In short

A forced (or driven) oscillator is a damped oscillator being pushed by a periodic external force F(t) = F_0 \sin(\omega t). After any initial transient dies out, the oscillator settles into a steady state in which it moves at the driving frequency \omega (not its own natural frequency \omega_0), with a well-defined amplitude and phase lag. The steady-state amplitude is

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (\gamma \omega)^2}},

where \omega_0 = \sqrt{k/m} is the natural frequency and \gamma = b/m is the damping coefficient. Resonance is the sharp peak in A(\omega) near \omega = \omega_0. For weak damping, the peak height is roughly F_0/(m \gamma \omega_0) and the peak's full width at half the maximum power is \gamma. The quality factor Q = \omega_0/\gamma measures how sharply tuned the resonance is — high Q (small damping) gives a tall narrow peak; low Q gives a broad flat one. Tuning a radio, a wineglass breaking at a soprano's high note, the Bhuj earthquake that collapsed selectively-tall buildings but left shorter ones standing, a child's swing being pumped at the right rhythm — all of these are resonance.

Push a child on a swing at a children's park in Bengaluru. If you time your pushes to arrive at the right moment of each swing — forward just as the child reaches the back — every push adds a little energy, and the swing climbs higher and higher with each cycle. If you push at random moments, some pushes add energy and some fight the motion; the swing barely moves. And if you push too fast or too slow, the swing fights you most of the time and stays small. There is one rhythm that works, and it is exactly the rhythm the swing would swing at on its own — its natural frequency.

This is resonance, and once you see it, you see it everywhere. Tune your grandfather's old Murphy transistor radio past the All India Radio band in Chennai. Static, static, static, then suddenly a clear voice speaking in Tamil. A twist of the dial and the voice fades, replaced by static again, then another station in Telugu. What your radio is doing: matching its internal tuned circuit's natural frequency to the broadcast frequency of one particular station. Only the station at that frequency resonates with the receiver; the others are there in the antenna but do not get through.

On 26 January 2001, a magnitude-6.9 earthquake struck Gujarat's Kutch region. In Bhuj and the surrounding towns, some tall buildings collapsed while shorter ones next to them stood. The reason — among many — was that earthquake ground motion contains a spectrum of frequencies, and a tall building's natural sway frequency is much lower than a short building's. Tall buildings resonated with the dominant low-frequency seismic waves; short buildings did not. A building that is not resonant at the dominant earthquake frequency barely moves; one that is resonant is shaken to catastrophic amplitude. Seismic engineering since Bhuj is, in large part, a study of how to detune buildings away from the earthquake frequencies that their region is prone to.

In the previous chapter on damped oscillations, you saw that an oscillator with friction eventually dies out: no external input, no lasting motion. But a damped oscillator being continuously driven does not die out — it settles into a steady oscillation at the driving frequency, and the amplitude of that oscillation depends sensitively on how close the driving frequency is to the oscillator's own natural frequency. That dependence — the shape of A(\omega) — is resonance, and this article derives it, explores it, and shows why it matters.

Setting up the equation of motion

Take a mass on a spring with damping (as in damped oscillations), and add a periodic external driving force F(t) = F_0 \sin(\omega t) pushing on it.

Driven damped spring-mass systemA wall on the left with a spring connecting to a block, a dashpot (damper) in parallel with the spring, and an external periodic force pushing the block horizontally. The spring has stiffness k, the damper has coefficient b, the block has mass m. The applied force F(t) = F₀ sin(ωt) is shown as an arrow. $k$ damper $b$ m $F(t) = F_0 \sin(\omega t)$
A driven damped spring-mass: the mass experiences a spring force $-kx$, a damping force $-b\dot{x}$ proportional to velocity, and a periodic external driving force $F_0 \sin(\omega t)$.

