Chapter 5 · Section 5.3

5.3 Free, Damped Motion

A plucked guitar string, a car going over a bump, a door closing on its hydraulic closer: all of them stop moving eventually. Energy is lost to friction, and in Section 5.1 we modelled that loss with a damping force $- c x^{'}$ . In this section we keep the damping but still allow no external force. We will see that damping does more than make the motion die out. Depending on how strong it is, it changes the character of the motion completely: the mass may oscillate with shrinking swings, or creep back to equilibrium without ever oscillating.

The equation in standard form

With $c > 0$ and $F (t) = 0$ , the equation of motion is

m x^{''} + c x^{'} + k x = 0.

Dividing by $m$ and introducing two constants with clear physical meaning,

β = \frac{c}{2 m} (1/s), ω_{0} = \frac{k}{m} (rad/s),

we get the standard form

x^{''} + 2 β x^{'} + ω_{0}^{2} x = 0.

(The factor $2$ is chosen so that the roots come out clean.) The characteristic equation is $r^{2} + 2 β r + ω_{0}^{2} = 0$ , and the quadratic formula gives

r = \frac{- 2 β \pm 4 β ^{2} - 4 ω _{0}^{2}}{2} = - β \pm β^{2} - ω_{0}^{2} .

Everything depends on the sign of $β^{2} - ω_{0}^{2}$ . Multiplying by $4 m^{2}$ , this is the sign of $c^{2} - 4 mk$ , the discriminant of $m r^{2} + cr + k$ .

Definition 5.3.1 (Critical damping and damping ratio)

The critical damping constant of a mass–spring system is

c_{cr} = 2 mk,

and the damping ratio is the dimensionless number

ζ = \frac{c}{c _{cr}} = \frac{c}{2 mk} = \frac{β}{ω _{0}} .

The motion is called

overdamped if $c > c_{cr}$ ( $ζ > 1$ ): two distinct real roots;
critically damped if $c = c_{cr}$ ( $ζ = 1$ ): one double real root;
underdamped if $0 < c < c_{cr}$ ( $0 < ζ < 1$ ): a pair of complex roots.

The damping ratio is how engineers talk about damping. "This structure has 2% damping" means $ζ = 0.02$ . It lets you compare a tiny sensor with a large building on the same scale.

Figure 5.3.1. The same mass (

m = 1

kg), spring (

k = 25

N/m) and start (

x_{0} = 1

v_{0} = 0

) with three dampers. Here

c_{cr} = 225 = 10

N·s/m. Too little damping gives oscillation; too much gives a slow crawl; the critically damped curve settles fastest without overshooting.

Case 1: overdamped motion ( $c > c_{cr}$ )

Now $β > ω_{0}$ , and the roots

r_{1} = - β + β^{2} - ω_{0}^{2}, r_{2} = - β - β^{2} - ω_{0}^{2}

are real and distinct. Both are negative: $r_{2}$ obviously, and $r_{1}$ because $β^{2} - ω_{0}^{2} < β^{2} = β$ . The general solution is

x (t) = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t}, r_{2} < r_{1} < 0.

Both exponentials decay, so $x (t) \to 0$ : the mass returns to equilibrium. It cannot oscillate. In fact it passes through equilibrium at most once. To see why, suppose $x (t) = 0$ with $c_{1}, c_{2}$ not both zero. Then $c_{1} e^{r_{1} t} = - c_{2} e^{r_{2} t}$ ; if one of the constants is zero, so is the other, so both are nonzero and $e^{(r_{1} - r_{2}) t} = - c_{2} / c_{1}$ . Since $r_{1} - r_{2} > 0$ , the left side is strictly increasing in $t$ , so this equation has at most one solution.

For large $t$ the term with the more negative exponent $r_{2}$ dies first, and the motion looks like $c_{1} e^{r_{1} t}$ . The "slow" root $r_{1}$ is close to zero when the damping is very heavy: imagine a spring in thick honey. Heavy damping therefore makes the return slower, not faster.

Case 2: critically damped motion ( $c = c_{cr}$ )

Now $β = ω_{0}$ and $r = - β$ is a double root. By Theorem 4.2.2,

x (t) = (c_{1} + c_{2} t) e^{- βt} .

