friction - Change in period of damped pendulum system

Suppose we have a simple pendulum damped by air resistance, proportional to the velocity of the pendulum. By using the small angle approximation of sin, we are able to solve a second order differential equation and arrive at the conclusion that the angle from the vertical, $\theta$, is equal to a trig function multiplied by a decaying exponential $$\theta(t) = A~\left(e^{-bt/2m}\right)\sin (ft + \omega)$$

It is evident that the amplitude of successive swings become smaller, yet the frequency of the oscillation $f$ remains constant, according to this.

Evidently wrong, how would one be able to quantify such a change in period of such a damped pendulum, as a function of time?

Answer

A damped oscillator (small angle) pendulum is characterised by the following equation of motion:

$$\ddot{x}+2\zeta\omega_0\dot{x}+\omega_0^2x=0\tag{1}$$

Where $\omega_0$ is the natural (undamped) angular velocity:

$$\omega_0=\sqrt{\frac{L}{g}}=2\pi f_0$$

And $\zeta$ is the damping ratio (with $c$ a constant):

$$\zeta=\frac{c}{2\sqrt{Lg}}$$

Underdamped oscillation occurs for $\zeta<1$, in which case the damped angular velocity is:

$$\omega_1=\omega_0\sqrt{1-\zeta^2}=2\pi f_1$$

So the frequency of the underdamped oscillator is smaller than the natural one but doesn't change in time.

The exponential decay function $e^{-\lambda t}$ is defined by:

$$\lambda=\omega_0\zeta$$

For the undamped oscillator, $(1)$ reduces to:

$$\ddot{x}+\omega_0^2x=0$$

Which is the equation of motion of the simple harmonic oscillator.