# Velocity of the bob of a pendulum without energy conservation

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?

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

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

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

$$\ddot \theta = \frac{-g}{l} \theta$$
$$\theta = A\cos (\sqrt{\frac{g}{l}}t)$$
Differentiate this with respect to time and multiply by $$l$$ to find the velocity.
• 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. Dec 15, 2019 at 16:33
• thanks a lot for the response, actually, I considered $\omega$ to be the angular velocity of the Bob, which it is not Dec 15, 2019 at 16:35