Again $x (t) \to 0$ , because the exponential beats the linear factor: $t e^{- βt} \to 0$ as $t \to \infty$ . The factor $c_{1} + c_{2} t$ is zero for at most one value of $t$ , so again the mass crosses equilibrium at most once. With $x (0) = x_{0}$ and $x^{'} (0) = v_{0}$ we get $c_{1} = x_{0}$ and $c_{2} = v_{0} + β x_{0}$ , so

x (t) = (x_{0} + (v_{0} + β x_{0}) t) e^{- βt} .

Three things can happen, depending on the start (Figure 5.3.2): the mass returns monotonically; or it first moves away, reaches one extreme position, and then returns; or it shoots through equilibrium once, reaches one extreme on the other side, and creeps back. For $x_{0} > 0$ , the crossing happens exactly when $v_{0} + β x_{0} < 0$ , i.e. when the mass is thrown toward equilibrium fast enough.

Figure 5.3.2. Critically damped motion

x = (1 + (v_{0} + 2) t) e^{- 2 t}

for three initial velocities. Overdamped motion shows the same three shapes. In no case is there more than one crossing.

Critical damping is the borderline: the smallest damping for which no oscillation occurs. Any decrease of $c$ , however small, produces oscillations. It also gives the fastest guaranteed decay: in every case the slowest part of the solution decays like $e^{- σ t}$ , where $σ$ is the smallest of $- Re r$ over the two roots, and $σ$ is largest, equal to $ω_{0}$ , exactly when $ζ = 1$ (see Figure 5.3.4). That is why door closers and many vehicle and instrument dampers are designed near critical damping. Measuring instruments are often tuned slightly below it, accepting a tiny overshoot for a quicker response.

Case 3: underdamped motion ( $0 < c < c_{cr}$ )

Now $β < ω_{0}$ and $β^{2} - ω_{0}^{2} < 0$ , so the roots are complex:

r = - β \pm i ω_{d}, ω_{d} = ω_{0}^{2} - β^{2} = ω_{0} 1 - ζ^{2} .

By Theorem 4.2.3, and then by exactly the amplitude–phase step of Section 5.2,

x (t) = e^{- βt} (c_{1} cos ω_{d} t + c_{2} sin ω_{d} t) = A e^{- βt} cos (ω_{d} t - φ),

with $A = c_{1}^{2} + c_{2}^{2}$ , $cos φ = c_{1} / A$ , $sin φ = c_{2} / A$ . The motion is a cosine oscillation whose amplitude $A e^{- βt}$ shrinks exponentially. The graph of $x$ lies between the two curves $\pm A e^{- βt}$ , called the envelope, and touches them once per half-cycle.

Definition 5.3.2 (Quasi-frequency, quasi-period, logarithmic decrement)

For underdamped motion $x = A e^{- βt} cos (ω_{d} t - φ)$ :

$ω_{d} = ω_{0}^{2} - β^{2}$ is the quasi-frequency (damped natural frequency), in rad/s;
$T_{d} = 2 π / ω_{d}$ is the quasi-period;
$δ = β T_{d} = \frac{2 π β}{ω _{d}} = \frac{2 π ζ}{1 - ζ ^{2}}$ is the logarithmic decrement.

The word "quasi" is there because the motion is not periodic: it never repeats exactly, since each swing is smaller than the one before. But the zeros of $x$ are equally spaced, $T_{d} /2$ apart, and so are the maxima, $T_{d}$ apart. Note that $ω_{d} < ω_{0}$ : damping slows the oscillation down, so $T_{d}$ is longer than the undamped period $2 π / ω_{0}$ . For light damping the difference is tiny; for $ζ = 0.1$ , $ω_{d} = ω_{0} 0.99 \approx 0.995 ω_{0}$ .

Why the logarithmic decrement is so useful

Shift time by one quasi-period. The cosine factor does not change, because $cos (ω_{d} (t + T_{d}) - φ) = cos (ω_{d} t - φ + 2 π)$ , but the exponential picks up a factor:

x (t + T_{d}) = A e^{- β (t + T_{d})} cos (ω_{d} t - φ) = e^{- β T_{d}} x (t) .

