I encountered the following differential equation when I tried to derive the equation of motion of a simple pendulum:
$\frac{\mathrm d^2 \theta}{\mathrm dt^2}+g\sin\theta=0$
How can I solve the above equation?
[Math] Solve $\theta”+g\sin(\theta)=0$
ordinary differential equations
Related Solutions
Let $\ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.
\begin{align*} \ddot{\theta}+\omega^{2} \sin \theta &= 0 \\ \omega &= \sqrt{\frac{mg\ell}{I}} \\ k &= \sqrt{\frac{E}{2mg \ell}} \end{align*}
$$ \begin{array}{|c|c|c|c|} \hline & k < 1 & k = 1 & k > 1 \\ \hline & & & \\ \displaystyle \sin \frac{\theta}{2} & k\operatorname{sn} (\omega t,k) & \tanh \omega t & \displaystyle \operatorname{sn} \left( k\omega t,\frac{1}{k} \right) \\ & & &\\ \theta & 2\sin^{-1} (k\operatorname{sn} (\omega t,k)) & 4\tan^{-1} e^{\omega t}-\pi & \displaystyle 2\operatorname{am} \left( k\omega t,\frac{1}{k} \right) \\ & & &\\ T & \displaystyle \frac{4K(k)}{\omega} & \infty & \displaystyle \frac{2K(\frac{1}{k})}{k\omega} \\ & & &\\ \hline \end{array}$$
For small bob, $I\approx m\ell^{2}$.
For $k<1$, amplitude $\alpha=2\sin^{-1} k$ and $k>1$ it's moving in complete circle.
For $k<<1$, $T=2\pi \sqrt{\frac{I}{m g\ell}} \left( 1+\frac{1}{4} \sin^{2} \frac{\alpha}{2}+\ldots \right)$
A plot of $T$ vs. $k$ with $\omega=1$ is shown below
$$\ddot\theta+k^2\sin\theta=0$$
can be rewritten
$$\dot\theta\ddot\theta+k^2\dot\theta\sin\theta=0$$
and after integration from $0$ to $t$,
$$\dot\theta^2-2k^2(\cos\theta-\cos\theta_0)=0.$$
Then we have a separable equation
$$\frac{\dot\theta}{\sqrt{\cos\theta-\cos\theta_0}}=\sqrt2k.$$ and by a second integration
$$\int_{\theta_0}^\theta\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}=\sqrt2kt.$$
This integral is a difficult one, requiring an Elliptic Integral of the First Kind:
$$\int_{\theta_0}^\theta\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}=\frac{2F\left(\frac\theta2,\csc^2\frac{\theta_0}2\right)}{\sqrt{1-\cos\theta_0}}.$$
For more tractability, one often assumes small angles so that $\cos\theta\approx1-\frac{\theta^2}2$ and
$$\int_{\theta_0}^\theta\frac{\sqrt2d\theta}{\sqrt{\theta_0^2-\theta^2}}=\sqrt2\arccos\frac\theta{\theta_0}.$$
Finally,
$$\theta=\theta_0\cos kt.$$
(This result can be directly obtained from the linearized equation $\ddot\theta+k^2\theta=0$, a much easier one.)
Best Answer
Use substitution : $\theta' =v$ ,therefore we have that :
$$\theta''=\frac{dv}{dt}\cdot \frac{dt}{d\theta}\cdot \frac{d\theta}{dt} \Rightarrow \theta''=\frac{dv}{d\theta}\cdot v \Rightarrow \theta''=v'\cdot v$$
where $v$ is function in terms of variable $\theta$ .So differential equation becomes :
$v' \cdot v +g \cdot \sin \theta=0$
which is separable differential equation .