Part 1 · The elementary oscillator

1.3 The damped oscillator

Source: Carboni’s notes for lesson 1, Oscillatore elementare, pp. 5–8

Objectives

Picture. The force–displacement cycle: a line that stores energy, an ellipse that loses it.
Recipe. The elastic and the dissipated energy of a cycle from its areas.
Vocabulary. Critical damping $C_{cr}$ , damping ratio $ξ$ , damped natural frequency $ω_{d}$ .
Recipe. Solve $M \ddot{U} + C \dot{U} + K U = 0$ and name its behaviour from $ξ$ .

Real systems lose energy: the undamped oscillator of §1.2 The free undamped oscillator would ring for ever, and nothing real does. The simplest model with a loss adds a dashpot (a piston in oil) in parallel with the spring.

The viscoelastic oscillator

Definition 1.3.1damped (viscoelastic) oscillatornotes L1, p. 5

A mass $M$ tied to a wall by a spring $K$ and a dashpot $C$ in parallel: $M \ddot{U} + C \dot{U} + K U = 0.$ $C$ (in N·s/m) is the damping coefficient: the dashpot resists with a viscous force $F_{V} = C \dot{U}$ , proportional to the velocity. The model is called linear viscoelastic.

The free-body diagram has three forces against the motion: the elastic force $F_{e l} = K U$ , the viscous force $F_{V} = C \dot{U}$ , and the inertial term $M \ddot{U}$ (Definition 1.1.5 explains why inertia can be drawn as a force).

Energy over a slow cycle

Before solving the equation, look at what the dashpot does to energy. Move the mass back and forth very slowly, imposing $U (t) = A sin (Ω t), Ω ≪ ω = K / M .$ $Ω$ is the frequency we impose (a forcing frequency), not the natural one.

Slow means no inertia. $\ddot{U} = - Ω^{2} U$ , so the inertial force $M \ddot{U} = - M Ω^{2} U$ is small next to $K U$ : $M Ω^{2} ≪ M ω^{2} = K$ .
So the force needed to drive the mass is just the sum of the elastic and viscous forces: $F_{t} = F_{e l} + F_{V} .$

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Figure 1.3.1 The three force–displacement plots of the notes (p. 6): elastic force, viscous force and their sum, which is the loop a test machine measures. Drag the point around a loop, or play it: it moves in step on all three. The shaded areas are the elastic energy

W_{E}

(green triangle) and the dissipated energy

W_{D}

(red ellipse).

What the three plots show (notes L1, p. 6):

Elastic force. $F_{e l} = K U$ is a straight line from $(- A, - K A)$ to $(A, K A)$ , travelled back and forth.
Viscous force. An ellipse, with horizontal semi-axis $A$ and vertical semi-axis $C Ω A$ .
Total force. The line and the ellipse added: a tilted ellipse reaching $\pm K A$ at $U = \pm A$ and crossing the force axis at $\pm C Ω A$ .

Why the viscous plot is an ellipsebeyond the lectures

With $U = A sin Ω t$ , the velocity is $\dot{U} = Ω A cos Ω t$ , so $F_{V} = C Ω A cos Ω t$ . Then $(\frac{U}{A})^{2} + (\frac{F _{V}}{C Ω A})^{2} = sin^{2} Ω t + cos^{2} Ω t = 1,$ the equation of an ellipse with semi-axes $A$ and $C Ω A$ .

Proposition 1.3.2energies of a cyclenotes L1, p. 6

The elastic energy is the area of the triangle under the line from $0$ to $A$ : $W_{E} = \frac{1}{2} K A^{2} .$ The dissipated energy per cycle is the area of the ellipse, $π$ times the product of its semi-axes: $W_{D} = π A \cdot C Ω A = π C Ω A^{2} .$

Proofbeyond the lectures

The area of the triangle with base $A$ and height $K A$ is $\frac{1}{2} K A^{2}$ ; it is also $\int_{0}^{A} K U d U$ . For $W_{D}$ , the work done against the viscous force over one cycle, of duration $2 π /Ω$ , is $W_{D} = \oint F_{V} d U = \int_{0}^{2 π /Ω} C \dot{U}^{2} d t = C Ω^{2} A^{2} \int_{0}^{2 π /Ω} cos^{2} Ω t d t = C Ω^{2} A^{2} \cdot \frac{π}{Ω} = π C Ω A^{2},$ using $d U = \dot{U} d t$ and the mean value $\frac{1}{2}$ of $cos^{2}$ over a period.

