Damping and Resonance

Beginner

A short pendulum swings faster than a long one, and a block on a stiff spring bounces faster than on a soft one. Everything that can swing or bounce has a natural frequency: its own rhythm, the number of back-and-forth trips it makes each second when it is left alone.

Step 1: damping. Start a swing and walk away: each trip is a little smaller than the one before. Air resistance and friction take some energy on every trip and turn it into heat, and they push back harder the faster the object moves. This gradual fading is called damping.

Step 2: how much damping? In the animation, three identical blocks hang on identical springs, and all three are pulled down and let go together. The left one moves in air (light damping) and bounces many times, each bounce smaller. The right one sits in thick syrup (heavy damping): it never overshoots, but it creeps back very slowly. The middle one, in a thinner liquid, has just enough damping to stop the bouncing. This is critical damping, and its “at rest” label appears first. Car shock absorbers and door closers are set close to it.

Step 3: resonance. Small pushes given in step with the natural rhythm all add energy at the right moment, so the motion grows. On the right, the top of a spring is shaken gently in the block's own rhythm, and soon the block bounces far more than the shaking. Damping is what stops the growth.

Check yourself: a washing machine often shakes hard for a few seconds while its drum speeds up, then calms down at full speed. Why?
Answer: on the way up, the drum's spin rate passes through the machine's natural frequency, so for a moment it is shaken in rhythm: resonance.

Intermediate

Two forces: the spring and the damper

More damping does not always bring a block to rest sooner. A block on a spring that is pulled and released feels two forces: the spring's restoring force $-kx$, pointing back towards equilibrium, and a damping force that opposes the motion, such as the drag of a liquid or of a shock absorber's oil. For slow motion in a liquid the damping force is nearly proportional to the velocity, so the net force is
$F=-kx-bv$

The damping constant $b$ is in kilograms per second (kg/s, the same as N·s/m). The two forces peak at different moments: the spring force at the turning points, where the block stops for an instant, and the damping force when the block moves fastest, which for light damping is close to equilibrium.

In the animation the gold arrows show $-bv$; they flip every time a block turns round. The right-hand block, in thick syrup with $b=30kg/s$, has the longest arrow although it moves the slowest: at its top speed, 0.077 m/s, the damping force $30\cdot 0.077\approx 2.3\mathrm{N}$ exactly balances the spring's pull.

Light damping: an exponential fade

With light damping the block keeps almost its natural rhythm, but each swing is smaller than the last by the same fraction, because a damping force that grows with the speed takes a fixed fraction of the remaining energy in every cycle. That is an exponential decay, $A=A_0{e}^{-\beta t}$, with
$\beta =\frac{b}{2m},{\omega }_0=\sqrt{\frac{k}{m}}$
The damping rate $\beta$ is in s⁻¹: after a time $1/\beta$ the amplitude is down to $1/e$, about 37%, and it halves every $\ln 2/\beta$. Here ${\omega }_0$ is the natural angular frequency. The full motion is a cosine inside the shrinking envelope:
$x=A_0{e}^{-\beta t}\cos ({\omega }_{d}t+\phi ),{\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$

The damper holds the block back a little, so the damped angular frequency ${\omega }_d$ is just below ${\omega }_0$. The quality factor measures how light the damping is:
$Q=\frac{{\omega }_0}{2\beta }=\frac{m{\omega }_0}{b}$
Each cycle multiplies the amplitude by about $e^{-\pi /Q}$, so after $Q$ cycles only $e^{-\pi }$, about 4%, is left. A tuning fork, with a $Q$ in the thousands, rings for seconds after a single tap.

Light, critical and heavy damping

As $b$ grows the decay speeds up, until at a critical value the block no longer overshoots:
$b_c=2\sqrt{km}=2m{\omega }_0$
There $\beta ={\omega }_0$ and ${\omega }_d$ drops to zero. Comparing $b$ with $b_c$ gives three kinds of motion:

• $b<b_c$, light damping: the block overshoots again and again, with swings that die away exponentially.
• $b=b_c$, critical damping: the quickest possible return to equilibrium without any overshoot.
• $b>b_c$, heavy damping: no overshoot either, but the damper resists so strongly that the block creeps back.

Too much damping is slow for the same reason a spoon is slow in honey: the damping force soon matches the spring's pull, which limits the speed to about $kx/b$. Triple $b$ and the block crawls back three times as slowly.

