Driven Oscillations and the Frequency Response

From a free spring to a shaken one

Every equation in this rung so far has had a zero on the right: m y'' + c y' + k y = 0, the spring-mass-damper left to ring down by itself. Pull it, release it, walk away. Now we refuse to walk away. We attach the support to a motor, or push the mass with a periodic hand, and the right-hand side stops being zero — it becomes a forcing function F(t) that we control. The honest equation of a driven oscillation is m y'' + c y' + k y = F(t), and the most important case by far is a sinusoidal drive, F(t) = F0 * cos(w t), where w is the frequency *we* choose to wiggle at.

It is worth pausing on what changed. The free spring had only one frequency it cared about: its own natural rhythm, the wn you met two guides back, set by the spring and mass alone (wn = sqrt(k/m)). A driven spring now juggles two frequencies — its own wn, baked into the left-hand side, and the imposed w, sitting on the right. The entire drama of this guide, and the resonance of the next, is the contest between those two numbers. When they nearly agree, something dramatic happens; when they disagree, the spring shrugs.

Because the right side is no longer zero, this is a nonhomogeneous equation, and you already know the rule from the second-order rung: the full solution splits as y = y_h + y_p. The complementary part y_h is exactly the free, damped motion you solved before — it carries the two arbitrary constants and the initial conditions. The particular part y_p is a single specific response to the forcing. We are about to see that this split is not just bookkeeping: the two pieces have completely different fates over time.

Transient dies, steady state survives

Here is the most clarifying fact about driven motion. For any real, damped spring (c > 0), the complementary solution y_h decays to zero — that is precisely what the previous guide's three damping regimes guarantee, since every characteristic root has a negative real part. So y_h is a transient: it remembers how you started, it dominates the first few seconds, and then it fades out and is gone forever. Whatever bump or kick you gave the system at t = 0 is, quite literally, forgotten.

What remains is y_p, the steady-state response. Once the transient has died, the spring settles into a clean, never-ending oscillation — and a beautiful thing is true of a linear system driven by cos(w t): the steady state oscillates at exactly the driving frequency w, not at the spring's own wn. The motor sets the beat; the spring eventually marches to it. The only freedom left is *how big* the response is and *how delayed* — an amplitude and a phase lag. Write it as y_p = A(w) * cos(w t - phi(w)), and the whole problem reduces to finding two functions of w: the gain A(w) and the lag phi(w).

How to find the steady state

Finding y_p is the undetermined-coefficients machinery from the nonhomogeneous rung, applied to a cosine forcing. The trick is that a cosine drive generally produces *both* a cosine and a sine in the response — the phase lag is exactly the sine creeping in. So you guess y_p = M * cos(w t) + N * sin(w t), substitute into m y'' + c y' + k y = F0 * cos(w t), and match the cos and sin coefficients separately. Two linear equations in M and N drop out, and solving them hands you the response.

1. Write the equation in clean form: m y'' + c y' + k y = F0 * cos(w t), with w the driving frequency you control.
2. Guess the steady state y_p = M * cos(w t) + N * sin(w t) — a cosine alone is not enough, because damping forces a phase shift.
3. Substitute, collect the cos(w t) and sin(w t) terms, and set each side's coefficients equal — two linear equations in M and N.
4. Solve for M and N, then repackage as a single wave: amplitude A = sqrt(M^2 + N^2), phase lag phi = atan2(N, M).

Grind that through and a clean formula emerges for the steady-state amplitude. With the natural frequency wn = sqrt(k/m), the gain is A(w) = (F0 / m) / sqrt((wn^2 - w^2)^2 + (c w / m)^2). Stare at the denominator and the whole story is already visible: the term (wn^2 - w^2)^2 vanishes when the drive matches the natural frequency, which would make A blow up — except the damping term (c w / m)^2 stands in the way, keeping it finite. That tension between a vanishing first term and a small-but-nonzero second term is exactly the seed of resonance.

Drive:        F(t) = F0 * cos(w t)        (w = driving frequency, chosen by us)
Natural:      wn = sqrt(k/m)              (set by the spring, fixed)

Steady-state amplitude:
              A(w) = (F0/m) / sqrt( (wn^2 - w^2)^2 + (c w / m)^2 )

Phase lag:    tan(phi) = (c w / m) / (wn^2 - w^2)

Limits:
  w -> 0      A -> F0/k          spring just follows the push, phi ~ 0
  w = wn      A = F0 / (c wn)     peak region, phi = 90 degrees
  w -> inf    A -> 0             too fast to follow, phi -> 180 degrees

Reading the frequency response curve

Now plot A(w) against w and you get the single most useful object in this rung: the frequency response curve. At very low w the spring is barely keeping pace with a slow push, and A settles to a constant F0/k — the static stretch you would get by just leaning on the spring. As w climbs toward the natural frequency wn, the denominator shrinks and the curve rises to a hump. Past the peak it falls away, and at very high w the mass simply cannot keep up — you are shaking it faster than it can respond — so A tails off toward zero. One bump, rising then falling: that shape is the fingerprint of a driven oscillator.

The phase lag phi(w) tells a parallel story, and it is just as physical. At low frequency the spring moves nearly in step with the push, phi near 0 — push right, mass goes right. Right at w = wn the lag is exactly a quarter cycle, phi = 90 degrees: the mass is fastest when you are pushing hardest, the condition for pumping in the most energy (this is how you learn to time your legs on a playground swing). At high frequency the mass lags by half a cycle, phi near 180 degrees — by the time it responds, you are already pushing the other way. The smooth swing of phi from 0 to 180 degrees as w sweeps up is the other half of the frequency response.

Honest edges and the bridge to resonance

Two honest caveats before we cross into resonance. First, this entire clean picture — one frequency in, one frequency out, a tidy gain and phase — is a gift of *linearity*. The moment the spring stiffens nonlinearly, or the damping depends on amplitude, the steady state can bend the peak over, jump between branches, or even respond at frequencies the drive never contained. The linear forced oscillation is the honest first model, not the final word, and real systems often deviate once you push them hard.

Second, a widespread myth deserves dismantling: that resonance requires zero damping. It does not. The amplitude formula has a finite peak for any positive c, and you will see in the next guide that the peak even sits *slightly below* wn, not exactly on it, once damping is present. Resonance is just the tall part of a real, damped curve — large but finite. Zero damping is the idealized limit where the peak runs to infinity (so-called pure resonance), a useful fiction that never quite occurs in nature because some dissipation is always there.

So you arrive at the next guide already holding the key. You know the steady state oscillates at the driving frequency, you know its amplitude is the frequency response curve, and you know that curve has a hump near wn whose sharpness is set by damping. The next guide pushes the drive right up to that hump (resonance) and then asks what happens when the drive sits *just beside* the natural frequency — where two nearly-equal frequencies interfere to produce the slow, throbbing swell of beats. Same equation, same machinery; we are only choosing where on the curve to stand.