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I am trying to derive an equation for angular frequency of a simple pendulum.

Since the torque on the bob is only due to the horizontal component of mg, I can say that,

$$ -mg(\sin\theta) l = \tau$$

$$ = I\frac{d^2\theta}{dt^2} \implies \frac{-g\sin\theta}{l}= \frac{d^2\theta}{dt^2} $$ $$ since, $$$$I = ml^2 $$

I am stuck here. How do I reach angular velocity from here? (Note: I am new to this. I only have a basic idea regarding this (the maths, I mean). Kindly go easy on me.

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    $\begingroup$ This is a non linear differential equation and can not be solved easily. The funny part is that you get $x=A\sin(wt)+B\cos(wt)$ precisely because earlier you didn't take a sine term to be a sine term, afterall. $\endgroup$
    – Physiker
    Jun 30, 2021 at 16:45

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Since you say that you're new to this, my answer will be quite basic.

The general formula for the angular velocity of a simple pendulum isn't simple to derive at all. However, it is possible to derive its angular frequency using the small angle approximation. In this approximation, the angle $\theta$ (in radians) is very small $$\theta \ll 1 \implies \sin \theta \approx \theta.$$

In this case, the differential equation becomes: $$\frac{\text{d}^2\theta}{\text{d} t^2} = - \frac{g}{l}\, \sin(\theta) \approx - \frac{g}{l}\ \theta.\label{1} \tag{1}$$

This is just a rewriting of a very well known equation (perhaps the most well known in Physics?), that of Simple Harmonic Motion:

$$\frac{\text{d}^2 x}{\text{d} t^2} = - \omega^2 x$$

There are many methods to show that the general solution to this equation can be written in terms of sins and cosines as $$x(t) = A \sin(\omega t) + B \cos(\omega t).$$ (You can plug this solution into the equation above, and see that it does indeed satisfy the equation. You can now see that the quantity I defined as $\omega$ above represents the angular frequency. I now leave it to you to look at Equation (\ref{1}) and figure out what the angular frequency is.

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