Worked example: the three blocks in the animation

All three blocks have $m=1\mathrm{kg}$ and $k=25N/m$, and each is pulled down 10 cm and released from rest:
${\omega }_0=\sqrt{25/1}=5rad/s,b_c=2\sqrt{25\cdot 1}=10kg/s$
so $f_0\approx 0.80\mathrm{Hz}$, one cycle every 1.26 s. The middle block is exactly critical; the right-hand one has three times $b_c$, the left-hand one a tenth of it. For the left block,
$\beta =\frac{1}{2\cdot 1}=0.5{\mathrm{s}}^{-1},Q=\frac{5}{2\cdot 0.5}=5$

After 3 s its amplitude is $10\mathrm{cm}\cdot e^{-0.5\cdot 3}\approx 2.2\mathrm{cm}$, the value on the blue $A$ readout when the $t$ readout, which starts at the release, shows 3 s. The amplitude halves every $\ln 2/0.5\approx 1.4\mathrm{s}$, each cycle multiplies it by $e^{-\pi /5}\approx 0.53$, and after 5 cycles, about 6.3 s, only 4% is left. The critical block stays within 4 mm of its dashed equilibrium line after about 1.0 s, the heavy one after about 3.8 s and the light one after about 6.4 s.

Resonance: pushes in step with the motion

Now drive the oscillator with a periodic force $F_0\cos \omega t$. After the start-up it oscillates steadily at the driving frequency, with the amplitude
$A(\omega )=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+(2\beta \omega {)}^2}}$
Far from ${\omega }_0$ the response is small; near it the amplitude rises steeply: this is resonance. At ${\omega }_0$ the spring force and the inertia cancel, and the drive only has to make up for the damping:
$A({\omega }_0)=\frac{F_0}{b{\omega }_0}=\frac{Q{F}_0}{k}$
A steady force would stretch the spring by $F_0/k$; applied in rhythm, it moves the block $Q$ times as far. At resonance the block lags a quarter cycle behind the push, so it moves fastest when the push is strongest, and every push does positive work.

On the right of the animation, the top of an identical, lightly damped spring is shaken by $Y=2\mathrm{cm}$ at exactly $f_0=0.80\mathrm{Hz}$. Moving the top of a spring by $y$ changes its pull by $ky$, so the shaking acts like a force $F_0=kY=25\cdot 0.02=0.5\mathrm{N}$, and
$A=\frac{F_0}{b{\omega }_0}=\frac{0.5}{1\cdot 5}=0.10\mathrm{m}$
which is $QY=10\mathrm{cm}$. The driven block's $A$ readout climbs there slowly, roughly as $QY(1-e^{-\beta t})$: about 6 cm after 2 s of shaking, 9 cm after 4.6 s and 9.9 cm at the end of the loop.

Worked example: a car with a worn shock absorber

One corner of a car puts about 400 kg on a suspension spring with $k=40000N/m$:
${\omega }_0=\sqrt{40000/400}=10rad/s,b_c=2\sqrt{40000\cdot 400}=8000kg/s$
so the body bounces at about 1.6 Hz. A healthy absorber, set somewhat below $b_c$, lets the body overshoot once after a bump. A worn one with $b=800kg/s$ gives $\beta =800/(2\cdot 400)=1.0{\mathrm{s}}^{-1}$ and $Q=10/(2\cdot 1.0)=5$, like the left block in the animation. A 4 cm bounce then halves every 0.7 s and is still $4\cdot e^{-2}\approx 0.5\mathrm{cm}$ after 2 s, three bounces later.

The road can drive the car too. Ripples spaced $d=10\mathrm{m}$ apart shake the wheel $v/d$ times per second, which matches the natural 1.59 Hz at $v\approx 16m/s$, about 57 km/h. At that speed the worn car heaves about $Q=5$ times as high as the ripples and seems to float.

Common mistakes

• More damping is not always faster: the right-hand block, at three times $b_c$, needs about 3.8 s to settle, the critical one only 1.0 s.
• Light damping changes the amplitude, not the rhythm: with $Q=5$ the period grows only from 1.257 s to 1.263 s.
• The time $1/\beta$ is not a half-life: with $\beta =0.5{\mathrm{s}}^{-1}$ the amplitude falls to 37% in 2 s and halves in 1.4 s; the energy, which goes as $A^2$, halves in 0.7 s.
• Resonance is not infinite: damping caps the amplitude at $Q{F}_0/k$, 10 cm for the shaken spring, where the drive's 0.125 W equals the damper's losses.
• The critical value is $2\sqrt{km}$, not $\sqrt{km}$: forgetting the 2 gives 5 kg/s instead of 10 kg/s for the blocks.

