Why is response of system same frequency as driving force frequency Super basic question: why does a system (to be definite, perhaps assume a collection of coupled harmonic oscillators) respond (in the steady-state, after transient effects have dissipated) with all members oscillating at the same frequency, albeit with different amplitude and phases. This seems to be a starting assumption for several calculations. Now, I mean it kinda makes sense, but I can't explain it to myself particularly eloquently and perhaps someone can give me an obvious/rigorous explanation. 
 A: Mathematically determine if the physical system is linear; i.e., can it be described by an ordinary differential equation (ODE), or by a system of ODEs.  Most simple mechanical and electronic systems can be modelled this way.
Now take any simple ODE, second order for example, and model the homogeneous ODE with Matlab, for example. Next add a driving function to the right hand side, making it an inhomogeneous ODE; a cosine function would work if your expressions are real, else a complex exponential.   
When this system is modeled you will see the driving frequency at work, and the corresponding responses. This example shows how linear systems respond to a driving frequency.  This is why linear algebra and ODEs are often studied together.
If you drive with multiple frequencies you will see additional phenomena appear. If the material has a nonlinear response the situation becomes richer and more complicated.
A: Technically in the absence of dissipation oscillations need not be limited to those of driving frequency, but driving frequency oscillations are always present under driving, and so are the focus of the discussion. If however dissipation is present, then oscillations that are not maintained by the driving input loose energy and die out in time, while the driven ones are supported by the constant energy input of the driving force. 
Say your system is described by some set of coordinates summed up in a vector ${\bf u}(t)$, the corresponding driving force is some ${\bf f} e^{i\omega t}$ (this can have just one non-zero component as you suggest) and the dynamics is non-dissipative and of the form
$$
A{\bf u}(t) = {\bf f} e^{i\omega t}
$$
where $A = A\left(\frac{d}{dt}\right)$ is a linear differential operator, usually 2nd order. Then the general solution for ${\bf u}(t)$ reads
$$
{\bf u}(t) = {\bf u}_0(t) + {\bf u}_1(t)
$$
where ${\bf u}_0(t)$ is any solution in the absence of driving,
$$
A{\bf u}_0(t) = 0
$$
and ${\bf u}_1(t)$ is some particular solution under driving,
$$
A{\bf u}_1(t) = {\bf f} e^{i\omega t}
$$ 
The former is of oscillatory type, while the latter can be found under the ansatz ${\bf u}_1(t) = {\bf u}^0_1e^{i\omega t}$ and will contribute driven oscillations. So in this case the general solution is a superposition of unperturbed and driven oscillations.
Now say we add the simplest dissipative contribution in the form of an identical friction force on all components, such that the dynamics becomes
$$
A{\bar{\bf u}} + \lambda {\bar{\bf u}} \equiv (A + \lambda){\bar{\bf u}} = {\bf f} e^{i\omega t}
$$
The general solution again accommodates a contribution ${\bar{\bf u}}_0(t)$ from the unperturbed dynamics in the absence of driving, but in this case ${\bar{\bf u}}_0(t)$ must satisfy
$$
(A + \lambda){\bar{\bf u}}_0(t) = 0
$$
and is in general a damped solution of the form
$$
{\bar{\bf u}}_0(t) = {\tilde{\bf u}}_0(t)e^{-\kappa(\lambda) t}
$$
The driven solution on the other hand remains an oscillatory one as before, ${\bar{\bf u}}_1(t) = {\tilde{\bf u}}^0_1e^{i\omega t}$ despite the dissipative term (just plug in ${\bar{\bf u}}_1(t)$ and see how the resulting equation is modified). Hence overall we are left with 
$$
{\bar{\bf u}} = {\tilde{\bf u}}_0(t)e^{-\kappa(\lambda) t} + {\tilde{\bf u}}^0_1e^{i\omega t}\;\; \rightarrow \;\; {\tilde{\bf u}}^0_1e^{i\omega t} \;\;\text{as}\;\;t\rightarrow \infty
$$
The unperturbed oscillations simply die out, only the driven ones persist. 
Note: Peter Diehr posted his answer while I was typing this, but didn't show on my editor. So lots of overlap.
A: Let me try a more general approach.
All forces are, one way or another, transported by particles at a given velocity. The strength of the force at any given time corresponds to the strength of the source at the time when it was emitted. If the force at a given point is oscillating, it just means the source was oscillating. And because all known forces do not depend directly on time of travel (they may only decrease by distance), then they must have the same frequency.
Now, if you have a moving source or a moving target, there may be changes in frequency due to the Doppler effect, or other tricky things. In addition, when considering multiple-particle interactions, there may also be complex behaviours leading to changes in frequency.
The next point is the consequence of that force on the studied system. If the system is linear (response proportional to force), then, almost by definition, the response will have the same frequency. The other answers were pretty good on that particular point.
Edit: actually some forces may depend on time of travel. Let's just focus on electromagnetic interaction here.
