# Solving $\frac{d^2}{dt^2}\theta = k\sin\theta$ for $t$

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.

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. – Mariano Suárez-Alvarez Jun 1 '12 at 18:43
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 solution – Santosh Linkha Jun 1 '12 at 18:45
"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 :) – Mariano Suárez-Alvarez Jun 1 '12 at 18:50
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? – Santosh Linkha Jun 1 '12 at 18:52
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. – Mariano Suárez-Alvarez Jun 1 '12 at 18:55

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 $$z''=f(z)$$
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 $\dfrac{\pi^{3/2}}{2 \Gamma(3/4)^2}$.