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Let $k$ be any constant, given $ \theta(0)=0$, $ \frac d{dt}\theta=0$ when $t=0$, t = ? if $ \theta = \frac \pi 2$ where $t$ represents time.

$ \frac{d^2}{dt^2}\theta = k\sin\theta $

How would I solve this problem in the simplest manner? This can be modeled with large angle pendulum or falling stick (of unifom thickness) falling from unstable equilibrium.

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Since we can calculate the time taken by the blob from $ \pi/2 \text{ to } 0 $ ( or $ \pi $ ) shouldn't we able to calculate the time theoretically? Correct me if I'm mistaken.

Let the parameter be $ \theta(0)=\pi/2$, $ \frac {d\theta}{dt}=0$ when $t=0$, t = ? if $ \theta = \pi \text{( or 0) } $

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    In any case, that equation does not have *elementary* solutions: its solutions are linear combinations of certain elliptic functions, so it is not obvious what *simplest* should mean in this context.2012-06-01
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    Yeah i know that .... but physically it is always possible to calculate the time. Mean when a stick falls or ... pendulum swings ... there is always finite time. I am not asking for general solution2012-06-01
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    "Physically it is always possible to calculate the time" does not mean anything really. "Physically", what you do is construct a pendulum and measure the time... If you are happy with "physical" solutions, then that is what you should do :)2012-06-01
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    does that mean can't i calculate the time that pendulum is at the bottom from horizontal theoretically or can't I calculate the time taken by stick to fall on the ground theoritically?2012-06-01
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    By the way, if you start with $\theta=\theta'=0$, then the solution is *identically* zero, that is, $\theta(t)=0$ for all $t$, so in particular, there is no such $t$... You have one of the two conditions wrong, I guess.2012-06-01
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    I am not sure either ... when a stick falls, t=0, you can assume $\theta = 0 \text{ or } \pi/2 \text{ but } \theta'=0 $2012-06-01
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    You misunderstand me: I *am* sure. The unique solution to the problem $$\theta''(t)=k\sin\theta(t), \qquad\theta(0)=0, \qquad\theta'(0)=0$$ is identically zero.2012-06-01
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    I am not a mathematician ... you must be right!! i asked similar question here. Don't misunderstand me ... since did have little understanding back then (about my problem) ... and don't understand the solution they presented me ... http://physics.stackexchange.com/questions/2372/how-long-it-will-take-for-a-tree-to-fall-on-a-ground2012-06-01
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    Well, they said the same thing as I did in different language: if you have a standing stick which is perfectly vertical and at rest, it just will not fall *in an ideal situation*. You should read about what an *unstable equilibrium* is. Wikipedia might be helpful and readable, for example.2012-06-01
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    I meant it for unstable equilibrium. i could say that i can make initial angular displacement for pendulum to be $ \pi/2$2012-06-01

2 Answers 2

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As a complement to the above multiplying both sides by $\frac{d \theta}{dt}dt=d\theta$ we obtain $$\frac{d^2 \theta}{dt^2}\frac{d \theta}{dt}dt=k\sin\theta d\theta$$ $$\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2=k\int_{\frac{\pi}{2}}^{\theta}\sin\theta d\theta$$ also referred to as the energy integral. $$\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2=-k\cos\theta$$ Now solve for $dt$ and integrate from $\frac{\pi}{2}$ to $\theta$ This is a general trick of lowering the order of equations of the form $$z''=f(z)$$

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    thanks for simple answer!!2012-06-02
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The solution of the initial value problem $\theta'' = k \sin \theta$, $\theta(0)= \pi/2$, $\theta'(0)=0$, is given implicitly (for $\pi/2 \le \theta(t) \le 3\pi/2$, i.e. the first swing of the pendulum) by $$ \int _{\pi/2 }^{\theta \left( t \right) }\!{\frac {1}{\sqrt {-2\,k \cos \left( s \right) }}}\ {ds}=t$$ The time to go from $\theta = \pi/2$ to $\theta=\pi$ is thus $$ \int_{\pi/2}^\pi \frac{1}{\sqrt{-2k\cos(s)}}\ ds = \frac{1}{\sqrt{k}} \int_{\pi/2}^\pi \frac{1}{\sqrt{-2\cos(s)}}\ ds$$ That last integral is non-elementary: its approximate value is $1.854074677$, and it can be expressed as ${\rm EllipticK}(1/\sqrt{2})$ in the convention used by Maple. Wolfram Alpha calls it $K(1/2)$. It can also be written as $\dfrac{\pi^{3/2}}{2 \Gamma(3/4)^2}$.