Free and Damped Mechanical Oscillations, differential equations lesson

Pull a mass on a spring away from its resting position and let go. With a little friction it swings back and forth in shrinking arcs, but with a strong shock absorber, as in a car’s suspension, it returns without repeated swings. Which of these happens, and how quickly does the motion die out? This lesson builds the equation of motion from the forces, sorts the motion into three regimes by the sign of one discriminant, solves a release exactly and uses mechanical energy to explain why the swings shrink.

Key terms

Mass-spring-damper model

Definition A body of mass $m$ moves along a line, attached to a spring of stiffness $k$ and to a damper with damping coefficient $c$ . Let $x(t)$ be its displacement at time $t$ , measured from the equilibrium position where the body can rest, and let primes denote derivatives with respect to $t$ . The spring force is $-kx$ (a linear spring, which is accurate for small stretches) and the damping force is $-cx'$ . Newton’s second law $mx'' = -kx - cx'$ becomes

\begin{gathered} mx'' + cx' + kx = 0, \\[0.4em] m > 0, \quad k > 0, \quad c \ge 0. \end{gathered}

In SI units $m$ is in kilograms, $k$ in newtons per meter and $c$ in newton seconds per meter, so every term is a force in newtons. The motion is called free because no outside force drives the body. The spring and damper forces still act. For a body hanging on a vertical spring, measuring $x$ from the stretched rest position removes gravity from the equation, because the static stretch $\Delta$ satisfies $k\Delta = mg$ .

In plain words Mass times acceleration equals a pull back toward equilibrium plus a drag against the motion. Moving both forces to the left side gives a second-order linear equation with constant coefficients.

Example A body of mass $2\ \text{kg}$ hung on a spring stretches it by $0.49\ \text{m}$ and comes to rest. With $g = 9.8\ \text{m/s}^2$ , the balance $k\Delta = mg$ gives $k = 2(9.8)/0.49 = 40\ \text{N/m}$ . Measured from that rest position and with a damper of $c = 8\ \text{N s/m}$ , the motion obeys

2x'' + 8x' + 40x = 0.

Each term is in newtons. For example, $8\ \text{N s/m}$ times a velocity in meters per second is a force in newtons.

Viscous damping

Definition Viscous damping is a resisting force proportional to the velocity and opposite to it,

F_{\text{damp}} = -c\,x', \qquad c \ge 0.

It models a dashpot, or slow motion through a fluid. Its power $F_{\text{damp}}\,x' = -c(x')^2$ is never positive, so it can only remove energy. Dry friction, whose size does not depend on the speed, and drag that grows like the square of the speed are different force laws, and they lead to nonlinear equations.

In plain words The faster the body moves, the harder the damper pushes back, and the push always points against the motion. When the body stops for an instant, the damping force is zero for that instant.

Example With $c = 2\ \text{N s/m}$ , a body moving at $x' = 0.5\ \text{m/s}$ feels a damping force of $-1\ \text{N}$ , and at $x' = -3\ \text{m/s}$ it feels $+6\ \text{N}$ . Doubling the speed doubles the force.

Underdamping and the damped angular frequency

Definition The trial $x = e^{rt}$ turns $mx'' + cx' + kx = 0$ into the characteristic equation $mr^2 + cr + k = 0$ , whose roots are

r = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}.

The system is underdamped when $c > 0$ and $c^2 < 4mk$ . The roots are then $-\alpha \pm i\omega_d$ , with decay rate $\alpha = c/(2m)$ and damped angular frequency

\omega_d = \sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2},

measured in radians per second. Every real solution is

\begin{aligned} x &= e^{-\alpha t}\bigl(A\cos\omega_d t \\ &\qquad + B\sin\omega_d t\bigr) \end{aligned}

with real constants $A$ and $B$ fixed by $x(0)$ and $x'(0)$ . A nonzero solution crosses equilibrium again and again, at time intervals of $\pi/\omega_d$ , inside an envelope that shrinks like $e^{-\alpha t}$ . The damped angular frequency is smaller than the natural angular frequency $\omega_0 = \sqrt{k/m}$ , which is the angular frequency of the undamped case $c = 0$ , where the envelope does not shrink at all.

In plain words Light damping keeps the oscillation but wraps it in a shrinking exponential envelope and slows its rhythm a little. The imaginary part of the roots sets the rhythm, and the real part sets how fast the swings die out.

