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.

  • 1
    $\begingroup$ 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

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.