forced (driven) oscillation
So far the system has been left alone to ring down. Now someone keeps pushing it rhythmically — a hand pumping a swing, a road repeatedly bumping a car, an alternating voltage driving a circuit. Forced oscillation is what happens when a steady, repeating external force is applied to a vibrating system. The question changes from 'how does it settle?' to 'how does it answer being shaken?'
Put a sinusoidal force on the right: m x'' + c x' + k x = F_0 cos(omega t), where omega is the driving frequency (not the system's own omega_0). The full solution splits into two parts. The transient is the free-damped part from before; it always dies away. The steady-state response is a forced oscillation x_p(t) = R cos(omega t - phi) that marches in step with the driver, oscillating at the driving frequency omega, with its own amplitude R and a phase lag phi behind the force. After the transient fades, only this steady-state survives.
The crucial insight is that the system responds at the frequency it is driven at, not at its natural frequency — but how strongly it responds depends on how close the driving frequency comes to omega_0. That dependence is the frequency-response curve, and its dramatic peak is resonance. This is why a car has a roughness that depends on speed, why a glass shatters at one particular sung note, and why every machine has speeds it should avoid.
x'' + 2 x' + 5 x = 10 cos(2t): the transient A e^(-t) cos(2t - psi) fades, leaving the steady forced oscillation, a cosine at the driving frequency 2 with a fixed amplitude and phase lag.
The lasting response oscillates at the driver's frequency 2, not the system's own.
A common confusion: the steady-state oscillates at the driving frequency omega, while the (vanishing) transient rings at the damped natural frequency omega_d. Early on you see both mixed together; only later does the pure driven response remain.