The same holds for $x^{'}$ , so if $t_{1}$ is a maximum, so are $t_{1} + T_{d}$ , $t_{1} + 2 T_{d}$ , and so on, and the values there form a geometric sequence. For any two successive maxima $x_{1}$ and $x_{2}$ :

\frac{x _{2}}{x _{1}} = e^{- β T_{d}}, ln \frac{x _{1}}{x _{2}} = β T_{d} = δ .

So each oscillation reduces the amplitude by the same factor $e^{- δ}$ . This gives a practical way to measure damping: record a free vibration, read off two successive peaks and the time between them, and compute $δ$ , then $ζ$ from

δ = \frac{2 π ζ}{1 - ζ ^{2}} ⟺ ζ = \frac{δ}{4 π ^{2} + δ ^{2}} .

(To get the second form, square the first: $δ^{2} (1 - ζ^{2}) = 4 π^{2} ζ^{2}$ , then solve for $ζ^{2}$ .) The maxima of $x$ are not exactly where $x$ touches the envelope; they come slightly earlier. That does not matter: the ratio of successive maxima is exactly $e^{- δ}$ either way.

Figure 5.3.3. Underdamped motion

x = e^{- 0.4 t} (cos 4 t + 0.1 sin 4 t)

, the solution of

x^{''} + 0.8 x^{'} + 16.16 x = 0

x (0) = 1

x^{'} (0) = 0

. The swings fit inside the envelope

\pm A e^{- 0.4 t}

(

A = 1.01

). Successive maxima are

T_{d} = 2 π /4

apart, and each is

e^{- 0.4 T_{d}} \approx 0.53

times the previous one.

One picture for all three cases

Keep $m$ and $k$ fixed and increase $c$ from $0$ . The roots of $m r^{2} + cr + k = 0$ move as in Figure 5.3.4. At $c = 0$ they sit at $\pm i ω_{0}$ (undamped, Section 5.2). As $c$ grows they move left along a circle of radius $ω_{0}$ (the product of the roots is always $k / m = ω_{0}^{2}$ , so $∣ r ∣ = ω_{0}$ while they are complex). They meet at $- ω_{0}$ when $c = c_{cr}$ . After that they split along the real axis: one heads toward $- \infty$ , the other creeps back toward $0$ , which is why very heavy damping is slow.

Figure 5.3.4. Roots of

r^{2} + c r + 25 = 0

c

grows. Complex roots (underdamped) lie on the circle

∣ r ∣ = 5

; they collide at

r = - 5

for

c = 10

(critical) and then separate along the real axis (overdamped). The real part of the slower root measures how quickly the motion dies out.

Method in a nutshell: free, damped motion

Compute $c_{cr} = 2 mk$ (or compare $c^{2}$ with $4 mk$ ) and classify.
Solve $m r^{2} + cr + k = 0$ , i.e. $r = - β \pm β^{2} - ω_{0}^{2}$ with $β = c / (2 m)$ , $ω_{0}^{2} = k / m$ .
Write the general solution:
- overdamped: $x = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t}$ ;
- critical: $x = (c_{1} + c_{2} t) e^{- βt}$ ;
- underdamped: $x = e^{- βt} (c_{1} cos ω_{d} t + c_{2} sin ω_{d} t)$ , $ω_{d} = ω_{0}^{2} - β^{2}$ .
Fit $x (0) = x_{0}$ , $x^{'} (0) = v_{0}$ (differentiate with the product rule).
Underdamped: write $x = A e^{- βt} cos (ω_{d} t - φ)$ and report $T_{d} = 2 π / ω_{d}$ and $δ = β T_{d}$ . Over/critical: find the crossing (if any) and the extreme value by setting $x = 0$ and $x^{'} = 0$ .

Try the three cases in the simulator. The three presets below use the same mass and spring as Figure 5.3.1 ( $m = 1$ , $k = 25$ , so $c_{cr} = 10$ ); change $c$ and watch the classification switch as you cross $10$ .

Worked examples

Example 5.3.1 (Underdamped)

A 2 kg mass hangs from a spring with $k = 58$ N/m and is attached to a damper with $c = 8$ N·s/m. It is pulled 20 cm below equilibrium and released from rest. Find $x (t)$ , the quasi-period and the logarithmic decrement. By what factor does the amplitude drop during one oscillation?

