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 Differential Equations Ordinary and Partial Differential Equations Math Forum

December 11th, 2016, 02:43 PM   #11
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 Originally Posted by topsquark Yes, I wound up doing a number of edits. Sorry for the confusion. By "vertical" I presume you mean when the pendulum is at its lowest level (ie. the string is straight up and down)? That would correspond to $\displaystyle \theta = 0$. So let $\displaystyle \theta (t) = 0$ and solve for t. Max height is the largest value of $\displaystyle \theta$. So you are finding the (absolute) maximum of $\displaystyle \theta$, which we know to be $\displaystyle \theta _0$. Find the first derivative and set it equal to 0, then solve for t. -Dan
Thank you very much for your help dan, if you could just solve for vertical 1st time t, that would completely satisfy everything i wanted to know, thank you very much December 11th, 2016, 04:33 PM   #12
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 Originally Posted by greally2014 Thank you very much for your help dan, if you could just solve for vertical 1st time t, that would completely satisfy everything i wanted to know, thank you very much
I dont know the last part about vertical height ahaha. could you show me? December 11th, 2016, 04:34 PM   #13
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 Originally Posted by topsquark Given $\displaystyle \theta (t) = A~cos( \omega t) + B~sin( \omega t)$ From the initial condition we have $\displaystyle \theta (t = 0) = \theta _0$ $\displaystyle \theta _0$ is just a constant which defines what $\displaystyle \theta$ is at t = 0. So in the general solution we have: $\displaystyle \theta (t = 0) = A~cos( \omega \cdot 0) + B~sin( \omega \cdot 0)$ $\displaystyle \theta (t = 0) = \theta _ 0 = A~cos( \omega \cdot 0) + B~sin( \omega \cdot 0)$ $\displaystyle \theta _ 0 = A \cdot 1 + B \cdot 0$ $\displaystyle \theta _ 0 = A$ -Dan
Could you show me time to reach vertical? December 11th, 2016, 04:50 PM   #14
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 Originally Posted by greally2014 Could you show me time to reach vertical?
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, 04:56 PM   #15
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 Originally Posted by topsquark 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
oh yes i did that earlier yes the angle is 90 degrees which means that root gravity over length by time is 90 degrees, but i still have unknowns December 11th, 2016, 05:41 PM   #16
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 Originally Posted by greally2014 oh yes i did that earlier yes the angle is 90 degrees which means that root gravity over length by time is 90 degrees, but i still have unknowns
What unknowns? It's a single variable equation!

$\displaystyle \theta (t = 0) = 0 = \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} ~ t \right )$

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 05:44 PM. Tags differential, equations, pendulum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rishav.roy10 Differential Equations 0 August 21st, 2013 05:59 AM giga.chalauri Differential Equations 2 April 3rd, 2013 10:34 AM thebigbeast Differential Equations 2 September 11th, 2012 09:27 AM FreaKariDunk Differential Equations 1 May 19th, 2011 02:02 PM jakeward123 Differential Equations 11 March 16th, 2011 08:45 AM

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