Newton's second law applied to the mass gives three forces:

So

m \ddot{x} = -k x - b \dot{x} + F_0 \sin(\omega t),

or equivalently

m\ddot x + b \dot x + k x = F_0 \sin(\omega t). \tag{1}

Divide through by m and introduce the natural frequency \omega_0^2 = k/m and the damping rate \gamma = b/m:

\ddot x + \gamma \dot x + \omega_0^2 x = \frac{F_0}{m} \sin(\omega t). \tag{2}

Why: the combinations \omega_0^2 = k/m and \gamma = b/m are the two parameters that describe the oscillator. \omega_0 is how fast it would oscillate if left alone and undamped; \gamma is how fast the amplitude would decay if left alone with damping and no drive. Every steady-state result from here on will be written in terms of \omega_0, \gamma, and the driving frequency \omega — not in terms of k, m, b separately.

Equation (2) is a second-order linear inhomogeneous differential equation. The word "inhomogeneous" refers to the right-hand side: it is not zero. The solution splits cleanly into two parts.

Transient and steady state

The general solution of equation (2) is the sum of two pieces:

x(t) = x_\text{transient}(t) + x_\text{steady}(t). \tag{3}

The transient. x_\text{transient}(t) is the general solution of the homogeneous equation — that is, equation (2) with the right-hand side set to zero. This is exactly the damped-oscillator equation, and from damped oscillations you already know the answer: x_\text{transient}(t) = A_0 e^{-\gamma t/2} \cos(\omega' t + \varphi) for some constants A_0 and \varphi set by initial conditions, with \omega' = \sqrt{\omega_0^2 - \gamma^2/4}. Crucially, this term has e^{-\gamma t/2} in it — it decays exponentially. After enough time (many times 1/\gamma), it is gone.

The steady state. x_\text{steady}(t) is any particular solution of the full inhomogeneous equation (2). Once the transient has died out, x(t) \to x_\text{steady}(t). The steady-state solution is the one that matters in the long run.

Physically: you start the driven oscillator, and for a while it wobbles in a complicated way as the initial conditions and the driving force both affect it. But after a time of order 1/\gamma, the memory of the initial conditions is gone. From then on, the oscillator moves in a simple, predictable, periodic way that depends only on the driving force, not on how the motion started. That long-term behaviour is the steady state.

This article is about the steady state. The transient is studied in the previous chapter; what is new here is the response to a sustained drive.

The steady-state solution — guess a sinusoid

Guess that the steady-state response oscillates at the same frequency as the drive, but with some amplitude A and phase lag \delta:

x_\text{steady}(t) = A \sin(\omega t - \delta). \tag{4}

The physical intuition: in the steady state, the drive paints the same "photograph" of the system every period 2\pi/\omega. If the system's response differed from one period to the next, it wouldn't be "steady." So the response must also be periodic with the same period as the drive. A linear equation with a sinusoidal forcing term gives a sinusoidal steady-state response.

Substitute the guess into equation (2). Compute derivatives:

\dot x_\text{steady} = A\omega \cos(\omega t - \delta), \qquad \ddot x_\text{steady} = -A\omega^2 \sin(\omega t - \delta).

Plug in:

-A\omega^2 \sin(\omega t - \delta) + \gamma A\omega \cos(\omega t - \delta) + \omega_0^2 A \sin(\omega t - \delta) = \frac{F_0}{m}\sin(\omega t).

Collect terms involving \sin(\omega t - \delta) and terms involving \cos(\omega t - \delta):

A(\omega_0^2 - \omega^2) \sin(\omega t - \delta) + \gamma A \omega \cos(\omega t - \delta) = \frac{F_0}{m}\sin(\omega t). \tag{5}

Now expand the right-hand side using \sin(\omega t) = \sin((\omega t - \delta) + \delta) = \sin(\omega t - \delta)\cos\delta + \cos(\omega t - \delta)\sin\delta:

A(\omega_0^2 - \omega^2)\sin(\omega t - \delta) + \gamma A\omega \cos(\omega t - \delta) = \frac{F_0}{m}\left[\cos\delta \sin(\omega t - \delta) + \sin\delta \cos(\omega t - \delta)\right].

For this to hold at every t, the coefficients of \sin(\omega t - \delta) and of \cos(\omega t - \delta) must match on both sides. Matching each:

A(\omega_0^2 - \omega^2) = \frac{F_0}{m}\cos\delta, \tag{6a}
\gamma A\omega = \frac{F_0}{m}\sin\delta. \tag{6b}

Why: two sinusoids of the same frequency (here \omega) but different phases are linearly independent — you can't write \sin(\omega t - \delta) as a multiple of \cos(\omega t - \delta). So the equation can only hold if the coefficient of each, on the left, equals the coefficient of the same thing on the right. That is what gives you two equations from one.