$W_{E}$ is stored while loading and given back while unloading: the line is travelled both ways and encloses no area.
$W_{D}$ is lost in every cycle, as heat in the dashpot.
The dissipated energy grows with the frequency $Ω$ : a viscous damper resists fast motion more.

Warningslip in the notes

In the left plot of p. 6 the positive end of the $U$ axis is labelled $- A$ , and there is a stray $- A$ on the force axis; the line ends at $U = A$ , force $K A$ . The plot and the energy formula write the damping coefficient as a small $c$ ; it is the same $C$ as elsewhere. The upper limit of the time interval, written $T_{g}$ , is the duration of the imposed motion (the subscript is hard to read).

Solving the damped equation

Same idea as for the undamped oscillator (Recipe 1.2.1): try an exponential.

Recipe 1.3.3solving the damped oscillatornotes L1, pp. 7–8

Try $U (t) = a e^{μ t}$ in $M \ddot{U} + C \dot{U} + K U = 0$ : $a (M μ^{2} + C μ + K) e^{μ t} = 0.$
Characteristic equation, divided by $M$ , with $K / M = ω^{2}$ : $μ^{2} + \frac{C}{M} μ + ω^{2} = 0.$
Roots: $μ = - \frac{C}{2 M} \pm (\frac{C}{2 M})^{2} - ω^{2} .$
Read the sign under the root: it decides whether the motion oscillates (below).

The sign under the root changes when $C /2 M = ω$ . That value of $C$ separates oscillating from non-oscillating motion.

Definition 1.3.4critical damping and damping rationotes L1, p. 7

The critical damping is $C_{cr} = 2 K M = 2 ω M,$ and the damping ratio (rapporto di smorzamento) is $ξ = \frac{C}{C _{cr}} = \frac{C}{2 ω M}, so \frac{C}{2 M} = ξ ω .$

(The two forms of $C_{cr}$ agree because $K M = K / M M = ω M$ .) In terms of $ξ$ the roots are $μ_{1, 2} = - ξ ω \pm ξ^{2} ω^{2} - ω^{2} = - ξ ω \pm - ω^{2} (1 - ξ^{2}) = - ξ ω \pm iω 1 - ξ^{2} .$

Warningslip in the notes

The last step on p. 7 drops the $\pm$ : it reads $- ξ ω + iω 1 - ξ^{2}$ . Both roots are needed, and p. 8 has the $\pm$ again.

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Figure 1.3.2 The free response for every damping ratio. Try the presets: undamped, light damping, critical, overdamped. The dashed curves are the envelope

\pm A e^{- ξ ω t}

; the dotted lines are one damped period

T_{d}

apart.

Critical damping, $ξ = 1$

$C = C_{cr} = 2 ω M$ . The root is double: $μ = - ξ ω = - \frac{C}{2 M} = - ω ⟹ U (t) = a e^{- \frac{C}{2 M} t} .$ The response decays without oscillating.

Warningslip in the notes

The notes write $μ = - ξ ω = \frac{C}{2 M}$ , dropping the minus in the middle term; the exponent that follows, $- \frac{C}{2 M} t$ , has it right.

Completing the critical casebeyond the lectures

At $ξ = 1$ the root is double, but a second-order equation needs two independent solutions to meet two initial conditions. The second one is $t e^{- ω t}$ (substitute it to check), so $U (t) = (a + b t) e^{- ω t}, a = U_{0}, b = V_{0} + ω U_{0} .$ Critical damping is the fastest return to rest without overshooting.

Underdamped, $ξ < 1$

$C < C_{cr}$ , so $ξ < 1$ and the roots are complex: $μ = - ξ ω \pm i ω_{d} .$

Definition 1.3.5damped natural frequencynotes L1, p. 8

$ω_{d} = ω 1 - ξ^{2}$ is the damped natural frequency (frequenza naturale smorzata), and $T_{d} = 2 π / ω_{d}$ the damped period.

