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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. –  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
    
Also, try to include tags other than than the homework one. I added a few for you. It helps your question be more specific if people are sorting through them by tags. –  Joe Jun 1 '12 at 18:48
    
thanks i'll from now onwards –  Santosh Linkha Jun 1 '12 at 18:49
    
"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
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2 Answers

up vote 1 down vote accepted

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!! –  Santosh Linkha Jun 2 '12 at 4:39
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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 $\dfrac{\pi^{3/2}}{2 \Gamma(3/4)^2}$.

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