Chapter 5 · Section 5.5

5.5 Resonance

Push a child on a swing at just the right moments and small pushes build up a large motion. Push at the wrong rhythm and nothing much happens. This is resonance: a system responds strongly to forcing at (or near) its own natural frequency. Engineers exploit it, for example to tune a radio receiver to one station, and fight it, for example to keep a machine running at a speed where its mounts do not shake violently. In this section we use the formulas of Section 5.4 to describe resonance precisely: with and without damping, exactly at and near the natural frequency.

Undamped resonance: growth without bound

Take $c = 0$ and force the system exactly at its natural frequency $ω_{0} = k / m$ :

m x^{''} + k x = F_{0} cos ω_{0} t, i.e. x^{''} + ω_{0}^{2} x = \frac{F _{0}}{m} cos ω_{0} t .

The usual trial $a cos ω_{0} t + b sin ω_{0} t$ fails, because it solves the homogeneous equation: substituting it gives $0$ on the left. The modification rule of Section 4.3 tells us to multiply by $t$ .

Trial. $x_{p} = t u (t)$ with $u = a cos ω_{0} t + b sin ω_{0} t$ . Note that $u^{''} = - ω_{0}^{2} u$ .
Differentiate the product. $x_{p}^{'} = u + t u^{'}$ and $x_{p}^{''} = 2 u^{'} + t u^{''}$ .
Substitute. $x_{p}^{''} + ω_{0}^{2} x_{p} = 2 u^{'} + t (u^{''} + ω_{0}^{2} u) = 2 u^{'} = - 2 a ω_{0} sin ω_{0} t + 2 b ω_{0} cos ω_{0} t .$
Match. $- 2 a ω_{0} = 0$ and $2 b ω_{0} = F_{0} / m$ , so $a = 0$ and $b = \frac{F _{0}}{2 m ω _{0}}$ .

Theorem 5.5.1 (Pure resonance)

The general solution of $m x^{''} + k x = F_{0} cos ω_{0} t$ , $ω_{0} = k / m$ , is

x (t) = c_{1} cos ω_{0} t + c_{2} sin ω_{0} t + \frac{F _{0}}{2 m ω _{0}} t sin ω_{0} t .

Every solution is unbounded: the last term oscillates between the lines $\pm \frac{F _{0}}{2 m ω _{0}} t$ , whose distance grows linearly with time.

Why does this happen physically? From the energy identity of Section 5.1, the force feeds in energy at the rate $F (t) x^{'}$ . In the resonant solution the displacement $t sin ω_{0} t$ lags the force $cos ω_{0} t$ by exactly a quarter cycle, so the velocity, whose main part is $t cos ω_{0} t$ , is in step with the force. The force therefore pushes in the direction of motion almost all the time, every cycle adds energy, and there is no damping to remove it. Of course real springs do not stretch forever: long before the amplitude is "infinite", the spring leaves its linear range, something breaks, or the small damping we ignored limits the growth. The model's message is still the right one: forcing an almost undamped system at its natural frequency is dangerous.

Figure 5.5.1. Pure resonance: the response

x = 0.1 t sin 6 t

of Example 5.5.1 started from rest. The swings grow linearly inside the lines

\pm 0.1 t

Near resonance: beats

Now let the forcing frequency be close to, but not equal to, $ω_{0}$ , still without damping, and start from rest: $x (0) = 0$ , $x^{'} (0) = 0$ . From Section 5.4,

x = c_{1} cos ω_{0} t + c_{2} sin ω_{0} t + \frac{F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} cos ω t .

The initial conditions give $c_{2} = 0$ and $c_{1} = - F_{0} / (m (ω_{0}^{2} - ω^{2}))$ , so

x (t) = \frac{F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} (cos ω t - cos ω_{0} t) .

The trigonometric identity $cos α - cos β = 2 sin \frac{β + α}{2} sin \frac{β - α}{2}$ turns the difference into a product:

x (t) = slowly varying amplitude \frac{2 F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} sin \frac{( ω _{0} - ω ) t}{2} \cdot fast oscillation sin \frac{( ω _{0} + ω ) t}{2} .