Divide (6b) by (6a) to eliminate A:

\tan \delta = \frac{\gamma \omega}{\omega_0^2 - \omega^2}. \tag{7}

Square and add (6a) and (6b) to eliminate \delta:

A^2\left[(\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2\right] = \frac{F_0^2}{m^2}\left[\cos^2\delta + \sin^2\delta\right] = \frac{F_0^2}{m^2}.

Solve for A:

\boxed{\; A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2}}. \;} \tag{8}

This is the steady-state amplitude as a function of driving frequency — the central formula of this chapter.

Why: equations (6a) and (6b) gave A\cos\delta and A\sin\delta. Squaring and adding uses \cos^2 + \sin^2 = 1 to eliminate \delta. Dividing gives \tan\delta to eliminate A. Together they decouple the two unknowns.

The resonance curve

Equation (8) says the steady-state amplitude depends on driving frequency \omega. Three features are worth studying.

At zero frequency (\omega \to 0)

The drive is just a constant push, F_0. The system settles to a displaced equilibrium where the spring force balances the push: kx = F_0, giving x = F_0/k. Check this from (8) at \omega = 0:

A(0) = \frac{F_0/m}{\sqrt{\omega_0^4}} = \frac{F_0/m}{\omega_0^2} = \frac{F_0}{m\omega_0^2} = \frac{F_0}{k}.

As expected.

At very high frequency (\omega \gg \omega_0)

The drive oscillates so fast that the mass can't keep up — inertia dominates. For \omega \gg \omega_0, equation (8) gives approximately

A(\omega) \approx \frac{F_0/m}{\omega^2}.

Why: in (\omega_0^2 - \omega^2)^2 the \omega^2 term dominates for \omega \gg \omega_0, and \gamma \omega \ll \omega^2 for modest damping, so the denominator becomes \sqrt{\omega^4} = \omega^2. The amplitude falls off as 1/\omega^2 — rapid decay with frequency.

At high frequencies the mass barely moves; the drive shoves it back and forth too fast for any real amplitude to build up.

Near \omega = \omega_0 — the resonance peak

The denominator of equation (8) is minimised — and so A is maximised — when (\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2 is smallest. For weak damping (\gamma much smaller than \omega_0), the minimum occurs approximately at \omega = \omega_0, where the first term (\omega_0^2 - \omega^2)^2 vanishes. Plug \omega = \omega_0 in:

A(\omega_0) = \frac{F_0/m}{\sqrt{0 + \gamma^2 \omega_0^2}} = \frac{F_0}{m \gamma \omega_0}. \tag{9}

With \omega_0 = \sqrt{k/m} and \gamma = b/m, this is A(\omega_0) = F_0 / (b\omega_0) — the peak amplitude at resonance.

Why: exactly at \omega = \omega_0, the only thing limiting the amplitude is the damping. A lightly damped oscillator (small \gamma) gets a huge amplitude at resonance; a heavily damped one gets only a modest one. As \gamma \to 0, the resonance peak grows without bound — which is, of course, mathematical only; no real system has exactly zero damping.

Compare the peak height to the zero-frequency value:

\frac{A(\omega_0)}{A(0)} = \frac{F_0/(m\gamma\omega_0)}{F_0/(m\omega_0^2)} = \frac{\omega_0}{\gamma} = Q.

The ratio is the quality factor Q = \omega_0/\gamma. For a precisely tuned oscillator (high Q), the peak amplitude is enormously larger than the static response. For a sloppy oscillator (low Q), the peak is only modestly enhanced.