Where you meet it

Near-critical damping is built into door closers, instrument needles and car suspensions; near the top of the Taipei 101 tower, a steel pendulum of about 660 tonnes swings against the building's sway. Resonance is put to work in quartz watches, whose crystal rings at 32 768 Hz, in musical instruments and in radio tuning. It must be avoided in structures: London's Millennium Bridge swayed sideways on its opening days in 2000, as walkers fell into step with it. The same equation governs circuits made of the parts in Capacitors and Electromagnetic Induction, and each mode in Standing Waves.

Advanced

The equation of free motion

Damping is any force that drains the energy of a vibration: air, oil, a magnet braking a metal plate. This section finds every possible free motion of a damped spring and block. It matters twice over: these motions are how a disturbed system settles, and they are also the transient that every driven oscillator has to shed before it reaches its steady state.

Take a block of mass $m$ on a spring of stiffness $k$, with a damper that pushes against the velocity with the force $-b\dot{x}$. Here $x$ is the displacement from equilibrium and $b$ is the damping coefficient. Newton's second law gives $m\ddot{x}=-kx-b\dot{x}$. Divide by $m$ and name two ratios, the damping rate $\beta =b/(2m)$ and the natural angular frequency ${\omega }_0=\sqrt{k/m}$. The equation takes its standard form
$\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=0$
The factor 2 is a convention that keeps the roots below free of halves.

The equation is linear, and its coefficients do not change with time. An exponential is the one function whose derivative is a multiple of itself, so try $x=e^{st}$ with a constant $s$, which may be complex. Each time derivative then just multiplies by $s$. Every term carries the same factor $e^{st}$, and dividing it out leaves the characteristic equation $s^2+2\beta s+{\omega }_0^2=0$. The quadratic formula gives its two roots:
$s_1=-\beta +\sqrt{{\beta }^2-{\omega }_0^2},s_2=-\beta -\sqrt{{\beta }^2-{\omega }_0^2}$
Because the equation is linear, any sum $c_1{e}^{s_{1}t}+c_2{e}^{s_{2}t}$ is also a solution, and the two constants are fixed by the initial position and velocity. So everything about the free motion is decided by where the two roots lie.

The animation uses $m=1\mathrm{kg}$ and $k=25N/m$, so ${\omega }_0=5rad/s$ (0.80 Hz), with three dampers, $b=1$, 10 and $30kg/s$, that is $\beta =0.5$, 5 and $15{\mathrm{s}}^{-1}$.

Three kinds of free motion

Whether a released block swings to and fro or just creeps home depends only on the sign of ${\beta }^2-{\omega }_0^2$ under the square root. A map of the roots shows the three cases at once, and also which damping brings the block back fastest.

Legend

$s_1$	Root with the plus sign: the upper blue dot, then on the real axis, moving back towards 0.	${\mathrm{s}}^{-1}$
$s_2$	Root with the minus sign: the lower blue dot, then on the real axis, moving away from 0.	${\mathrm{s}}^{-1}$
$\beta$	Damping rate $b/(2m)$: minus the real part of the pair, its distance left of the vertical axis.	${\mathrm{s}}^{-1}$
${\omega }_d$	Damped angular frequency $\sqrt{{\omega }_0^2-{\beta }^2}$: the imaginary part of $s_1$, the height of the upper dot.	$rad/s$
${\omega }_0$	Natural angular frequency $\sqrt{k/m}$: the radius of the arc, and $-{\omega }_0$ is where the two roots meet.	$rad/s$

Figure 1 is the complex plane of $s$: real part along the horizontal axis, imaginary part up the vertical one. One number sorts the cases, the damping ratio $\zeta =\beta /{\omega }_0$. It equals $b/b_c$, where $b_c=2\sqrt{km}$ is the critical damping coefficient, the value of $b$ that makes $\beta ={\omega }_0$. In the animation $b_c=10kg/s$, so the three blocks have $\zeta =0.1$, 1 and 3.

Light damping, $\zeta <1$. The number under the square root is negative, so the roots are the complex pair $-\beta \pm i{\omega }_d$, the two blue dots, with the damped angular frequency ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. Since $e^{(-\beta \pm i{\omega }_d)t}=e^{-\beta t}{e}^{\pm i{\omega }_{d}t}$, the real part $-\beta$ sets the decay and the imaginary part $\pm {\omega }_d$ sets the oscillation. Combining the two exponentials so that $x$ is real gives
$x=C{e}^{-\beta t}\cos ({\omega }_{d}t+\phi )$
an oscillation inside an envelope that shrinks as $e^{-\beta t}$; the constants $C$ and $\phi$ come from the initial position and velocity.