When $ω \approx ω_{0}$ , the first factor changes slowly and the second oscillates fast, at about the natural frequency. The result is a fast vibration whose amplitude swells and fades periodically: beats (Figure 5.5.2). The amplitude envelope $sin \frac{( ω _{0} - ω ) t}{2}$ repeats with the beat period

T_{beat} = \frac{2 π}{∣ ω _{0} - ω ∣} .

You hear beats when two guitar strings are almost in tune: the sound throbs, slower and slower as the strings approach the same pitch. Musicians tune by making the beats disappear.

Figure 5.5.2. Beats in Example 5.5.2 (

ω_{0} = 10

ω = 9

): the motion fills the slowly varying envelope

\pm 0.2 sin (t /2)

. The swing repeatedly builds up to

0.2

m and dies away, every

2 π \approx 6.3

Beats become resonance

Let $ω \to ω_{0}$ in the beats formula. For fixed $t$ , $sin \frac{( ω _{0} - ω ) t}{2} \approx \frac{( ω _{0} - ω ) t}{2}$ and $ω_{0}^{2} - ω^{2} = (ω_{0} - ω) (ω_{0} + ω)$ , so

x (t) \to \frac{2 F _{0}}{m \cdot 2 ω _{0}} \cdot \frac{t}{2} \cdot sin ω_{0} t = \frac{F _{0}}{2 m ω _{0}} t sin ω_{0} t,

exactly the resonant solution from rest. The beat period becomes infinite: the first swelling never stops.

Damped resonance

With damping, Theorem 5.4.2 says the steady-state amplitude is

M (ω) = \frac{F _{0}}{( k - m ω ^{2} ) ^{2} + c ^{2} ω ^{2}} .

The denominator never vanishes when $c > 0$ , so the amplitude stays finite. But it can still be large. For which forcing frequency is it largest?

Finding the peak

$M$ is largest where the expression under the square root is smallest. That expression depends on $ω$ only through $u = ω^{2} \geq 0$ :

D (u) = (k - m u)^{2} + c^{2} u .

Differentiate. $D^{'} (u) = - 2 m (k - m u) + c^{2} = 2 m^{2} u - 2 mk + c^{2}$ .
Critical point. $D^{'} (u) = 0$ at $u_{r} = \frac{k}{m} - \frac{c ^{2}}{2 m ^{2}} .$ Since $D^{''} (u) = 2 m^{2} > 0$ , $D$ is a parabola opening upward and $u_{r}$ is its minimum.
Is it allowed? We need $u_{r} > 0$ , i.e. $c^{2} < 2 mk$ . If instead $c^{2} \geq 2 mk$ , then $D^{'} (u) > 0$ for all $u > 0$ , $D$ increases, and $M$ simply decreases as $ω$ grows: there is no peak.
Peak value. At $u_{r}$ we have $k - m u_{r} = c^{2} / (2 m)$ , so $D (u_{r}) = \frac{c ^{4}}{4 m ^{2}} + c^{2} (\frac{k}{m} - \frac{c ^{2}}{2 m ^{2}}) = c^{2} (\frac{k}{m} - \frac{c ^{2}}{4 m ^{2}}) = c^{2} ω_{d}^{2},$ where $ω_{d} = k / m - c^{2} / (4 m^{2})$ is the quasi-frequency of Section 5.3.

Theorem 5.5.2 (Resonance frequency and peak amplitude)

Let $m, k, c > 0$ and $F (t) = F_{0} cos ω t$ . If $c^{2} < 2 mk$ (equivalently $ζ < 1/ 2 \approx 0.707$ ), the steady-state amplitude $M (ω)$ is largest at the resonance frequency

ω_{r} = \frac{k}{m} - \frac{c ^{2}}{2 m ^{2}} = ω_{0} 1 - 2 ζ^{2},

and the largest amplitude is

M_{m a x} = \frac{F _{0}}{c ω _{d}} = \frac{F _{0}}{c \frac{k}{m} - \frac{c ^{2}}{4 m ^{2}}} .