The full curve — tall and narrow, or short and broad

The resonance curve A(\omega) has its peak near \omega_0 and falls off on both sides. The width of the peak and its height both depend on Q:

Interactive: steady-state amplitude versus driving frequency, with adjustable dampingA plot of the resonance curve A(ω) = (F₀/m) / √((ω₀² − ω²)² + γ² ω²) versus ω/ω₀, with ω₀ = 1 and F₀/m = 1 for simplicity. A draggable handle controls the damping γ. Small damping gives a tall narrow peak at ω/ω₀ = 1; large damping gives a short broad peak or no peak at all. driving frequency $\omega / \omega_0$ amplitude $A$ (arbitrary units) 0 1 2 3 4 5 1 2 3 $\omega=\omega_0$ drag the dark handle (damping)
The steady-state amplitude $A$ versus normalised driving frequency $\omega/\omega_0$. Drag the dark handle to change the damping rate $\gamma/\omega_0$. Small damping gives a tall narrow resonance peak at $\omega = \omega_0$; large damping flattens the curve completely. The quality factor $Q = \omega_0/\gamma$ controls both the peak height and the width.

Watch the transient die out

Another way to see the driven oscillator: start it from rest, apply the driving force, and watch it build up to the steady-state amplitude over the first few cycles.

Animated: driven oscillator at resonance — transient buildupA damped oscillator driven exactly at its natural frequency, starting from rest. The equation is x(t) = (0.3/ω₀) (1 − e^(−γt/2)) sin(ω₀ t) — a buildup from zero toward the steady-state amplitude 0.3/ω₀, which is the amplitude at resonance. The plot shows x versus t. Oscillation grows from zero amplitude, reaches its full steady-state amplitude after several periods, and then continues at constant amplitude indefinitely. time $t$ displacement $x$
A damped oscillator driven at its natural frequency ($\omega = \omega_0 = 2\pi$ rad/s) with damping rate $\gamma/2 = 0.25$/s, starting from rest. The amplitude builds up over the first few oscillations toward the steady-state value $F_0/(b\omega_0)$ and then stays constant. The envelope $1 - e^{-\gamma t/2}$ governs the buildup. Click replay to watch again.

The amplitude grows from zero because the drive is pumping energy in at exactly the right rhythm, and the initial displacement is zero. The damping takes energy out at a rate proportional to amplitude. At first, the drive wins and the amplitude grows; eventually, the two rates balance, and the amplitude settles at A(\omega_0) = F_0/(b\omega_0). That balance is the steady state.

The phase lag

Equation (7) gave \tan\delta = \gamma\omega/(\omega_0^2 - \omega^2). The phase \delta — the amount by which the response lags behind the drive — is itself interesting.

This quarter-period lag at resonance is the secret of pumping a swing. To push a swing effectively, you push when it is already moving the way you want to push it — that is, when the velocity is in the forward direction. If you push when the swing is at its extreme (position maximum), you are pushing when the velocity is zero; no energy goes in. If you push when the swing is at the centre (position zero) moving forward, you add energy every push.

Energy absorption and the power-versus-frequency curve

A different way to characterise resonance is by energy. The instantaneous power delivered by the drive to the oscillator is P(t) = F(t)\dot{x}(t). Averaging over one full cycle (using \langle \sin(\omega t)\cos(\omega t - \delta)\rangle = \tfrac{1}{2}\sin\delta):

\langle P \rangle = \tfrac{1}{2}\, F_0\, A\, \omega\, \sin\delta. \tag{P1}

From equation (6b), A\sin\delta = \gamma\omega A^2 \cdot (m/F_0). Substituting,

\langle P \rangle = \tfrac{1}{2}\, m\, \gamma\, \omega^2\, A^2 = \tfrac{1}{2}\, b\, \omega^2\, A^2.

This exactly equals the power dissipated by the damping force, \langle b\dot{x}^2\rangle = \tfrac{1}{2}b\omega^2 A^2 — energy in equals energy out, as it must be in the steady state.

Plugging in the resonance amplitude (equation 8):

\langle P(\omega) \rangle = \frac{1}{2}\cdot \frac{b\, \omega^2\, (F_0/m)^2}{(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2}.

The function \langle P(\omega)\rangle peaks sharply at \omega = \omega_0 (exactly — not shifted by damping). The full-width-at-half-maximum of this power curve is \gamma in angular frequency, for weak damping. Equivalently, \Delta\omega/\omega_0 = 1/Q: the fractional width of the power resonance equals the reciprocal of the quality factor.