Solution

Classify. $c_{cr} = 2 2 \cdot 58 = 2116 \approx 21.5$ N·s/m. Since $c = 8 < 21.5$ , the motion is underdamped, with $ζ = 8/21.5 \approx 0.37$ .
Standard form. Divide by $2$ : $x^{''} + 4 x^{'} + 29 x = 0$ . So $β = 2$ and $ω_{0}^{2} = 29$ .
Roots. $r = - 2 \pm 4 - 29 = - 2 \pm 5 i$ . So $ω_{d} = 5$ rad/s and $x = e^{- 2 t} (c_{1} cos 5 t + c_{2} sin 5 t) .$
Initial conditions. $x (0) = c_{1} = 0.2$ . By the product rule, $x^{'} = e^{- 2 t} [(- 2 c_{1} + 5 c_{2}) cos 5 t + (- 2 c_{2} - 5 c_{1}) sin 5 t],$ so $x^{'} (0) = - 2 c_{1} + 5 c_{2} = 0$ and $c_{2} = 0.4/5 = 0.08$ .
Solution. $x (t) = e^{- 2 t} (0.2 cos 5 t + 0.08 sin 5 t)$ . In amplitude–phase form, $A = 0.04 + 0.0064 \approx 0.215$ and $φ = arctan (0.08/0.2) \approx 0.381$ (first quadrant): $x (t) \approx 0.215 e^{- 2 t} cos (5 t - 0.381) m .$
Quasi-period and decrement. $T_{d} = 2 π /5 \approx 1.257$ s, $δ = β T_{d} = 2 \times 1.257 \approx 2.51$ .
Amplitude drop. Each oscillation multiplies the amplitude by $e^{- δ} \approx 0.081$ : after one cycle only about 8% of the swing is left. This system is quite strongly damped, even though it still oscillates.

Example 5.3.2 (Critically damped, with overshoot)

The mass and spring of Example 5.1.1 ( $m = 0.5$ kg, $k = 50$ N/m) are fitted with a damper of critical strength. The mass is pulled 4 cm down and thrown upward at 1 m/s. Does it pass through equilibrium? If so, when, and how far above equilibrium does it go?

Solution

Critical damping. $c = c_{cr} = 2 0.5 \times 50 = 10$ N·s/m. Standard form: $x^{''} + 20 x^{'} + 100 x = 0$ , $β = 10$ .
Solution. The double root is $r = - 10$ . Using $x = (x_{0} + (v_{0} + β x_{0}) t) e^{- βt}$ with $x_{0} = 0.04$ , $v_{0} = - 1$ : $x (t) = (0.04 + (- 1 + 0.4) t) e^{- 10 t} = (0.04 - 0.6 t) e^{- 10 t} .$
Crossing. $x = 0$ when $0.04 - 0.6 t = 0$ , i.e. at $t = 1/15 \approx 0.067$ s. (As predicted: $v_{0} + β x_{0} = - 0.6 < 0$ .)
Extreme position. Differentiate: $x^{'} = - 0.6 e^{- 10 t} - 10 (0.04 - 0.6 t) e^{- 10 t} = (- 1 + 6 t) e^{- 10 t},$ which is zero at $t = 1/6$ s. There $x (\frac{1}{6}) = (0.04 - 0.1) e^{- 5/3} \approx - 0.06 \times 0.189 \approx - 0.0113 m .$
Answer. The mass passes through equilibrium at $t \approx 0.067$ s, rises to about $1.13$ cm above equilibrium at $t \approx 0.167$ s, and then returns to equilibrium without crossing again. Critical damping prevents oscillation, not overshoot.

Example 5.3.3 (Overdamped)

A 1 kg mass on a spring with $k = 12$ N/m moves in a heavy oil with $c = 8$ N·s/m. It is pulled 10 cm down and released from rest. Find $x (t)$ and estimate how long it takes to get within 1 cm of equilibrium.

