The Student Room Group
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Differential Equation banter

Scroll to see replies

Original post by Gome44
While the normal pendulum equation is solvable with elliptic integrals, the quadratically damped pendulum is not unfortunately.

Only numerical methods can be used to evaluate t at certain theta (incidentally I took this from one of my Matlab projects)


Oh :0

My bad. I used Runge-Kutta for both in my program anyway.
Original post by Gome44
You have the differential equation below. Prove the thing next to that. I can't do it so don't ask me for help, have fun xx

Edit: I now know how to do it. Alas, I still will not help to be evil


This is fun. Let's see if we can get started. First of all, let
Unparseable latex formula:

y = \dot_{\theta}

. Then we're going to look at

ddθ(y2)=1θ˙ddt(θ˙2)\displaystyle \frac{d}{d\theta}(y^2) = \frac{1}{\dot{\theta}} \frac{d}{dt}(\dot{\theta}^2)

If you crunch through this for a bit, you'll get something like

ddθ(y2)=−y2−2sin⁡θ\displaystyle \frac{d}{d\theta}(y^2) = -y^2 -2\sin \theta

which looks not too unpleasant. Who wants to take it up from here?
Original post by Gregorius
This is fun. Let's see if we can get started. First of all, let
Unparseable latex formula:

y = \dot_{\theta}

. Then we're going to look at

ddθ(y2)=1θ˙ddt(θ˙2)\displaystyle \frac{d}{d\theta}(y^2) = \frac{1}{\dot{\theta}} \frac{d}{dt}(\dot{\theta}^2)

If you crunch through this for a bit, you'll get something like

ddθ(y2)=−y2−2sin⁡θ\displaystyle \frac{d}{d\theta}(y^2) = -y^2 -2\sin \theta

which looks not too unpleasant. Who wants to take it up from here?


This is a really good idea. That's why I made it in post 4 :wink:

But on a more serious note: what are the initial conditions?
Original post by Gome44
Not the way I've been told to do it, but it looks like that would work :smile:


What was your method?
Reply 24
Avatar for Gome44
Gome44
OP
10
Original post by atsruser
What was your method?


Actually I've just tried my method and it is essentially the same, just a different substitution (v= exp(-theta/2)w where v=d(theta)/dt). :smile:
Original post by atsruser
This is a really good idea. That's why I made it in post 4 :wink:


Doh! Sorry. :colondollar:
Original post by Gregorius
Doh! Sorry. :colondollar:


I was hiding my light under a spoiler-shaped bushel which is why you missed my display of calculus-related brilliance, I guess.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you