Simple pendulum

Figure 1.7: (a) and (b)

The simple possible shown in Figure 1.7(a) is possibly familiar to all of us. A mass $m$ is suspended by a rigid rod of length $l$, the rod is assumed to be massless. The gravitations potential energy of the mass is

$$V(\theta)=m g l [1 - \cos \theta]\,.$$ (1.19)

For small $\theta$ we may approximate whereby the potential is

$$V(\theta)= \frac{1}{2} m g l \theta^2$$ (1.20)

which is the SHO potential. Here gives the torque not the force. The pendulum's equation of motion is

$$I \ddot{\theta} = - m g l \theta$$ (1.21)

where $I= m l^2$ is the moment of inertia. This can be written as

which allows us to determine the angular frequency

$$\omega_0=\sqrt{\frac{g}{l}}$$ (1.23)