Example Take $m = 1\ \text{kg}$ , $c = 2\ \text{N s/m}$ and $k = 5\ \text{N/m}$ , and release the body from $x(0) = 2\ \text{m}$ with $x'(0) = 0$ . Here $c^2 = 4 < 20 = 4mk$ , and $r^2 + 2r + 5 = (r + 1)^2 + 4$ has the roots $-1 \pm 2i$ , so $\alpha = 1$ per second and $\omega_d = 2$ radians per second. The condition $x(0) = 2$ gives $A = 2$ . The velocity of $e^{-t}(A\cos 2t + B\sin 2t)$ at $t = 0$ is $-A + 2B$ , so $x'(0) = 0$ gives $B = 1$ , not $B = 0$ , because the decaying factor also contributes to the velocity. The motion is

x = e^{-t}\left(2\cos 2t + \sin 2t\right),

with velocity $x' = -5e^{-t}\sin 2t$ and acceleration $x'' = e^{-t}(5\sin 2t - 10\cos 2t)$ . In $x'' + 2x' + 5x$ the cosine terms give $-10 + 10 = 0$ and the sine terms give $5 - 10 + 5 = 0$ , so the equation holds. The body turns around whenever $\sin 2t = 0$ , first at $t = \pi/2$ , where $x = -2e^{-\pi/2} \approx -0.416\ \text{m}$ .

Critical damping

Definition The system is critically damped when $c^2 = 4mk$ , that is, when $c = 2\sqrt{mk}$ . The characteristic equation then has the repeated root $r = -c/(2m) = -\omega_0$ , where $\omega_0 = \sqrt{k/m}$ , and every solution is

x = (A + Bt)\,e^{-\omega_0 t}.

A nonzero solution vanishes at most once, at the single time where $A + Bt = 0$ if there is one, so there is no oscillation. A large enough initial velocity toward equilibrium can still carry the body across it once.

In plain words Critical damping is the borderline between swinging and creeping back. The body returns without repeated swings, and with less damping it would start to oscillate.

Example For $m = 1\ \text{kg}$ and $k = 4\ \text{N/m}$ the critical value is $c = 2\sqrt{4} = 4\ \text{N s/m}$ , and $x'' + 4x' + 4x = 0$ has the double root $-2$ . A release from $x(0) = 1\ \text{m}$ with $x'(0) = 0$ gives $A = 1$ and $-2A + B = 0$ , so

x = (1 + 2t)\,e^{-2t}.

This stays positive for $t \ge 0$ and is about $0.017\ \text{m}$ at $t = 3\ \text{s}$ .

Overdamping

Definition The system is overdamped when $c^2 > 4mk$ . The characteristic roots are then real, distinct and negative,

r_{1,2} = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m},

and every solution is $x = C_1e^{r_1t} + C_2e^{r_2t}$ . Both roots are negative because $\sqrt{c^2 - 4mk} < c$ . A nonzero solution vanishes at most once, because $C_1e^{r_1t} = -C_2e^{r_2t}$ can hold at only one time. The root closer to zero can be written as

r_1 = -\frac{2k}{c + \sqrt{c^2 - 4mk}},

and it approaches $0$ as $c$ grows. So a very strong damper makes every solution that contains $e^{r_1t}$ creep back slowly.

In plain words The damper is so strong that the body oozes back to equilibrium. More damping than the critical amount does not speed up the return, because the slower of the two exponentials gets even slower.

Example Keep $m = 1\ \text{kg}$ and $k = 4\ \text{N/m}$ , but take $c = 5\ \text{N s/m}$ , which exceeds the critical value $4$ . Then $r^2 + 5r + 4 = (r + 1)(r + 4)$ , with roots $-1$ and $-4$ . A release from $x(0) = 1\ \text{m}$ with $x'(0) = 0$ requires $C_1 + C_2 = 1$ and $-C_1 - 4C_2 = 0$ , so

x = \tfrac{4}{3}e^{-t} - \tfrac{1}{3}e^{-4t}.

At $t = 3\ \text{s}$ this is about $0.066\ \text{m}$ , nearly four times the critically damped value from the same release.

Mechanical energy

Definition The mechanical energy of the oscillator is the kinetic energy of the body plus the potential energy stored in the spring,

E(t) = \tfrac{1}{2}m(x')^2 + \tfrac{1}{2}kx^2,

measured in joules. Along every solution of $mx'' + cx' + kx = 0$ ,

\begin{aligned} E' &= x'\left(mx'' + kx\right) \\[0.4em] &= -c\,(x')^2 \le 0. \end{aligned}

So the energy never increases, and the energy lost between times $0$ and $T$ is the integral of $c(x')^2$ over $[0, T]$ . With $c = 0$ the energy is constant. With $c > 0$ it strictly decreases along every nonzero solution, because the velocity of such a solution is zero only at isolated instants.