If $c^{2} \geq 2 mk$ , $M (ω)$ decreases for all $ω > 0$ and never exceeds $F_{0} / k$ .

Three frequencies now carry similar names; keep them apart:

ω_{r} = ω_{0}^{2} - 2 β^{2} < ω_{d} = ω_{0}^{2} - β^{2} < ω_{0} = k / m, β = \frac{c}{2 m} .

$ω_{0}$ is the undamped natural frequency, $ω_{d}$ the frequency of free damped oscillation, and $ω_{r}$ the forcing frequency that produces the largest steady-state response. For light damping all three are very close, and engineers often just say "resonance occurs at the natural frequency".

In dimensionless form (divide by the static deflection $F_{0} / k$ ) the peak is

\frac{M _{m a x}}{F _{0} / k} = \frac{1}{2 ζ 1 - ζ ^{2}} \approx \frac{1}{2 ζ} for small ζ .

With $ζ = 0.02$ , a value typical of lightly damped structures, the resonant response is about 25 times the static one.

Amplitude and phase response curves

A plot of $M / (F_{0} / k)$ against $ω / ω_{0}$ is called an amplitude response curve (or frequency response curve). Figure 5.5.3 shows it for several damping ratios. Note how the peak grows and sharpens as $ζ$ decreases, how it moves slightly to the left of $ω / ω_{0} = 1$ , and how it disappears for $ζ \geq 1/ 2$ . Far above resonance all curves fall off together, whatever the damping.

Figure 5.5.3. Amplitude response curves. The dotted curve joins the peaks, located at

ω / ω_{0} = 1 - 2 ζ^{2}

with height

1/ (2 ζ 1 - ζ^{2})

. For

ζ \geq 1/ 2

there is no peak.

The phase lag also changes across resonance. From Theorem 5.4.2, $tan θ = c ω / (k - m ω^{2})$ with $0 < θ < π$ . At $ω = ω_{0}$ the denominator is zero, so $θ = π /2$ exactly, for every damping. For light damping the lag jumps quickly from nearly $0$ to nearly $π$ as $ω$ passes $ω_{0}$ (Figure 5.5.4). Measuring where the lag equals $π /2$ is in fact a common laboratory way to locate $ω_{0}$ .

Figure 5.5.4. Phase response curves for the same damping ratios. All pass through

θ = π /2

ω = ω_{0}

Fighting resonance in practice

Theorem 5.5.2 lists the three remedies an engineer has: move the natural frequency away from the operating frequency (change $k$ or $m$ ), add damping (increase $c$ , which lowers $M_{m a x}$ ), or reduce the force $F_{0}$ (for instance by balancing a rotor). Machines that must pass through resonance on start-up are run up quickly, so that large amplitudes have no time to build up.

Method in a nutshell: resonance questions

Compute $ω_{0} = k / m$ and, if $c > 0$ , $ζ = c / (2 mk)$ .
Undamped, $ω = ω_{0}$ : $x_{p} = \frac{F _{0}}{2 m ω _{0}} t sin ω_{0} t$ (for $F_{0} cos ω_{0} t$ ); for $F_{0} sin ω_{0} t$ use $x_{p} = - \frac{F _{0}}{2 m ω _{0}} t cos ω_{0} t$ .
Undamped, $ω \approx ω_{0}$ , from rest: $x = \frac{F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} (cos ω t - cos ω_{0} t)$ ; beat period $2 π /∣ ω_{0} - ω ∣$ , largest swing $\frac{2 F _{0}}{m ∣ ω _{0}^{2} - ω ^{2} ∣}$ .
Damped: check $c^{2} < 2 mk$ . If yes, $ω_{r} = k / m - c^{2} / (2 m^{2})$ and $M_{m a x} = F_{0} / (c ω_{d})$ ; if not, no peak and $M \leq F_{0} / k$ .