So a wineglass with Q = 1000 has a power-resonance curve that is 0.1% wide — you have to hit its frequency to one part in a thousand for it to absorb energy effectively. A radio tuner has Q \sim 100, enough to pull one station out of many. A car suspension has Q \sim 1 and absorbs roadway bumps across an enormous range of frequencies.

Worked examples

Example 1: A driven spring-mass at three frequencies

A 0.5 kg block on a spring with k = 50 N/m and damping b = 0.4 N·s/m is driven by an external force F(t) = (2.0\text{ N})\sin(\omega t). Find the steady-state amplitude at three driving frequencies: (a) \omega = 5 rad/s, (b) \omega = 10 rad/s (= \omega_0), (c) \omega = 20 rad/s.

Step 1. Compute \omega_0 and \gamma.

\omega_0 = \sqrt{k/m} = \sqrt{50/0.5} = \sqrt{100} = 10 \text{ rad/s}.
\gamma = b/m = 0.4/0.5 = 0.8 \text{ /s}.

The quality factor Q = \omega_0/\gamma = 10/0.8 = 12.5 — a lightly damped system.

Why: compute the two derived parameters once; then the resonance formula uses only \omega_0, \gamma, and F_0/m = 2.0/0.5 = 4.0 m/s².

Step 2. (a) \omega = 5 rad/s — below resonance.

Denominator: (\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2 = (100 - 25)^2 + (0.8)^2(25) = 75^2 + 16 = 5625 + 16 = 5641.

A = \frac{4.0}{\sqrt{5641}} = \frac{4.0}{75.1} \approx 0.0533 \text{ m} = 5.33 \text{ cm}.

Step 3. (b) \omega = 10 rad/s — at resonance.

Denominator: (100 - 100)^2 + (0.8)^2(100) = 0 + 64 = 64.

A = \frac{4.0}{\sqrt{64}} = \frac{4.0}{8} = 0.500 \text{ m} = 50.0 \text{ cm}.

Step 4. (c) \omega = 20 rad/s — above resonance.

Denominator: (100 - 400)^2 + (0.8)^2(400) = (-300)^2 + 256 = 90{,}000 + 256 = 90{,}256.

A = \frac{4.0}{\sqrt{90{,}256}} = \frac{4.0}{300.4} \approx 0.01331 \text{ m} = 1.33 \text{ cm}.

Why: at \omega = 20 \gg \omega_0 = 10, the (\omega_0^2 - \omega^2)^2 term dominates — the denominator is huge. At \omega = 10 = \omega_0, that term is zero, and only the damping term \gamma^2\omega^2 remains — the denominator is tiny. At \omega = 5, the first term is still moderate (inertia is not dominant yet), but larger than at resonance.

Result: A(5) \approx 5.3 cm, A(10) = 50.0 cm, A(20) \approx 1.3 cm.

Three driving frequencies comparedA bar chart showing three bars: below-resonance (5.3 cm at ω = 5 rad/s), on-resonance (50.0 cm at ω = 10 rad/s), and above-resonance (1.3 cm at ω = 20 rad/s). The on-resonance bar is dramatically taller than the other two. 0 10 20 30 40 50 amplitude (cm) $\omega=5$ 5.3 $\omega=\omega_0=10$ 50.0 $\omega=20$ 1.3
The steady-state amplitude at three driving frequencies. At resonance, the amplitude is nearly 10 times larger than at half or double the natural frequency.

What this shows: The same driving force — 2.0 newtons amplitude — produces wildly different responses depending on frequency. At resonance, the amplitude is nearly ten times what it is just a factor-of-two away. This is the essence of resonance. If you are trying to maximise the response of an oscillator to a given drive, match the drive frequency to \omega_0. If you are trying to minimise it (as in seismic design), detune away.

Example 2: Tuning an AIR radio

A simple tuned LC circuit in an old Indian AIR radio has inductance L = 250\ \muH and variable capacitor C. The tuned circuit behaves like an oscillator with natural angular frequency \omega_0 = 1/\sqrt{LC} (which you will derive formally when you meet alternating current), driven by the electromagnetic signal from the broadcast antenna. The circuit's quality factor is Q = 80.

