My Math Forum Differential equations on pendulum

 Differential Equations Ordinary and Partial Differential Equations Math Forum

December 11th, 2016, 02:43 PM   #11
Newbie

Joined: Dec 2016
From: Ireland

Posts: 10
Thanks: 0

Quote:
 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
Newbie

Joined: Dec 2016
From: Ireland

Posts: 10
Thanks: 0

Quote:
 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
Newbie

Joined: Dec 2016
From: Ireland

Posts: 10
Thanks: 0

Quote:
 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,093
Thanks: 853

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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
Newbie

Joined: Dec 2016
From: Ireland

Posts: 10
Thanks: 0

Quote:
 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,093
Thanks: 853

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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.

 Thread Tools Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top