Solution

Classify. $c_{cr} = 212 \approx 6.93 < 8$ : overdamped.
Roots. $r^{2} + 8 r + 12 = (r + 2) (r + 6) = 0$ , so $r_{1} = - 2$ , $r_{2} = - 6$ , and $x = c_{1} e^{- 2 t} + c_{2} e^{- 6 t}$ .
Initial conditions. $c_{1} + c_{2} = 0.1$ and $- 2 c_{1} - 6 c_{2} = 0$ . The second gives $c_{1} = - 3 c_{2}$ ; then $- 2 c_{2} = 0.1$ , so $c_{2} = - 0.05$ and $c_{1} = 0.15$ : $x (t) = 0.15 e^{- 2 t} - 0.05 e^{- 6 t} .$
No crossing. $x = 0$ would need $e^{4 t} = 1/3$ , i.e. $t < 0$ . So for $t > 0$ the mass stays below equilibrium and approaches it monotonically.
Settling. For $t > 1$ the second term is below $0.05 e^{- 6} \approx 0.0001$ m, negligible. So $x \approx 0.15 e^{- 2 t} = 0.01$ gives $t \approx \frac{1}{2} ln 15 \approx 1.35 s .$ Compare: with critical damping ( $c = 212 \approx 6.93$ ) the same start, $x = (0.1 + 0.1 12 t) e^{- 12 t}$ , gets within 1 cm after about $1.12$ s (try it in the simulator).

Example 5.3.4 (Designing a damper)

A 4 kg instrument is mounted on a spring with $k = 900$ N/m. (a) What damping constant makes the mounting critically damped? (b) What damping constant makes each free oscillation exactly half as large as the previous one?

Solution

(a) $c_{cr} = 2 4 \times 900 = 2 \times 60 = 120$ N·s/m.
(b) Translate the requirement. Successive maxima in ratio $\frac{1}{2}$ means $e^{- δ} = \frac{1}{2}$ , i.e. $δ = ln 2 \approx 0.693$ .
Damping ratio. $ζ = \frac{δ}{4 π ^{2} + δ ^{2}} = \frac{0.693}{39.48 + 0.480} \approx \frac{0.693}{6.321} \approx 0.110.$
Damping constant. $c = ζ c_{cr} \approx 0.110 \times 120 \approx 13.2$ N·s/m.
Check. $β = c / (2 m) \approx 1.645$ , $ω_{d} = 225 - 2.71 \approx 14.91$ , $T_{d} \approx 0.421$ s, and $e^{- β T_{d}} \approx e^{- 0.693} = 0.5$ .

Example 5.3.5 (Identifying a system from a measurement)

One corner of a car (effective mass 300 kg) is pushed down and released. A sensor records successive maximum displacements of 8.0 mm and 5.0 mm, 0.4 s apart. Estimate the suspension's spring constant, damping constant and damping ratio.

Solution

Logarithmic decrement. $δ = ln (8.0/5.0) = ln 1.6 \approx 0.470$ .
Decay rate and quasi-frequency. The maxima are one quasi-period apart, so $T_{d} = 0.4$ s. Then $β = \frac{δ}{T _{d}} \approx 1.175 s^{- 1}, ω_{d} = \frac{2 π}{T _{d}} \approx 15.71 rad/s .$
Undamped natural frequency. From $ω_{d}^{2} = ω_{0}^{2} - β^{2}$ : $ω_{0}^{2} = 246.7 + 1.38 \approx 248.1 s^{- 2}$ .
Spring and damper. $k = m ω_{0}^{2} \approx 300 \times 248.1 \approx 7.44 \times 1 0^{4} N/m, c = 2 m β \approx 705 N \cdotp s/m .$
Damping ratio. $ζ = β / ω_{0} \approx 1.175/15.75 \approx 0.075$ . Such light damping lets the car bounce several times after a bump; it is a typical symptom of worn shock absorbers.

Exercises

In Exercises 1–3 classify the motion of $m x^{''} + c x^{'} + k x = 0$ and give the general solution.

