Answer to Question #161653 in Classical Mechanics for Asadkhan

Question

What is the lagrange's equation of simple pendulum?

Answer 1

Let the length of the pendulum be l, then we can represent the co-ordinates of the motion of the pendulum as per the given question

$x(\theta)= l\sin\theta$

$y(\theta) = -l\cos\theta$

now, differentiating with respect to t,

$\frac{d\theta}{dt}=\dot{\theta}$

$\frac{d(x(\theta))}{dt}=l\cos\theta \frac{d\theta}{dt}$

$\Rightarrow \frac{d(x(\theta))}{dt}=l\dot{\theta} \cos\theta$

$\frac{d(y(\theta))}{dt}=l\sin\theta \frac{d\theta}{dt}$

$\Rightarrow \frac{d(y(\theta))}{dt}=l\dot{\theta}\sin\theta$

Now, applying the conservation of energy

$L=T-V=KE-PE$

$=\frac{ml\dot{\theta^2}}{2}-mgl(1-\cos\theta)$

We know that the Lagrange equation for the generalized co-ordinate

$\frac{d}{dt}(\frac{(\partial L)}{d\dot{\theta}})-\frac{\partial L}{\partial{\theta}}=0$

$\Rightarrow \ddot{\theta}+\frac{g}{l}\sin(\theta)=0$

Hence, natural frequency of the simple pendulum

$\omega = \sqrt{\frac{g}{l}}$

$T=2\pi \sqrt{\frac{l}{g}}$