Now look at the right triangle, drawn for $\zeta =0.6$. Its horizontal leg is $\beta$, its vertical leg is ${\omega }_d$, and its hypotenuse is $\sqrt{{\beta }^2+{\omega }_d^2}={\omega }_0$. So a complex root always lies at the distance ${\omega }_0$ from the origin. As $b$ grows, the pair moves along the blue arc of radius ${\omega }_0$, the upper root down and the lower one up, towards the real axis.

Critical damping, $\zeta =1$. The two roots meet at $-{\omega }_0$. A double root supplies only one exponential, $e^{-{\omega }_{0}t}$; the second independent solution is $t{e}^{-{\omega }_{0}t}$, as substituting it into the equation confirms.

Heavy damping, $\zeta >1$. The square root is now real and smaller than $\beta$, so both roots are real and negative. As $b$ grows, $s_1$ turns back towards 0 and $s_2$ runs off to the left. Nothing oscillates: the block creeps back.

The expert $x(t)$ graph of the animation uses the blue, green and purple of Figure 1 and shows the three blocks released from rest at $-10\mathrm{cm}$; call this release distance $A_0=10\mathrm{cm}$. Blue ($\zeta =0.1$, $s=-0.5\pm 4.97i{\mathrm{s}}^{-1}$) overshoots to $+7.3\mathrm{cm}$ half a period after the release, at 0.63 s. Green ($\zeta =1$) follows $x=-A_0(1+{\omega }_{0}t)e^{-{\omega }_{0}t}$ and is within 4 mm of zero after about 1.0 s. Purple ($\zeta =3$) has the roots $-0.86$ and $-29.1{\mathrm{s}}^{-1}$. Its fast term, $e^{-29.1t}$, is gone within a tenth of a second. The slow root, close to $-{\omega }_0^2/(2\beta )$ when the damping is strong, leaves a creep with a time constant of 1.2 s.

Which damping brings the block home fastest? A sum of exponentials dies only as fast as its slowest term, the one whose root lies nearest the vertical axis in Figure 1. Below critical damping that distance is $\beta$, which grows with $b$ as the pair moves left along the arc. Above critical damping it is $|s_1|$, which shrinks as $b$ grows, because $s_1$ drifts back towards 0. The two trends meet at $\zeta =1$: there the nearest root is as far left as it can get, and the decay rate reaches its largest value, ${\omega }_0$. If a small overshoot is acceptable, slightly less damping is quicker still. At $\zeta =0.75$ the block stays within 4 mm of zero after 0.64 s, which is why instrument needles are often tuned a little below critical.

Energy and the meaning of Q

The roots say how fast the motion decays; the energy says why. It also gives the one number engineers quote for any oscillator, the quality factor $Q$.

Legend

$x$	Displacement of the block from equilibrium.	$\mathrm{m}$
$t$	Time since the release.	$\mathrm{s}$
$E$	Stored energy, kinetic plus spring: $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2$.	$\mathrm{J}$
$E_0$	Energy at the release, all in the spring: $\frac{1}{2}k{A}_0^2$.	$\mathrm{J}$
$e^{-2\beta t}$	Decay factor of the dashed curve $E_0{e}^{-2\beta t}$, which the energy touches at every turning point.
$dE/dt$	Rate of change of the energy, $-b{\dot{x}}^2$: largest in size where the block moves fastest, zero at a turning point.	$\mathrm{W}$
$T_d$	Damped period $2\pi /{\omega }_d$: one full cycle, the green span.	$\mathrm{s}$

Figure 2 follows the light block of the animation ($\zeta =0.1$), released from rest at $x=-A_0$: its position $x(t)$ on top and its stored energy $E$ below, on one time axis.

To read the lower panel, first find how fast $E$ changes. Multiply the equation of motion, $m\ddot{x}+b\dot{x}+kx=0$, by $\dot{x}$. The product $m\ddot{x}\dot{x}$ is the time derivative of $\frac{1}{2}m{\dot{x}}^2$, and $kx\dot{x}$ is that of $\frac{1}{2}k{x}^2$. So the inertia and spring terms together make the derivative of $E=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2$, and what is left reads
$\frac{dE}{dt}=-b{\dot{x}}^2$
The damper removes energy at the rate $b{\dot{x}}^2$. That rate is never negative, and it is zero whenever the block stops for an instant.