The two exponentials combine, as in Recipe 1.2.1, into $U (t) = e^{- ξ ω t} (a sin (ω_{d} t) + b cos (ω_{d} t)) = A e^{- ξ ω t} cos (ω_{d} t - ϕ), A = a^{2} + b^{2}, ϕ = arctan \frac{a}{b} .$

Key observation 1.3.6an oscillation inside an envelopenotes L1, p. 8

The underdamped response is an oscillation at $ω_{d}$ inside the decaying envelope $\pm A e^{- ξ ω t}$ . Successive maxima, at $t_{n}, t_{n + 1}, \dots$ , are one damped period apart, $t_{n + 1} = t_{n} + T_{d}$ , and each is lower than the one before.

Warningslip in the notes

p. 8 writes $A = a + b^{2}$ ; the square on $a$ is missing, as the undamped case on p. 3 shows. The phase is $arctan (a / b)$ , not $arctan (b / a)$ (see the slip in §1.2 The free undamped oscillator).

Remarkbeyond the lectures

Damping slows the oscillation a little: $ω_{d} < ω$ . For $ξ = 0.05$ , $ω_{d} = 0.9987 ω$ , so for light damping $ω_{d} ≃ ω$ .
Initial conditions: with $U (0) = U_{0}$ and $\dot{U} (0) = V_{0}$ , $b = U_{0}$ and $a = (V_{0} + ξ ω U_{0}) / ω_{d}$ .
Overdamped, $ξ > 1$ : both roots $μ = - ξ ω \pm ω ξ^{2} - 1$ are real and negative. The response decays without oscillating, more slowly than at $ξ = 1$ , because one root is closer to zero than $- ω$ .

$ξ$	roots	response
$0$	$\pm iω$	oscillates for ever at $ω$
$0 < ξ < 1$	$- ξ ω \pm i ω_{d}$	oscillates at $ω_{d}$ , decaying as $e^{- ξ ω t}$
$1$	$- ω$ , double	returns to rest, fastest without overshoot
$> 1$	two real, negative	returns to rest, slowly

Exercises

Exercise 1.3.1

A damped oscillator has $M = 2$ kg, $K = 800$ N/m, $C = 8$ N·s/m. Find $ω$ , $C_{cr}$ , $ξ$ , $ω_{d}$ and $T_{d}$ .

Solution

$ω = 800/2 = 20$ rad/s.
$C_{cr} = 2 ω M = 80$ N·s/m, so $ξ = 8/80 = 0.1$ .
$ω_{d} = 20 0.99 = 19.90$ rad/s, $T_{d} = 2 π / ω_{d} = 0.3157$ s.

Exercise 1.3.2

The same oscillator is driven slowly with $U = A sin Ω t$ , $A = 1$ cm, $Ω = 2$ rad/s. Find $W_{E}$ and $W_{D}$ . Was the motion really slow?

Solution

$W_{E} = \frac{1}{2} K A^{2} = \frac{1}{2} \cdot 800 \cdot 1 0^{- 4} = 0.04$ J.
$W_{D} = π C Ω A^{2} = π \cdot 8 \cdot 2 \cdot 1 0^{- 4} = 5.03 \times 1 0^{- 3}$ J.
$Ω/ ω = 0.1$ , so the inertial force is $M Ω^{2} / K = 1%$ of the elastic one: slow enough.

Exercise 1.3.3

Which damping coefficient makes the oscillator of Exercise 1.3.1 return to rest as fast as possible without overshooting? Write its response from $U_{0} = 1$ cm, $V_{0} = 0$ .

Solution

Critical damping: $C = C_{cr} = 80$ N·s/m.
$U (t) = (a + b t) e^{- 20 t}$ with $a = U_{0} = 1$ cm and $b = V_{0} + ω U_{0} = 20$ cm/s: $U (t) = (1 + 20 t) e^{- 20 t}$ cm.

1.3 The damped oscillator

Objectives

The viscoelastic oscillator

Energy over a slow cycle

Solving the damped equation

Critical damping, ξ=1

Underdamped, ξ<1

Exercises

Critical damping, $ξ = 1$

Underdamped, $ξ < 1$