Amplitude of oscillations in non-resonant forced vibrations I am somewhat confused about the amplitude of forced vibrations at non-resonance driving frequencies.
If I was to assume that there was no / negligible damping present, then at resonance, the amplitude of the harmonic oscillator would continue to grow with each oscillation?
If this is the case, then is this also true for driving forces that are not at resonance, or would the oscillator reach a constant amplitude?
Thanks
 A: Yes, for the harmonic oscillator, the amplitude diverges with time at resonance, as already discussed in Undamped Resonance of a Classical Harmonic Oscillator.
In general, what happens away from resonance depends both on the system (e.g., harmonic oscillator or anharmonic?), on the forcing (very high or low frequency, non-periodic?), and possibly on the starting point of the system. For example, the movement can be chaotic for the cubic (anharmonic) potential
$$ U(x) = \alpha x + \beta x^3, $$
which is the well-known Duffing oscillator.
Generally, the amplitude away from resonance will not diverge for small but nonzero damping, but will also not be necessarily constant; it'll grow during some oscillations, and decrease with others. When the forcing injects energy (i.e, in the instants its work is positive - when the force is in the same direction as the oscillation speed), then the amplitude increases, while decreasing otherwise (negative work done by forcing, which happens when it acts against the momentary movement).
The sinusoidally forced harmonic oscillator will, after a transient behavior, oscillate at the forcing frequency at constant amplitude. If the forcing is periodic, but not sinusoidal, or the oscillator not harmonic, then the steady-state amplitude is constant, but the movement is described by an infinite sum of sines (at multiples of the forcing frequency, i.e., a Fourier series). For example, Keith Fratus notes gives, for an arbitrarily forced harmonic oscillator, the following solution:
$$ x(t) = \sum_{n=1}^\infty \frac{\sqrt{a_n^2+b_n^2}}{\sqrt{(\omega^2-n^2\lambda^2)^2+4\beta^2n^2\lambda^2}}\cos{(n\lambda t - \delta_n-\phi_n)}, $$
where $\lambda$ is the angular frequency of the forcing, $\omega$ the oscillator frequency, and $\beta$ the damping constant.
Related discussions are A conceptual doubt regarding Forced Oscillations and Resonance and Physical reason behind having greater amplitude when driving frequency$ < $ natural frequency than that when driving frequency $>$ natural frequency. 
A: You are correct that for resonant frequencies, the amplitude will grow with every oscillation (until something breaks, runs out of space, or cannot apply force to it).
For forces not at resonance, the amplitude will not really "reach a constant", nor will it grow infinitely.
The amplitude will grow and shrink, creating a second periodic oscillation with a longer period (see beat in acoustics).
To visualize what is happening, consider the case where the frequency of the driving force is near the resonant frequency, and the driving force is also a sinusoidal wave.  When in phase, the two forces act together to form positive interference, graphically it's the equivalent of adding the two waves together and getting a bigger wave (see superposition for more information on the subject).
If this is allowed to keep happening, the wave will keep growing forever. Thankfully, if the frequencies aren't exactly the same, it won't happen.
Instead, because the frequencies are different, the driving force will become out of phase with the oscillator over time (the closer the frequency to resonant, the slower this process).  After some time, it will form destructive interference and the amplitude will greatly decrease.
The cycle then repeats, with the phase going back towards resonance.  This will form the "beat" pattern (as demonstrated graphically in the link earlier).