Now look at the lower panel. Energy drains only while the block moves, so $E$ falls in steps. Each dot marks a turning point, where $\dot{x}=0$ and all the energy sits in the spring; there the slope of $E$ is zero and the curve is flat. The tangent marked $dE/dt$ touches the steepest part of the first fall, where the block moves fastest, just before it passes equilibrium. At a turning point $|x|=A_0{e}^{-\beta t}$, so the energy there is $\frac{1}{2}k{A}_0^2{e}^{-2\beta t}$. The flat parts therefore sit exactly on the dashed curve $E_0{e}^{-2\beta t}$, with $E_0=\frac{1}{2}k{A}_0^2=\frac{1}{2}\cdot 25\cdot 0.10^2=0.125\mathrm{J}$. Between turning points $E$ wanders about 10% above and below that curve. From one turning point to the next, half a damped period $T_d/2$, the energy is multiplied by $e^{-\beta T_d}\approx 0.53$.

Smoothing out the steps gives a simple law. Over one cycle of light damping the motion is nearly $A\cos {\omega }_{0}t$ with a slowly shrinking amplitude $A$. The average of ${\dot{x}}^2$ over the cycle is then $\frac{1}{2}{\omega }_0^2{A}^2$, while $E=\frac{1}{2}m{\omega }_0^2{A}^2$. So the average of $b{\dot{x}}^2$ is $(b/m)E=2\beta E$, and
$\frac{dE}{dt}=-2\beta E$
The energy decays as $e^{-2\beta t}$. Energy goes as the square of the amplitude, so the amplitude decays half as fast, as $e^{-\beta t}$, just as the roots said.

The quality factor is defined as $Q={\omega }_0/(2\beta )$. With it the law reads $dE/dt=-{\omega }_{0}E/Q$. The block moves through one radian of phase in the time $1/{\omega }_0$, so in that time it loses the fraction $1/Q$ of its energy: $Q$ is the stored energy divided by the energy lost per radian. A high-$Q$ oscillator rings for many cycles; the light block has only $Q=5$.

A familiar rule says the fraction lost per cycle is $2\pi /Q$. Here that would be 126%, which is impossible: the rule is only the first term of the exact $1-e^{-2\pi /Q}$, good for large $Q$ alone. The exponential law itself holds exactly from one cycle to the next. Over one damped period $T_d=2\pi /{\omega }_d=1.263\mathrm{s}$, the green span in Figure 2, the energy keeps the fraction $e^{-2\beta T_d}\approx 0.28$, so 72% goes in the first cycle.

The driven oscillator

Now push the block with a periodic force $F_0\cos \omega t$. This is the situation of a loudspeaker cone, a bridge in gusty wind or an electron in a light wave, and the questions are how large the response is and how far it lags. The equation of motion gains a right-hand side:
$\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=(F_0/m)\cos \omega t$
Because the equation is linear, its general solution is any one particular solution plus the free motion found above. The free part dies away, as $e^{-\beta t}$ for light damping; it is called the transient. What remains is the steady state, which oscillates at the drive frequency $\omega$, not at ${\omega }_0$.

Complex numbers find the steady state quickly. Write the force as the real part of $F_0{e}^{i\omega t}$ and look for $x$ as the real part of $\hat{x}{e}^{i\omega t}$, where the complex amplitude $\hat{x}$ is a constant still to be found. Taking the real part commutes with adding, with differentiating and with multiplying by real numbers, so one can solve the complex equation and take the real part at the end. Each $d/dt$ acting on $e^{i\omega t}$ becomes a factor $i\omega$. Multiplying by $i$ is a quarter turn in the complex plane, so the velocity leads the displacement by 90°. The differential equation becomes algebra:
$({\omega }_0^2-{\omega }^2+2i\beta \omega )\hat{x}=F_0/m$

The trick needs linearity: a term in $x^2$ would mix the real and imaginary parts, and the real part of the complex answer would no longer solve the real equation. Linearity also gives superposition, so a force made of many frequencies produces the sum of the separate responses.

The phasor triangle: amplitude and lag

The complex equation says that two arrows add up to a third. Drawing them turns the amplitude and the phase lag into plain geometry.