$m = 1$ , $c = 9$ , $k = 14$ .
Answer
overdamped; $x = c_{1} e^{- 2 t} + c_{2} e^{- 7 t}$
$m = 9$ , $c = 12$ , $k = 4$ .
Answer
critically damped ( $c^{2} = 144 = 4 mk$ ); $x = (c_{1} + c_{2} t) e^{- 2 t /3}$
$m = 0.5$ , $c = 1$ , $k = 13$ .
Answer
underdamped; $x = e^{- t} (c_{1} cos 5 t + c_{2} sin 5 t)$
Solve the system of Exercise 3 with $x (0) = 0.1$ , $x^{'} (0) = 0$ . Give the amplitude–phase form, the quasi-period and the logarithmic decrement.
Answer
$x = e^{- t} (0.1 cos 5 t + 0.02 sin 5 t) \approx 0.102 e^{- t} cos (5 t - 0.197)$ ; $T_{d} = 2 π /5 \approx 1.257$ s; $δ = 2 π /5 \approx 1.257$
Find the critical damping constant for a 3 kg mass on a spring with $k = 27$ N/m.
Answer
$c_{cr} = 18$ N·s/m
A 1 kg mass with $k = 9$ N/m and $c = 10$ N·s/m is struck at equilibrium, receiving a downward velocity of 0.4 m/s. Find $x (t)$ , the time of the largest displacement and its value.
Answer
$x = 0.05 (e^{- t} - e^{- 9 t})$ ; maximum at $t = \frac{1}{8} ln 9 \approx 0.275$ s, $x_{m a x} \approx 0.0338$ m
Solve $x^{''} + 12 x^{'} + 36 x = 0$ , $x (0) = 0.05$ , $x^{'} (0) = - 0.6$ . Does the mass cross equilibrium? When?
Answer
$x = (0.05 - 0.3 t) e^{- 6 t}$ ; yes, once, at $t = 1/6$ s
Two successive maxima of a free damped vibration are 12 mm and 9 mm, recorded 0.25 s apart. Find $δ$ , $β$ and $ζ$ .
Answer
$δ = ln \frac{4}{3} \approx 0.288$ , $β = δ /0.25 \approx 1.15 s^{- 1}$ , $ζ \approx 0.0457$
A 2 kg mass hangs on a spring with $k = 50$ N/m. (a) For which $c$ is the motion underdamped? (b) Can damping make the quasi-period equal to 1 s? (c) Find $c$ so that the quasi-period is 1.5 s.
Answer
(a) $0 < c < 20$ N·s/m; (b) no: $T_{d} > 2 π / ω_{0} = 2 π /5 \approx 1.257$ s always; (c) $β = 25 - (4 π /3)^{2} \approx 2.73$ , $c = 4 β \approx 10.9$ N·s/m
Show that the zeros of an underdamped motion $x = A e^{- βt} cos (ω_{d} t - φ)$ are spaced exactly $T_{d} /2$ apart.
Answer
$x = 0$ iff $cos (ω_{d} t - φ) = 0$ iff $ω_{d} t - φ = \frac{π}{2} + nπ$ ; consecutive solutions differ by $π / ω_{d} = T_{d} /2$ .
For critical damping with $x_{0} > 0$ , show that the mass crosses equilibrium at some $t > 0$ if and only if $v_{0} < - β x_{0}$ .
Answer
$x = (x_{0} + (v_{0} + β x_{0}) t) e^{- βt}$ vanishes at $t = - x_{0} / (v_{0} + β x_{0})$ , which is positive iff $v_{0} + β x_{0} < 0$ .
A quarter-car model has $m = 350$ kg and $k = 35000$ N/m. Find the damping constant that gives $ζ = 0.3$ , and the resulting quasi-frequency.
Answer
$c_{cr} = 7000$ N·s/m, $c = 2100$ N·s/m; $ω_{0} = 10$ rad/s, $ω_{d} = 10 0.91 \approx 9.54$ rad/s

The equation in standard form

Case 1: overdamped motion (c>ccr​)

Case 2: critically damped motion (c=ccr​)

Case 3: underdamped motion (0<c<ccr​)

Why the logarithmic decrement is so useful

One picture for all three cases

Worked examples

Exercises

Case 1: overdamped motion ( $c > c_{cr}$ )

Case 2: critically damped motion ( $c = c_{cr}$ )

Case 3: underdamped motion ( $0 < c < c_{cr}$ )