(a) What capacitance C is needed to tune to AIR Chennai's medium-wave station at f_\text{AIR} = 720 kHz? (b) What is the frequency width \Delta f of the resonance? (c) Why does this \Delta f matter for separating this station from its neighbours?

Step 1. Convert to angular frequency.

\omega_0 = 2\pi f_\text{AIR} = 2\pi \times 720{,}000 = 4.524 \times 10^6 \text{ rad/s}.

Step 2. Solve \omega_0 = 1/\sqrt{LC} for C.

C = \frac{1}{\omega_0^2 L} = \frac{1}{(4.524 \times 10^6)^2 \times 250 \times 10^{-6}}.

Compute: \omega_0^2 = 2.047 \times 10^{13}; \omega_0^2 L = 2.047 \times 10^{13} \times 2.5 \times 10^{-4} = 5.117 \times 10^9. So

C = \frac{1}{5.117 \times 10^9} \approx 1.95 \times 10^{-10} \text{ F} = 195 \text{ pF}.

Why: LC-circuit resonance works just like the mechanical oscillator, with L playing the role of mass (inertia of the current) and 1/C playing the role of spring stiffness (restoring force per unit charge). The resonance frequency is \omega_0 = 1/\sqrt{LC}, and setting this equal to the desired broadcast frequency fixes C.

Step 3. Compute the resonance width.

Q = \omega_0/\gamma = 80, so \gamma = \omega_0/80. The FWHM (in angular frequency) is \Delta\omega = \gamma = \omega_0/80. Converting to ordinary frequency:

\Delta f = \frac{f_\text{AIR}}{Q} = \frac{720}{80} = 9 \text{ kHz}.

Why: the resonance width at half-maximum-power is \gamma in angular frequency, equivalently f_0/Q in ordinary frequency. High-Q circuits have narrow resonances — better selectivity.

Step 4. Compare with medium-wave station spacing.

Indian AM stations on the medium-wave band are typically separated by 9 kHz (this is in fact the international Region 1 standard). So the resonance width happens to equal the inter-station spacing. The radio can therefore separate adjacent stations — but only just.

Result: C \approx 195 pF, \Delta f = 9 kHz. A higher Q circuit would give a narrower peak and better station separation, at the cost of needing more precise tuning.

What this shows: Tuning a radio is exactly matching \omega_0 to the broadcast frequency. Selectivity — the ability to pick out one station from its neighbours — is controlled by the quality factor Q. A low-Q radio would pick up two or three adjacent stations as a muddle; a high-Q one needs very fine tuning but gives a clean single station. Modern digital radio uses superheterodyne techniques to get enormous effective Q without mechanical precision, but the underlying physics is the same forced-oscillator resonance.

Example 3: A tall building during an earthquake

A ten-storey apartment building in Ahmedabad sways side-to-side as a damped oscillator. Its natural period is T_0 = 1.2 s (so \omega_0 = 2\pi/T_0 \approx 5.24 rad/s) and its quality factor is Q = 15. Suppose the dominant seismic ground motion during the 2001 Bhuj earthquake had a period of 1.1 s (an angular frequency \omega = 5.71 rad/s) — close to the building's resonance.

(a) By what factor is the building's sway amplitude increased relative to the amplitude it would have at zero frequency (a slow, quasi-static ground drift)? (b) How many times larger is the sway at \omega = 5.71 rad/s compared to a building with natural period T_0 = 0.3 s (for which \omega_0 = 20.9 rad/s, far from the driving frequency)?

Step 1. Set up the ratio A(\omega)/A(0) for each building.

From equation (8), with F_0/m the same in each case (same earthquake ground motion):

\frac{A(\omega)}{A(0)} = \frac{\omega_0^2}{\sqrt{(\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2}},

using A(0) = F_0/(m\omega_0^2).

Step 2. (a) The tall building.

\omega_0^2 = 5.24^2 = 27.46. \omega^2 = 5.71^2 = 32.60. (\omega_0^2 - \omega^2)^2 = (27.46 - 32.60)^2 = (-5.14)^2 = 26.42.

\gamma = \omega_0/Q = 5.24/15 = 0.349/s. \gamma^2\omega^2 = 0.122 \times 32.60 = 3.98.