Legend

$F_0/m$	The drive, force per unit mass, along the real axis.	$m/{\mathrm{s}}^2$
$\hat{x}$	Complex amplitude of the displacement, at the angle $-\delta$ from the drive; drawn at its own scale.	$\mathrm{m}$
$({\omega }_0^2-{\omega }^2)\hat{x}$	Spring and inertia terms together: a real multiple of $\hat{x}$, so it points along $\hat{x}$.	$m/{\mathrm{s}}^2$
$(2\beta \omega )i\hat{x}$	Damping term: $\hat{x}$ turned a quarter turn ahead by the factor $i$ and scaled by $2\beta \omega$.	$m/{\mathrm{s}}^2$
$\delta$	Lag of the displacement behind the drive: the angle between $\hat{x}$ and $F_0/m$.	$\mathrm{rad}$

Figure 3 draws the equation in the complex plane, for $\omega =0.75{\omega }_0$ and $\beta =0.25{\omega }_0$. The drive $F_0/m$ lies along the real axis. The gold arrow is $\hat{x}$, drawn at its own scale. The spring term ${\omega }_0^2\hat{x}$ and the inertia term $-{\omega }^2\hat{x}$ are both real multiples of $\hat{x}$, so together they make the green arrow $({\omega }_0^2-{\omega }^2)\hat{x}$, which points along $\hat{x}$. The damping term $2\beta \omega i\hat{x}$ carries the extra factor $i$, so the orange arrow stands at a right angle to the green one. Green plus orange must equal the drive. The drive is therefore the hypotenuse of a right triangle, and $\hat{x}$, lying along the green leg, lags behind the drive by the angle $\delta$ between that leg and the hypotenuse.

Solving the equation for $\hat{x}$ gives
$\hat{x}=\frac{F_0/m}{{\omega }_0^2-{\omega }^2+2i\beta \omega }$
Its modulus is the amplitude $A$, and its argument is $-\delta$, where $\delta$ is the lag of the displacement behind the force. The triangle gives both without further algebra. Its legs measure $({\omega }_0^2-{\omega }^2)A$ and $2\beta \omega A$ and its hypotenuse is $F_0/m$, so Pythagoras gives the amplitude and the ratio of the legs gives the lag:
$A(\omega )=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+(2\beta \omega {)}^2}}$
$\tan \delta =\frac{2\beta \omega }{{\omega }_0^2-{\omega }^2}$
As $\omega$ rises through ${\omega }_0$, $\delta$ rises from 0 to $180^{\circ }$. Figure 3 is drawn below resonance, where the green leg is positive and $\delta$ is under $90^{\circ }$. Above ${\omega }_0$ the factor ${\omega }_0^2-{\omega }^2$ turns negative, the green arrow points against $\hat{x}$, and $\delta$ passes $90^{\circ }$.

Reading the resonance curve

The two formulas above contain the whole steady response. Plotting them against the drive frequency shows where the oscillator responds strongly, how sharp that response is, and what the phase does on the way through.

Legend

$A$	Steady amplitude of the displacement.	$\mathrm{m}$
$\delta$	Lag of the displacement behind the drive.	$\mathrm{rad}$
$\omega$	Angular frequency of the drive.	$rad/s$
${\omega }_0$	Natural angular frequency: just right of the amplitude peak, where $\delta =\pi /2$.	$rad/s$
${\omega }_1$	Lower half-power frequency, where $\delta =\pi /4$.	$rad/s$
${\omega }_2$	Upper half-power frequency, where $\delta =3\pi /4$.	$rad/s$
$\Delta \omega$	Full width at half power: ${\omega }_2-{\omega }_1=2\beta ={\omega }_0/Q$.	$rad/s$
$F_0/k$	Static stretch: the amplitude for a drive that changes very slowly.	$\mathrm{m}$
$Q{F}_0/k$	Amplitude at ${\omega }_0$: $Q$ times the static stretch.	$\mathrm{m}$
$\pi /2$	A quarter turn: the lag at ${\omega }_0$.	$\mathrm{rad}$
$\pi$	A half turn: the lag far above ${\omega }_0$, where the block moves against the drive.	$\mathrm{rad}$

Figure 4 sketches the steady response of a block with $Q=3$: the amplitude $A(\omega )$ on top and the lag $\delta (\omega )$ below, on the same frequency axis. The resonance panel of the animation plots $A(\omega )$ for the shaken spring, with $Q=5$ and $Q=1$. There the top of the spring is moved by $Y=2\mathrm{cm}$, which acts on the block like a force with $F_0/m={\omega }_0^{2}Y$, so $F_0/k=Y$.

Read the curves from left to right. Far below ${\omega }_0$ the spring term dominates: $A\approx F_0/k$, the static stretch (2 cm in the animation), $\delta \approx 0$, and the block follows the shaker. Far above ${\omega }_0$ inertia dominates: $A\approx F_0/(m{\omega }^2)$ shrinks below the dashed $F_0/k$ level, $\delta$ approaches $\pi$, and the block moves against the drive. At ${\omega }_0$, just right of the peak, spring and inertia cancel and only damping balances the drive: $A=Q{F}_0/k$ and $\delta =\pi /2$. The velocity is then in phase with the force, so the drive does positive work all through the cycle. In the animation the tall $Q=5$ curve reaches 10 cm, while the low grey $Q=1$ curve gives only 2 cm at ${\omega }_0$.

