Does it make sense to talk about a frequency when we deal with damped oscillations? I'm solving a problem in which there is a damping force in the form $F = -bv$.
The question asks for the "frequency of oscillation", but since it is a damped oscillation I am confused, because I think that in this case one can't define a value for the frequency.  
 A: If the motion is being driven
Well, after dividing by various constants your differential equation is going to look like:
$$ \ddot f + 2\lambda \dot f + \omega_0^2 f = g(t).$$
If $g(t) = A \cos(\Omega t)$ for example, then you will find that $f(t)$ has (angular) frequency $\Omega$ as well. This is easiest to see under a Fourier transform, $$f[\omega] = \frac{g[\omega]}{-\omega^2 + 2i \lambda\omega + \omega_0^2}.$$ We can treat cosines and sines via the Dirac $\delta$-function, and we find that the inverse transform for  $g[\omega] = \delta(\omega - \Omega)$ is going to be $$f(t) = \frac{e^{i\Omega t}}{-\Omega^2 + 2i\lambda\Omega + \omega_0^2}.$$ The term on the bottom describes a phase shift and amplitude rescaling due to $\Omega$ being on or off the resonance peak; the term on the top says that it oscillates with the driving frequency $\Omega$ regardless.
If the motion is not being driven
Of course the above are "particular solutions" $f_P$ and to use them you also need the "homogeneous solution" $f_H$ to the equation
$$ \ddot f + 2\lambda \dot f + \omega_0^2 f = 0.$$Then you will see that the general solution is $f = f_P + f_H$ due to the linearity of derivatives. But there is a trick very similar to the Fourier transform which is to just assume that $f(t) = A~e^{st}$ for some $s$. In this case we get $$s^2 + 2\lambda s + \omega_0^2 = 0$$which we can solve to find $s = -\lambda \pm \sqrt{\lambda^2 - \omega_0^2}$.
This also has an interpretation as having an oscillation with a special frequency, if $\lambda^2 < \omega_0^2.$ Let $\tilde\omega^2 = \omega_0^2 - \lambda^2$ be a positive number, then this trick tells us that our homogeneous solution family looks like, $$f_H(t) = A e^{-\lambda t} e^{i\tilde\omega t} + B e^{-\lambda t} e^{-i\tilde\omega t}. $$ You can recast this as you like; for example $A e^{-\lambda t} \cos(\tilde\omega t + \varphi)$ works.
In any case, we can clearly see that the sinusoidal part of an "underdamped" system oscillates with frequency $\tilde\omega = \sqrt{\omega_0^2 - \lambda^2}$, but it does go to zero in an exponential "envelope." This $\tilde \omega$ could also be a reasonable definition of the frequency of the damped oscillator.
A: The period of oscillation can be defined from the times at which the oscillator passes through the equilibrium position. This assumes that it does so several times - ie the damping is significantly less than the critical value. It is harder to identify or measure the period from the maxima or minima, since the amplitude is constantly decreasing. 
The frequency measured between zero-crossings is slightly less than the natural (undamped) frequency, but for a damped harmonic oscillator (restoring force proportional to displacement) it remains constant as the amplitude of oscillation decreases. This is not quite the case for the simple pendulum because the restoring force is not strictly proportional to displacement; it is not because of the damping.
The underdamped frequency is $\omega_1=\sqrt{\omega_0^2-\gamma^2}$ where $\omega_0$ is the natural frequency and $\gamma=b/2m$ is the decay constant. See Eqn #71 on the Farside Physics website.
A: Damped simple harmonic motion comes in three flavors: "light", "critical", and "heavy". A lightly damped system will oscillate (with a frequency close to the undamped resonance frequency). See this diagram:

Clearly, lightly damped motion has a frequency.
If the equation of motion is of the form
$$\ddot x + \beta \dot x + \omega_0^2 x = 0$$
Then when $\beta = 2\omega_0$, you have critical damping. If it is less, you have light damping; if it's greater, you have heavy damping.
See for example Wolfram Alpha article on critical damping
A: When there is damping, your energy is being dissipated. The energy of the system is not represented by frequency. What is happening is that your amplitude is reducing, and your frequency remains the same. Hence yes, it does make sense to talk about frequency.