Explore the amplitude response with the widget below. It starts with the system of Example 5.5.3; it also draws the curves for half and twice your damping constant. Change $c$ and watch the peak rise, sharpen and shift toward $ω_{0}$ as $c$ decreases, and vanish once $c^{2} \geq 2 mk$ .

The spring simulator shows the time response. The preset is the undamped resonance of Example 5.5.1; then try $ω = 5.5$ to see beats, or add a little damping to see the growth level off.

Worked examples

Example 5.5.1 (Pure resonance)

A 0.25 kg mass hangs from a spring with $k = 9$ N/m; damping is negligible. It is pulled 2 cm down and released from rest while a force $0.3 cos 6 t$ N acts. Find $x (t)$ . The spring stays linear for displacements up to 0.3 m; roughly when does the motion exceed this?

Solution

Resonance? $ω_{0} = 9/0.25 = 6 = ω$ . Yes.
Equation. $x^{''} + 36 x = 1.2 cos 6 t$ , $x (0) = 0.02$ , $x^{'} (0) = 0$ .
Particular solution. By Theorem 5.5.1 with $F_{0} / m = 1.2$ : $x_{p} = \frac{1.2}{2 \cdot 6} t sin 6 t = 0.1 t sin 6 t$ .
Initial conditions. $x = c_{1} cos 6 t + c_{2} sin 6 t + 0.1 t sin 6 t$ . Then $x (0) = c_{1} = 0.02$ . Also $x_{p}^{'} (t) = 0.1 sin 6 t + 0.6 t cos 6 t$ , so $x_{p}^{'} (0) = 0$ and $x^{'} (0) = 6 c_{2} = 0$ , giving $c_{2} = 0$ .
Solution. $x (t) = 0.02 cos 6 t + 0.1 t sin 6 t m .$
Leaving the linear range. After the first second the growing term dominates; the swings are about $0.1 t$ m in size. They reach $0.3$ m at about $t \approx 3$ s, i.e. after fewer than three oscillations (the period is $2 π /6 \approx 1.05$ s). Beyond that the model is no longer valid.

Example 5.5.2 (Beats)

A 1 kg mass on a spring with $k = 100$ N/m (no damping) starts at rest at equilibrium. A force $1.9 cos 9 t$ N acts. Find $x (t)$ , write it as a product, and give the beat period and the largest displacement.

Solution

Frequencies. $ω_{0} = 10$ , $ω = 9$ : close, but different.
Solution. By the beats formula with $F_{0} / m = 1.9$ and $ω_{0}^{2} - ω^{2} = 100 - 81 = 19$ : $x (t) = \frac{1.9}{19} (cos 9 t - cos 10 t) = 0.1 (cos 9 t - cos 10 t) .$
Product form. With $\frac{ω _{0} - ω}{2} = \frac{1}{2}$ and $\frac{ω _{0} + ω}{2} = \frac{19}{2}$ : $x (t) = 0.2 sin \frac{t}{2} sin \frac{19 t}{2} .$
Beat period and largest swing. $T_{beat} = 2 π /∣10 - 9∣ = 2 π \approx 6.28$ s. The envelope reaches $0.2$ m, twice the amplitude $0.1$ m of the forced term alone: the free and forced oscillations alternately add up and cancel (Figure 5.5.2).

Example 5.5.3 (Damped resonance)

For $x^{''} + 2 x^{'} + 27 x = 4 cos ω t$ ( $m = 1$ kg, $c = 2$ N·s/m, $k = 27$ N/m, $F_{0} = 4$ N), find the resonance frequency, the largest steady-state amplitude and the steady-state solution at resonance. Compare with the amplitude at $ω = ω_{0}$ and with the static deflection.

