Velocity of the bob of a pendulum without energy conservation

Question

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?

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.