Denominator: \sqrt{26.42 + 3.98} = \sqrt{30.40} = 5.51.

\frac{A(\omega)}{A(0)} = \frac{27.46}{5.51} \approx 4.98.

The tall building's sway amplitude is almost five times what it would be in a slow ground drift — approaching but not quite reaching the peak value of Q = 15, because \omega is close to but not exactly \omega_0.

Step 3. (b) The short building.

\omega_0 = 20.9 rad/s, so \omega_0^2 = 436.8. (\omega_0^2 - \omega^2)^2 = (436.8 - 32.6)^2 = 404.2^2 = 163{,}378.

\gamma in this case = \omega_0/Q = 20.9/15 = 1.39/s (assuming the same Q). \gamma^2\omega^2 = 1.94 \times 32.60 = 63.2.

Denominator: \sqrt{163{,}378 + 63} = \sqrt{163{,}441} = 404.3.

\frac{A(\omega)}{A(0)} = \frac{436.8}{404.3} \approx 1.08.

The short building's sway is only 8% above the quasi-static level — the earthquake barely excites it.

Step 4. Compare.

\frac{(A/A_0)_\text{tall}}{(A/A_0)_\text{short}} = \frac{4.98}{1.08} \approx 4.6.

Why: matched tuning (tall building at \omega_0 \approx \omega) amplifies the response nearly 5×; detuned (short building with \omega_0 \gg \omega) gives essentially no amplification. The same earthquake, same ground amplitude — but completely different building response.

Result: The tall building's sway is about 5× its static response, nearly 5× larger than the short building's sway for the same ground shaking.

What this shows: In an earthquake, which buildings suffer depends not just on the ground motion but on how closely their natural frequency matches the dominant seismic frequency. A building detuned from the earthquake spectrum barely moves; a building tuned to the earthquake spectrum shakes itself apart. This is why post-Bhuj Indian seismic codes (IS 1893) now require buildings in high-seismic-zone regions to have structural modifications — tuned mass dampers, base isolators, or simple stiffening — that shift their natural frequencies away from the expected earthquake spectrum. Seismic engineering is the engineering of detuning.

Common confusions

If you have the steady-state amplitude formula, the resonance curve, and the meaning of Q, you have the core physics of driven oscillators. What follows adds: the complex-number method for solving equation (2) in one line, an exact treatment of the peak position, the driven oscillator's energy balance, and a glimpse at nonlinear resonance.

The complex-number method — resonance in one line

Equation (2) can be solved in half the work if you allow the driving force to be complex:

F(t) = F_0 e^{i\omega t},

with the understanding that the physical drive is the real part, F_0\cos(\omega t). Guess a complex response x(t) = X e^{i\omega t}, substitute:

-\omega^2 X e^{i\omega t} + i\gamma\omega X e^{i\omega t} + \omega_0^2 X e^{i\omega t} = \frac{F_0}{m} e^{i\omega t}.

Cancel e^{i\omega t}:

X = \frac{F_0/m}{\omega_0^2 - \omega^2 + i\gamma\omega}. \tag{10}

This complex amplitude X encodes both real amplitude and phase. Its modulus:

|X| = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2}},

which is exactly equation (8). Its phase:

\arg(X) = -\arctan\left(\frac{\gamma\omega}{\omega_0^2 - \omega^2}\right),

which is -\delta from equation (7) (the minus because X multiplies e^{i\omega t} while the sinusoidal form \sin(\omega t - \delta) corresponds to e^{i(\omega t - \delta)} = e^{-i\delta}e^{i\omega t}).

The complex method makes the derivation one line. It also makes the analogy with AC circuits explicit: the denominator \omega_0^2 - \omega^2 + i\gamma\omega is the impedance of the oscillator, dividing force into displacement — exactly analogous to electrical impedance dividing voltage into current.

Exact peak position

Maximise A(\omega)^2 with respect to \omega. Let u = \omega^2. Then

A^2 = \frac{(F_0/m)^2}{(\omega_0^2 - u)^2 + \gamma^2 u}.

