
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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December 11th, 2016, 11:43 AM  #1 
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  Differential equations on pendulum
Hi guys my professor is starting diff equations with us and gave us this question to look up online and try to see how it works, no matter where i look i cannot find the solution and i have never done these kind of diff equations before, only basic ones about population growth and stuff. They are usually e^kt, so very basic. Here is the question its in the attachment. Took me ages to type properly haha. Anyways I would appreciate the answer as I have no clue and I just need to try and understand it. As you sol it could you please explain why you are doing what you are doing XD. Thanks so much guys. I had to register on this forum just to try and get the solution. Last edited by greally2014; December 11th, 2016 at 12:10 PM. Reason: To add an attatchment 
December 11th, 2016, 12:18 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,764 Thanks: 707 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
The equation for $\displaystyle \theta (t) $ should be $\displaystyle \theta = \theta _0~cos \left ( \sqrt{ \frac{g}{L}} t \right )$. $\displaystyle \theta = \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} t \right )$ $\displaystyle \frac{d \theta }{dt} = \theta _ 0 ~ \sqrt{\frac{g}{t}}~sin \left ( \sqrt{\frac{g}{L}} t \right )$ $\displaystyle \frac{d^2 \theta }{dt^2} = \theta _0 \frac{g}{t}~ cos \left ( \sqrt{ \frac{g}{L}} t \right )$ Now plug that into your differential equation and see if it works. Dan Addendum: You should also check that the given $\displaystyle \theta (t)$ matches the initial conditions. Last edited by topsquark; December 11th, 2016 at 12:24 PM.  
December 11th, 2016, 12:45 PM  #3 
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  I think i get it...
So if i understand, you know that θ(0) = θo and you know what θ(t) is for all t. in other words, θ = θo, and we have θ(t)? So we want an expression for θ using both. So why do you multiply θ(t) by θo? the rest is basic differentiation But when you are deriving sin and cos, you are multiplying by the derivative inside that brackets of sin and cos which i get, but how is the derivative of the root of the gravity over length multiplied by t = the root of gravity over t? and what do you mean when you say plug into diff equation, i have never done these before my job is to try and understand like my teacher wants, we arent expected to be able to do this 
December 11th, 2016, 01:05 PM  #4  
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  I think i get it... Quote:
and you know what θ(t) is for all t. in other words, θ = θo, and we have θ(t)? So we want an expression for θ using both. So why do you multiply θ(t) by θo? the rest is basic differentiation But when you are deriving sin and cos, you are multiplying by the derivative inside that brackets of sin and cos which i get, but how is the derivative of the root of the gravity over length multiplied by t = the root of gravity over t? and what do you mean when you say plug into diff equation, i have never done these before my job is to try and understand like my teacher wants, we arent expected to be able to do this  
December 11th, 2016, 01:16 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,764 Thanks: 707 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$ so you need to verify that $\displaystyle \left ( \theta _0 ~ \frac{g}{L} ~ cos \left ( \sqrt{ \frac{g}{L}}~t \right ) \right ) + \frac{g}{L} \left ( \theta _0 ~ cos \left ( \sqrt{\frac{g}{L}} ~ t \right ) \right ) = 0$ Since you are starting off and the question gives you $\displaystyle \theta (t)$ then you don't need to know how to solve it. Your instructor will get there. But since you are curious: $\displaystyle \frac{d^2 \theta }{dt^2} + \frac{g}{L} \theta = 0$ This is a second order linear homogeneous differential equation with constant coefficients. One method for solution uses the characteristic equation: $\displaystyle m^2 + \frac{g}{L} = 0$ (Use a factor of m for each $\displaystyle \frac{d^m \theta }{dt^m}$) The solution for m is $\displaystyle i \sqrt{ \frac{g}{L} }$. We have two forms we can use for this solution: Either $\displaystyle \theta = A e^{i \omega t} + B e^{i \omega t}$ or $\displaystyle \theta = A ~ cos( \omega t) + B ~ sin(\omega t)$ where $\displaystyle \omega = \sqrt{\frac{g}{L}}$ in each case. Both forms give the same result but since your question uses cosine we'll use the second form. Now for the initial conditions. We know that $\displaystyle \theta (0) = \theta _0$ so $\displaystyle \theta (0) = A~cos( \omega \cdot 0 ) + B~sin( \omega \cdot 0) = \theta _0$ $\displaystyle A \cdot 1 + B \cdot 0 = \theta _0$ Thus $\displaystyle A = \theta _0$. To get B we use $\displaystyle \frac{d \theta }{dt} = \theta _0 \omega ~ sin( \omega t) + B \omega ~cos( \omega t)$ where I've substituted $\displaystyle A = \theta _0$. So $\displaystyle \frac{d \theta}{dt} (0) = \theta _0 \omega ~ sin(\omega \cdot 0) + B \omega ~cos( \omega \cdot 0) = 0$ $\displaystyle \theta _0 \omega \cdot 0 + B \omega \cdot 1 = 0$ or B = 0. Thus the solution is $\displaystyle \theta (t) = \theta _0 ~ cos( \omega t)$ It looks complicated but you'll get plenty of experience with these in your class. It's not so bad once you've done it a couple of times. Dan Last edited by topsquark; December 11th, 2016 at 01:27 PM.  
December 11th, 2016, 01:26 PM  #6  
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  Quote:
 
December 11th, 2016, 01:33 PM  #7  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,764 Thanks: 707 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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  
December 11th, 2016, 02:08 PM  #8  
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  Quote:
All im doing is deriving what im given? and subbing it in to make sure its correct? after that im just using theta (t)? ok that seems cool  
December 11th, 2016, 02:14 PM  #9  
Newbie Joined: Dec 2016 From: Ireland Posts: 10 Thanks: 0  Quote:
Quote:
 
December 11th, 2016, 02:42 PM  #10  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,764 Thanks: 707 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \theta (t) = A~cos( \omega t) + B~sin( \omega t)$ 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  

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