The width of the peak is measured with power. In the steady state the mean power taken from the drive equals the mean power the damper dissipates, $P=\frac{1}{2}b{\omega }^2{A}^2$, which is 0.125 W at resonance for $Q=5$. Writing ${\omega }^2{A}^2=(F_0/m{)}^2/[({\omega }_0^2-{\omega }^2{)}^2/{\omega }^2+4{\beta }^2]$ shows that $P$ peaks exactly at ${\omega }_0$ and falls to half where $|{\omega }_0^2-{\omega }^2|=2\beta \omega$. These are the frequencies ${\omega }_1$ and ${\omega }_2$ of the figure; for $Q=5$ they are 4.52 and 5.52 rad/s (0.72 and 0.88 Hz). There the two legs of the triangle in Figure 3 are equal, so $\tan \delta =\pm 1$: the lag curve passes $\pi /4$ at ${\omega }_1$ and $3\pi /4$ at ${\omega }_2$. The full width is
$\Delta \omega ={\omega }_2-{\omega }_1=2\beta =\frac{{\omega }_0}{Q}$
so $Q$ also measures the sharpness of the peak, ${\omega }_0/\Delta \omega$. The amplitude peak itself sits slightly below ${\omega }_0$, at $\sqrt{{\omega }_0^2-2{\beta }^2}\approx 4.95rad/s$, where it reaches 10.05 cm rather than 10 cm.

The dot on the animation's curve shows the start-up. The shaker starts from its centre and moves as $Y\sin {\omega }_{0}t$, while the block starts from rest. The steady displacement lags the drive by a quarter turn, which makes it $-QY\cos {\omega }_{0}t$, and at first the transient cancels it, so $x\approx -QY(1-e^{-\beta t})\cos {\omega }_{0}t$. The amplitude climbs on the time scale $1/\beta =2\mathrm{s}$ and reaches 90% of the peak only after $\ln 10/\beta \approx 4.6\mathrm{s}$. A narrow peak and a long ring-down are one fact seen twice: the width $2\beta$ times the decay time $1/\beta$ is always 2, so a sharply tuned oscillator is slow to respond.

Worked problem: shaking at the wrong rhythm

Shake the same spring ($F_0/m=0.5m/{\mathrm{s}}^2$, $\beta =0.5{\mathrm{s}}^{-1}$, ${\omega }_0^2=25{\mathrm{s}}^{-2}$, $b=1kg/s$) at 0.50 Hz and at 1.00 Hz, either side of its natural 0.80 Hz. Find the steady amplitude, the lag and the mean power in each case.

Below resonance, at 0.50 Hz: $\omega =2\pi f=\pi \approx 3.14rad/s$, so ${\omega }_0^2-{\omega }^2=25-9.87=15.13{\mathrm{s}}^{-2}$ and $2\beta \omega =3.14{\mathrm{s}}^{-2}$. Then
$A=\frac{0.5}{\sqrt{15.13^2+3.14^2}}=0.032\mathrm{m},\tan \delta =\frac{3.14}{15.13},\delta \approx 12^{\circ }$
Above resonance, at 1.00 Hz: $\omega =2\pi \approx 6.28rad/s$, so ${\omega }_0^2-{\omega }^2=25-39.48=-14.48{\mathrm{s}}^{-2}$ and $2\beta \omega =6.28{\mathrm{s}}^{-2}$. The negative denominator puts $\delta$ between $90^{\circ }$ and $180^{\circ }$:
$A=\frac{0.5}{\sqrt{14.48^2+6.28^2}}=0.032\mathrm{m},\tan \delta =\frac{6.28}{-14.48},\delta \approx 157^{\circ }$

Both amplitudes are about 3.2 cm, a third of the resonant 10 cm. The difference is in the phase: at 0.50 Hz the block rides along with the shaker, while at 1.00 Hz it moves almost against it. The mean power $\frac{1}{2}b{\omega }^2{A}^2$ differs too: about 5 mW at 0.50 Hz and 20 mW at 1.00 Hz, against 125 mW at resonance. The amplitudes are equal but the frequency has doubled, so the velocity amplitude $\omega A$ at 1.00 Hz is twice as large and the power four times as large.

