A small angle pendulum (SHM) has angular amplitude $\theta_0$, length $l$. I want to find the velocity of the bob at the lowest position.

The pendulum starts with the Bob aligned at $\theta_0$. The acceleration at any point is $g\sin{\theta}$, where $\theta$ is the angle it makes with the equilibrium position. For small angular displacement $\text{d}\theta$, $\text{d}\theta=\omega\text{d} t \Rightarrow \text{d}t=\dfrac{\text{d}\theta}{\omega}$.

So the final velocity should be $\displaystyle\int a \text{d}t =\int_{\theta_0}^0 \dfrac{g}{\omega}\sin\theta\text{d}\theta$. Here, we can approximate $\omega=\sqrt{\dfrac{g}{l}}$ to be constant due to small angle.

But this does not give the correct answer... what am I thinking or doing wrong?

  • $\begingroup$ When you use the small angle approximation, remember that this also applies to $\sin(\theta)$. $\endgroup$ Commented Dec 15, 2019 at 12:16
  • $\begingroup$ @AndersSandberg, it would give $\dfrac{g\theta^2}{2\omega}$, which isn't correct either... $\endgroup$
    – Riz222
    Commented Dec 15, 2019 at 12:34

1 Answer 1


The usual way to do it is by setting up and solving a differential equation for $\theta$.

$$a=-g\sin \theta \approx -g\theta$$

This leads to

$$\ddot \theta = \frac{-g}{l} \theta$$

The solution to this equation is

$$\theta = A\cos (\sqrt{\frac{g}{l}}t)$$

Differentiate this with respect to time and multiply by $l$ to find the velocity.

  • $\begingroup$ If you calculate the integral in your question it gives $-\frac{g}{\omega}\cos \theta$ which is approximately $-\frac{g}{\omega}$ using the small angle approximation. $\endgroup$
    – bemjanim
    Commented Dec 15, 2019 at 16:33
  • $\begingroup$ thanks a lot for the response, actually, I considered $\omega$ to be the angular velocity of the Bob, which it is not $\endgroup$
    – Riz222
    Commented Dec 15, 2019 at 16:35
  • $\begingroup$ no problemo, yeah i make that mistake a lot $\endgroup$
    – bemjanim
    Commented Dec 15, 2019 at 16:37

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