Period of a simple pendulum accounting for friction

Question

The period of a simple pendulum is $$T=2\pi\sqrt{\ell/g},$$ but no where in there do I see that it accounts for friction. Does it somehow account for friction, and if not, how could you do that?

Answer 1

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since you are asking for a period, we will assume that the pendulum oscilates. Then you've got a solution with angular frequency $$\omega=\sqrt{\frac{g}{l}-\frac{b^2}{4m^2}}$$Therefore, the period of this osscilator is $$T=\frac{2\pi}{\sqrt{\frac{g}{l}-\frac{b^2}{4m^2}}}$$That shows how the period changes compared to the undamped pendulum.

Answer 2

When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator.

The form they give is

$$\omega = \omega_0\sqrt{1 - \zeta^2}$$

Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.