The start-up matters as well. Started from rest, the free ringing at ${\omega }_d=4.97rad/s$ adds to the steady motion and beats against it with the period $2\pi /|\omega -{\omega }_d|$, 3.4 s and 4.8 s here. With the shaker starting from its centre, as in the animation, the early swings therefore reach 4.1 and 4.5 cm, more than the steady 3.2 cm. After $3/\beta =6\mathrm{s}$ the transient is down to $e^{-3}$, about 5%.

Beyond linear damping

Everything so far assumed a damping force proportional to the velocity. Real friction often is not, and the shape of the decay tells which law is at work.

Dry friction is a constant force $\mu N$ against the motion, with $\mu$ the friction coefficient and $N$ the normal force. It keeps the period at $2\pi /{\omega }_0$ but cuts the amplitude by the same $2\mu N/k$ every half-swing. The reason: over a half-swing from $A_1$ to $A_2$, the spring energy lost, $\frac{1}{2}k(A_1^2-A_2^2)$, equals the work $\mu N(A_1+A_2)$ done against friction. The envelope is a straight line, and the block stops dead after a finite time, once the spring force can no longer beat friction. With 0.1 N of friction the animation's block would lose 8 mm per half-swing and stop after 12 of them, about 7.5 s.

Drag growing as $v^2$ kills large swings fast and lets small ones linger, so the amplitude falls roughly as $1/(1+t/t_1)$, with a time $t_1$ set by the starting amplitude. A practical test: plot the logarithm of the amplitude against time. Only linear damping gives a straight line.

The spring can be nonlinear too. A stiffening spring rings faster at large amplitude, so its resonance curve leans towards high frequencies, and as the drive frequency is swept the amplitude can jump between two steady values.

Energy can also enter without a periodic force. Pumping a swing by standing and squatting twice per cycle changes a property of the oscillator, its effective length, instead of pushing it: this is parametric resonance. A bowed string or a clock escapement feeds energy in step with the motion itself. That acts as negative damping, which grows the swing until nonlinearity limits it. The Tacoma Narrows Bridge, destroyed in 1940, failed by such a self-excited flutter in a steady wind, not by simple resonance with a periodic force.

Limits of validity

The linear damper is exact for an electrical resistor, nearly so for an eddy-current brake, and good for slow motion through a viscous liquid (Drag and Terminal Velocity). A real spring is linear only for small stretches (Elasticity: Hooke's Law, Stress and Strain). The resonant amplitude also takes time to build: the drive must last many times $1/\beta =2Q/{\omega }_0$, which is 1.4 s for a tuning fork with $Q\approx 2000$ at 440 Hz.

Damping never comes alone. The molecular collisions that drain the energy also kick the block at random, so an oscillator in equilibrium at temperature $T$ keeps an average energy $k_{B}T$, where $k_B$ is Boltzmann's constant. Half of that sits in the spring, so the mean of $\frac{1}{2}k{x}^2$ is $\frac{1}{2}{k}_{B}T$, and the block jitters by $\sqrt{k_{B}T/k}$ (root mean square). For the 25 N/m spring at 300 K that is $1.3\times 10^{-11}\mathrm{m}$, far below anything visible. For an atomic-force-microscope cantilever with $k=0.1N/m$ it is 0.2 nm, a floor on what the instrument can sense. This link between dissipation and random force is the fluctuation–dissipation theorem. Quantum effects come in steps of $\hslash {\omega }_0\approx 5\times 10^{-34}\mathrm{J}$; against the block's 0.125 J they are utterly negligible, and they matter only for molecules, atoms and very cold micro-resonators.

Where this leads

With the charge $q$ in place of $x$, the same equation describes a series circuit, $L\ddot{q}+R\dot{q}+q/C=V_0\cos \omega t$. The inductance $L$, the resistance $R$ and $1/C$ play the parts of $m$, $b$ and $k$, so ${\omega }_0=1/\sqrt{LC}$ and $Q={\omega }_{0}L/R$. A radio that separates medium-wave stations 10 kHz apart near 1 MHz needs a width ${\omega }_0/Q$ no larger than the spacing, so a $Q$ of order 100; see Capacitors, Electromagnetic Induction and Electromagnetic Waves.

Light drives the electrons bound in atoms in the same way. Below their ultraviolet resonances the response grows with frequency, which is why glass bends violet more than red (Prisms, Dispersion and Colour); at a resonance, energy is absorbed, as by the ions of a salt crystal in the infrared. The same curve, each time with its own ${\omega }_0$ and $Q$, fits atmospheric tides, the modes in Standing Waves and the nuclei in an MRI scanner. In quantum physics, width times decay time returns as the link between the width of a spectral line and the lifetime of the state that emits it.