Wibbly wobbly timeywimey stuff.  When the pendulum is vertical we have $\displaystyle \theta = 0$. So $\displaystyle \theta (t) = \theta _0 ~ cos \left ( \sqrt{ \frac{g}{L} }~ t \right ) = 0$ Solve this for t. Divide both sides by $\displaystyle \theta _0$ then take the inverse cosine. (Hint: For what angle is cos(angle) = 0?) Dan 
December 11th, 2016, 05:56 PM  #15  
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December 11th, 2016, 06:41 PM  #16  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,443 Thanks: 531 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \theta (t = 0) = 0 = \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} ~ t \right )$ Dividing both sides by $\displaystyle \theta _0$: $\displaystyle cos \left ( \sqrt{\frac{g}{L}} ~ t \right ) = 0$ Take the inverse cosine, noting that we are betwen 0 and $\displaystyle \pi$ radians: $\displaystyle \sqrt{\frac{g}{L}}~t = \frac{\pi}{2}$ (I'll stick with angles in radians as that is the usual convention so make sure to convert to degrees if you have to.) $\displaystyle \frac{g}{L} t^2 = \frac{\pi ^2}{4}$ $\displaystyle t^2 = \frac{\pi ^2 L}{4g}$ $\displaystyle t = \sqrt{\frac{\pi ^2 L}{4g}} = \frac{\pi}{2} \sqrt{\frac{L}{g}}$ Dan Last edited by topsquark; December 11th, 2016 at 06:44 PM.  

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