Solution

Is there a peak? $c^{2} = 4 < 2 mk = 54$ . Yes. ( $ζ = 2/ (227) \approx 0.19$ .)
Resonance frequency. $ω_{r} = 27 - \frac{4}{2} = 25 = 5 rad/s .$ For comparison, $ω_{0} = 27 \approx 5.196$ and $ω_{d} = 27 - 1 = 26 \approx 5.099$ .
Peak amplitude. $M_{m a x} = \frac{F _{0}}{c ω _{d}} = \frac{4}{2 26} \approx 0.392 m .$ Check directly: at $ω = 5$ , $(27 - 25)^{2} + (2 \cdot 5)^{2} = 4 + 100 = 104$ and $4/ 104 = 4/ (226)$ . ✓
Steady state at resonance. With $k - m ω^{2} = 2$ and $c ω = 10$ , Theorem 5.4.2 gives $x_{ss} = \frac{4}{104} (2 cos 5 t + 10 sin 5 t) = \frac{1}{13} (cos 5 t + 5 sin 5 t),$ with lag $θ = arctan 5 \approx 1.37$ rad, slightly less than $π /2$ (since $ω_{r} < ω_{0}$ ).
Comparisons. At $ω = ω_{0}$ : $M = 4/ 4 \cdot 27 \approx 0.385$ m, only slightly less than the peak. Static deflection: $F_{0} / k = 4/27 \approx 0.148$ m, so the peak magnification is $0.392/0.148 \approx 2.65$ , matching $1/ (2 ζ 1 - ζ^{2})$ .
A few more values.

$ω$ (rad/s) 0 3 4 5 6 8

$M$ (m) 0.148 0.211 0.294 0.392 0.267 0.099

The amplitude rises to the peak at $ω = 5$ and falls quickly beyond it. Try these numbers in the widget above.

$ω$ (rad/s)	0	3	4	5	6	8
$M$ (m)	0.148	0.211	0.294	0.392	0.267	0.099

Example 5.5.4 (How much damping is enough?)

A 50 kg machine sits on mounts of total stiffness $k = 20000$ N/m. Its rotating parts produce a vertical force of amplitude 100 N at a frequency that sweeps through all values during start-up. (We assume, for simplicity, that the force amplitude does not depend on the frequency.) Find the smallest damping constant for which the steady-state amplitude never exceeds 1 cm.

Solution

Data. $ω_{0}^{2} = k / m = 400$ , so $ω_{0} = 20$ rad/s. Static deflection $F_{0} / k = 0.005$ m $= 0.5$ cm, already within the limit. The danger is the resonance peak.
Case $c^{2} \geq 2 mk = 2 \times 1 0^{6}$ . No peak; $M \leq F_{0} / k = 0.5$ cm. Acceptable.
Case $c^{2} < 2 \times 1 0^{6}$ . The worst amplitude is $M_{m a x} = F_{0} / (c ω_{d})$ with $ω_{d}^{2} = 400 - c^{2} / (4 \cdot 5 0^{2}) = 400 - c^{2} /1 0^{4}$ . We need $\frac{100}{c 400 - c ^{2} /1 0 ^{4}} \leq 0.01 ⟺ c^{2} (400 - \frac{c ^{2}}{1 0 ^{4}}) \geq 1 0^{8} .$
Solve for $u = c^{2}$ . Multiply by $1 0^{4}$ and rearrange: $u^{2} - 4 \times 1 0^{6} u + 1 0^{12} \leq 0$ . The roots are $u = \frac{4 \times 1 0 ^{6} \pm 16 \times 1 0 ^{12} - 4 \times 1 0 ^{12}}{2} = (2 \pm 3) \times 1 0^{6},$ and the inequality holds between them. Together with $u < 2 \times 1 0^{6}$ , this case allows $(2 - 3) \times 1 0^{6} \leq c^{2} < 2 \times 1 0^{6}$ .
Combine both cases. The requirement holds exactly when $c^{2} \geq (2 - 3) \times 1 0^{6}$ : $c \geq 1000 2 - 3 \approx 518 N \cdotp s/m .$
Check. For $c = 518$ : $ζ = 518/ (2 50 \cdot 20000) \approx 0.26$ , $ω_{r} = 400 - c^{2} /5000 \approx 18.6$ rad/s, and $M_{m a x} = 100/ (518 400 - 26.8) \approx 0.0100$ m. ✓