In plain words Energy passes back and forth between the moving body and the stretched spring, and the damper drains it at the rate $c(x')^2$ . At a turning point the drain stops for an instant, and it resumes as soon as the body moves again.

Example For the release $x = e^{-t}(2\cos 2t + \sin 2t)$ with $m = 1$ , $c = 2$ and $k = 5$ , all the energy starts in the spring, $E(0) = \tfrac{1}{2}(5)(2^2) = 10\ \text{J}$ . It drains at the rate $-E' = 2(x')^2 = 50e^{-2t}\sin^2 2t$ watts. At the first turning point, $t = \pi/2$ , the velocity is zero and $x = -2e^{-\pi/2}$ , so

\begin{aligned} E(\pi/2) &= \tfrac{1}{2}(5)\left(2e^{-\pi/2}\right)^2 \\[0.4em] &= 10e^{-\pi} \approx 0.43\ \text{J}. \end{aligned}

More than 95 percent of the energy is gone after the first swing.

Check yourself

A body of mass $2\ \text{kg}$ hangs on a spring with $k = 8\ \text{N/m}$ and a damper with $c = 4$ , $8$ or $10\ \text{N s/m}$ . Classify the motion in each case, and give the damped angular frequency where there is one.

Show the answer

The threshold is $4mk = 64$ . For $c = 4$ we have $c^2 = 16 < 64$ , so the motion is underdamped, with

\begin{aligned} \omega_d &= \sqrt{\frac{8}{2} - \left(\frac{4}{4}\right)^2} \\[0.4em] &= \sqrt{3} \approx 1.73 \end{aligned}

radians per second. For $c = 8$ we have $c^2 = 64$ , so the motion is critically damped, with the double root $-2$ . For $c = 10$ we have $c^2 = 100 > 64$ , so the motion is overdamped, and $2r^2 + 10r + 8 = 2(r + 1)(r + 4)$ gives the roots $-1$ and $-4$ .

With primes denoting derivatives in $t$ , solve $x'' + 2x' + 10x = 0$ with $x(0) = 1$ and $x'(0) = 0$ . A classmate answers $x = e^{-t}\cos 3t$ . What went wrong?

Show the answer

The roots of $r^2 + 2r + 10 = (r + 1)^2 + 9$ are $-1 \pm 3i$ , so $x = e^{-t}(A\cos 3t + B\sin 3t)$ . The condition $x(0) = 1$ gives $A = 1$ . The velocity at $0$ is $-A + 3B$ , so $x'(0) = 0$ gives $B = \tfrac{1}{3}$ , and

x = e^{-t}\left(\cos 3t + \tfrac{1}{3}\sin 3t\right).

Its velocity is $x' = -\tfrac{10}{3}e^{-t}\sin 3t$ , which is zero at $t = 0$ . The classmate set $B = 0$ and forgot that the factor $e^{-t}$ also changes the velocity. That answer has $x'(0) = -1$ , so it describes a body pushed toward equilibrium, not one released at rest.

The critically damped equation $x'' + 4x' + 4x = 0$ is started at $x(0) = 1$ with $x'(0) = -6$ . Does the body cross equilibrium, and does that contradict the claim that critical damping has no oscillation?

Show the answer

The double root is $-2$ , so $x = (A + Bt)e^{-2t}$ . Here $A = 1$ , and $x'(0) = -2A + B = -6$ gives $B = -4$ , so

x = (1 - 4t)\,e^{-2t}.

The body crosses equilibrium at $t = \tfrac{1}{4}$ . Its velocity $x' = e^{-2t}(8t - 6)$ is zero at $t = \tfrac{3}{4}$ , where it reaches its farthest point on the other side, $x = -2e^{-3/2} \approx -0.446$ . After that it returns toward $0$ from below without crossing again, since $1 - 4t < 0$ for every $t > \tfrac{1}{4}$ . There is no contradiction. A strong push toward equilibrium can carry a critically damped body across once, but it never oscillates.

Free and Damped Mechanical Oscillations

Key terms

Mass-spring-damper model

Viscous damping

Underdamping and the damped angular frequency

Critical damping

Overdamping

Mechanical energy

Check yourself

How this lesson connects

Builds on

Used later in