The motion of the simple pendulum is described by the following differential equation

$$\frac{d^2 \theta}{dt^2}+\frac{g \sin \theta }{l}=0$$

Multiply through $$\frac{2d \theta}{dt}$$

and integrate and apply initial condition $\theta=\theta_0$ and $\dfrac{d \theta}{dt}=0$

Then, separate the resulting equation in the variable to obtain

$$ \frac{d \theta}{\sqrt{\cos\theta- \cos \theta_0}}=\pm \sqrt{\frac{2g}{l}}dt $$

My work

$$\frac{2 d \theta}{d t} \frac{d^2 \theta}{d t^2}+\frac{2 d \theta}{dt}\frac{g \theta }{l}=0$$

I got stuck here.I don't think I can proceed any further. Can someone show me some hint

  • 1
    $\begingroup$ Rather It is a non- linear pendulum $\endgroup$ – Narasimham Jul 13 '17 at 6:17


can be rewritten


and after integration from $0$ to $t$,


Then we have a separable equation

$$\frac{\dot\theta}{\sqrt{\cos\theta-\cos\theta_0}}=\sqrt2k.$$ and by a second integration


This integral is a difficult one, requiring an Elliptic Integral of the First Kind:


For more tractability, one often assumes small angles so that $\cos\theta\approx1-\frac{\theta^2}2$ and



$$\theta=\theta_0\cos kt.$$

(This result can be directly obtained from the linearized equation $\ddot\theta+k^2\theta=0$, a much easier one.)

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I'm kind of confused about what the question is exactly. But anyway, starting from $$ \frac{d^2\theta}{dt^2}+\frac{g}{l}\sin \theta=0 $$ and multiplying by $2d\theta/dt$ you get $$ 2\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}+\frac{2g}{l}\frac{d\theta}{dt}\sin \theta=0 $$ Note that $$\frac{d}{dt}\left[\frac{d\theta}{dt}\right]^2=2\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}$$ So $$ \frac{d}{dt}\left[\frac{d\theta}{dt}\right]^2=-\frac{d\theta}{dt}\frac{2g}{l}\sin \theta $$ I leave the rest to you. (Hint integrate the above equation over $t$ given the initial conditions.)

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  • $\begingroup$ Thanks . Now I see it! $\endgroup$ – Crazy Jul 13 '17 at 5:27

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