Exercises

Solve $x^{''} + 81 x = 0.9 cos 9 t$ , $x (0) = 0$ , $x^{'} (0) = 0$ .
Answer
$x = 0.05 t sin 9 t$
A 2 kg mass on a spring with $k = 32$ N/m, no damping, is pulled 10 cm down and released from rest while a force $1.6 sin 4 t$ N acts. Find $x (t)$ .
Answer
$x = 0.1 cos 4 t + 0.025 sin 4 t - 0.1 t cos 4 t$
Solve $x^{''} + 36 x = 1.1 cos 5 t$ from rest at equilibrium. Write the answer as a product and give the beat period and largest displacement.
Answer
$x = 0.1 (cos 5 t - cos 6 t) = 0.2 sin \frac{t}{2} sin \frac{11 t}{2}$ ; $T_{beat} = 2 π$ ; largest displacement $0.2$
For $m = 1$ , $c = 2$ , $k = 17$ , find the resonance frequency and the peak steady-state amplitude in terms of $F_{0}$ .
Answer
$ω_{r} = 15 \approx 3.87$ rad/s; $ω_{d} = 4$ , $M_{m a x} = F_{0} /8$
Show that the system $x^{''} + 3 x^{'} + 4 x = F_{0} cos ω t$ has no resonance peak. What is the largest steady-state amplitude?
Answer
$c^{2} = 9 \geq 2 mk = 8$ ; $D = (4 - ω^{2})^{2} + 9 ω^{2}$ has $d D / d ω = 4 ω^{3} + 2 ω > 0$ , so $M$ decreases; it approaches $F_{0} /4$ as $ω \to 0$
A system has $m = 0.5$ kg and $k = 10$ N/m. Find the damping constant for which the resonance frequency is 4 rad/s.
Answer
$20 - c^{2} /0.5 = 16$ , so $c = 2 \approx 1.41$ N·s/m
A structure has damping ratio $ζ = 0.05$ . By what factor does its resonant steady-state amplitude exceed the static deflection?
Answer
$1/ (2 ζ 1 - ζ^{2}) \approx 10.01$
Show that for every $c > 0$ the steady-state response at $ω = ω_{0}$ lags the force by exactly $π /2$ , and that its amplitude is $F_{0} / (c ω_{0})$ .
Answer
At $ω = ω_{0}$ , $k - m ω^{2} = 0$ , so $cos θ = 0$ , $sin θ = 1$ , and $M = F_{0} / c^{2} ω_{0}^{2}$ .
A washing-machine drum (40 kg) sits on mounts with $k = 16000$ N/m and $c = 160$ N·s/m and is shaken by a force of amplitude 80 N. Find $ζ$ , the resonance frequency and the largest steady-state amplitude.
Answer
$ζ = 0.1$ ; $ω_{r} = 392 \approx 19.8$ rad/s; $M_{m a x} = 80/ (160396) \approx 0.0251$ m
Starting from $x = \frac{F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} (cos ω t - cos ω_{0} t)$ , compute $lim_{ω \to ω_{0}} x (t)$ with l'Hôpital's rule (differentiate numerator and denominator with respect to $ω$ ).
Answer
$\frac{F _{0}}{m} \cdot \frac{- t sin ω _{0} t}{- 2 ω _{0}} = \frac{F _{0}}{2 m ω _{0}} t sin ω_{0} t$
An undamped 1 kg system with $ω_{0} = 5$ rad/s is driven from rest by $F_{0} cos 5 t$ . After 10 s the swings are 0.5 m in size. Find $F_{0}$ .
Answer
$\frac{F _{0}}{2 \cdot 5} \cdot 10 = 0.5$ , so $F_{0} = 0.5$ N