Set d(A^2)/du = 0. The denominator is D(u) = u^2 - (2\omega_0^2 - \gamma^2)u + \omega_0^4. Minimising the denominator: D'(u) = 2u - (2\omega_0^2 - \gamma^2) = 0, giving

u_\text{peak} = \omega_\text{peak}^2 = \omega_0^2 - \frac{\gamma^2}{2}.

For \gamma \ll \omega_0, \omega_\text{peak} \approx \omega_0 - \gamma^2/(4\omega_0) — a tiny downward shift. Exact resonance occurs slightly below \omega_0, but for high-Q oscillators the distinction is usually irrelevant. For \gamma^2 \ge 2\omega_0^2, u_\text{peak} \le 0, meaning there is no peak in the positive-frequency range — the amplitude is monotonically decreasing with \omega. This is the over-damped regime; no resonance at all.

Energy balance in the steady state

In the steady state, the drive does work on the oscillator and the damping dissipates that work. Averaged over a cycle:

\text{Power in} = \langle F(t)\dot{x}(t)\rangle = \tfrac{1}{2} F_0 A\omega \sin\delta.
\text{Power out} = \langle b\dot{x}^2\rangle = \tfrac{1}{2} b A^2 \omega^2.

Energy conservation requires these to be equal. Setting them equal:

F_0 \sin\delta = b A \omega.

But from equation (6b), \gamma A\omega = (F_0/m)\sin\delta, i.e. bA\omega = F_0\sin\delta. The two expressions match exactly — the "guess-a-sinusoid" method automatically produces a solution satisfying energy balance.

This energy-balance view also makes the peak power transfer obvious: power is \tfrac{1}{2}F_0 A\omega\sin\delta, and at \omega = \omega_0 (where \delta = \pi/2), \sin\delta = 1 is maximum. Off-resonance, some of the drive's energy is out of phase with the velocity and does not get absorbed.

The half-power bandwidth

Power absorbed as a function of \omega is proportional to A^2 \omega^2 = \omega^2 (F_0/m)^2/[(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2]. The peak is at \omega = \omega_0 where it equals (F_0/m)^2/\gamma^2. "Half power" means the power drops to half its peak value. Setting

\frac{(F_0/m)^2}{(\omega_0^2 - \omega^2)^2/\omega^2 + \gamma^2} = \frac{1}{2} \cdot \frac{(F_0/m)^2}{\gamma^2}

gives (\omega_0^2 - \omega^2)^2/\omega^2 = \gamma^2, so \omega_0^2 - \omega^2 = \pm\gamma\omega near resonance. Solving for \omega:

\omega \approx \omega_0 \pm \gamma/2.

The full-width-at-half-maximum (FWHM) of the power curve is \gamma. The quality factor Q = \omega_0/\gamma is therefore the ratio of the peak frequency to the width — a high-Q resonance is sharp; a low-Q resonance is broad. This is the operational definition of Q and the reason tuned circuits, atomic transitions, and gravitational waves are all described by their Q.

Why resonance is everywhere — a universality argument

Any linear system with a natural frequency, damping, and a periodic driving force obeys an equation of the form (2). The formula (8) is therefore not a special result about a spring-mass; it is the generic behaviour of any driven, damped linear oscillator. The electron circuit in a radio, the displaced mass in a seismograph, the electrons in a crystal lattice absorbing infrared radiation, the nuclear spins in an MRI magnet absorbing radio-frequency photons, the air column in a flute's bore — all of these, when driven, show an A(\omega) curve peaked near their own \omega_0 and broadened by their own \gamma. Resonance is not a special trick of springs. It is a feature of every linear driven damped system in physics.

Nonlinear resonance — the bent peak

Real oscillators have slightly nonlinear spring forces (e.g., a rubber band, or a pendulum at moderate amplitudes where \sin\theta \ne \theta). Nonlinearity makes the natural frequency depend on amplitude, and then the resonance peak "bends" — the peak position shifts with amplitude. At large drive strengths this produces hysteresis: sweeping \omega from low to high gives a different curve than sweeping from high to low. This is the Duffing oscillator, and it is the bridge from this chapter to the physics of nonlinear dynamics. The bent-peak resonance curve is why real musical instruments (which are mildly nonlinear) can produce subharmonics and "pitch bend" effects that linear resonance cannot.

Where this leads next