Damped Oscillations

Resistance adds a term

In simple harmonic motion, $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$d²x/dt² = −ω²x and the swings never die away. Add a resistance that is proportional to the speed and opposes the motion, such as a damper or the drag of a liquid. It contributes $-k\frac{dx}{dt}$−k dx/dt to the acceleration, with k a positive constant, so $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x-k\frac{dx}{dt}$d²x/dt² = −ω²x − k dx/dt.

Rearranged, $\frac{d^{2}x}{d{t}^2}+k\frac{dx}{dt}+{\omega }^{2}x=0$d²x/dt² + k dx/dt + ω²x = 0. With a dot for each derivative with respect to time, $\frac{dx}{dt}$dx/dt is written ẋ and $\frac{d^{2}x}{d{t}^2}$d²x/dt² is written ẍ. The auxiliary equation gains a middle term: $m^2+km+{\omega }^2=0$m² + k m + ω² = 0, with roots $m=-\frac{k}{2}\pm \frac{1}{2}\sqrt{k^2-4{\omega }^2}$m = −k/2 ± ½√(k² − 4ω²). The sign of the discriminant $k^2-4{\omega }^2$k² − 4ω² decides what kind of motion this is.

To compare the three kinds, take ${\omega }^2=4$ω² = 4 throughout, and release the particle from rest at x = 1, so x = 1 and $\frac{dx}{dt}=0$dx/dt = 0 at t = 0.

Light damping

When $k^2<4{\omega }^2$k² < 4ω² the roots are complex: $m=-\frac{k}{2}\pm qi$m = −k/2 ± qi, with $q=\sqrt{{\omega }^2-\frac{k^2}{4}}$q = √(ω² − k²/4). So $x=e^{-\frac{kt}{2}}\left(A\cos qt+B\sin qt\right)$x = e^(−kt/2)(A cos qt + B sin qt). The motion still swings, at the rate q, which is a little slower than $\omega$ω, and the factor $e^{-\frac{kt}{2}}$e^(−kt/2) shrinks every swing.

With k = 2, the equation $\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+4x=0$d²x/dt² + 2 dx/dt + 4x = 0 has discriminant 4 − 16 = −12, so $m=-1\pm \sqrt{3}i$m = −1 ± √3i. Then $x=e^{-t}\left(A\cos \sqrt{3}t+B\sin \sqrt{3}t\right)$x = e^(−t)(A cos √3t + B sin √3t). The start gives A = 1, and $\frac{dx}{dt}=-A+\sqrt{3}B=0$dx/dt = −A + √3B = 0 at t = 0, so $B=\frac{1}{\sqrt{3}}$B = 1/√3.

Light damping: $x=e^{-t}\left(\cos \sqrt{3}t+\left(\frac{1}{\sqrt{3}}\right)\sin \sqrt{3}t\right)$x = e^(−t)(cos √3t + (1/√3) sin √3t) from rest at x = 1. It crosses the center at $t\approx 1.21$t ≈ 1.21 and swings out to −0.16 at $t=\frac{\pi }{\sqrt{3}}\approx 1.81$t = π/√3 ≈ 1.81. The dashed curves are $\pm \left(\frac{2}{\sqrt{3}}\right)e^{-t}$±(2/√3)e^(−t), the shrinking envelope, since $\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$√(1 + 1/3) = 2/√3.

Heavy damping

When $k^2>4{\omega }^2$k² > 4ω² the roots $m_1$m₁ and $m_2$m₂ are real and different. They add to −k, which is negative, and multiply to ${\omega }^2$ω², which is positive, so both are negative. So $x=A{e}^{m_{1}t}+B{e}^{m_{2}t}$x = A e^(m₁t) + B e^(m₂t) is two shrinking exponentials, and there is no swing at all.

With k = 5, $\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+4x=0$d²x/dt² + 5 dx/dt + 4x = 0 has auxiliary equation $m^2+5m+4=\left(m+1\right)\left(m+4\right)=0$m² + 5m + 4 = (m + 1)(m + 4) = 0, so $x=A{e}^{-t}+B{e}^{-4t}$x = A e^(−t) + B e^(−4t). The start gives A + B = 1 and −A − 4B = 0, so $A=\frac{4}{3}$A = 4/3, $B=-\frac{1}{3}$B = −1/3, and $x=\frac{1}{3}\left(4e^{-t}-e^{-4t}\right)$x = ⅓(4e^(−t) − e^(−4t)).

The $e^{-4t}$e^(−4t) part dies away quickly and leaves $e^{-t}$e^(−t), the root nearer 0, to set the pace. So the return is slow: x is still 0.18 at t = 2.

Critical damping

When $k^2=4{\omega }^2$k² = 4ω², that is $k=2\omega$k = 2ω, the auxiliary equation has the repeated root $m=-\frac{k}{2}=-\omega$m = −k/2 = −ω, and $x=\left(A+Bt\right)e^{-\omega t}$x = (A + B t)e^(−ωt).

With k = 4, $\frac{d^{2}x}{d{t}^2}+4\frac{dx}{dt}+4x=0$d²x/dt² + 4 dx/dt + 4x = 0 has m = −2 twice, so $x=\left(A+Bt\right)e^{-2t}$x = (A + B t)e^(−2t). The start gives A = 1, and $\frac{dx}{dt}=B-2A=0$dx/dt = B − 2A = 0 at t = 0, so B = 2 and $x=\left(1+2t\right)e^{-2t}$x = (1 + 2t)e^(−2t). Here 1 + 2t stays positive, so x never crosses the center.

Less damping than this overshoots the center, and more damping makes the return slower. Critical damping is the fastest return that never crosses the center: from this start it stays within 0.05 of the center from t = 2.37 on, against 2.64 for k = 2 and 3.28 for k = 5.

The gold curve is critical damping, $x=\left(1+2t\right)e^{-2t}$x = (1 + 2t)e^(−2t). The dashed curve is heavy damping, $x=\frac{1}{3}\left(4e^{-t}-e^{-4t}\right)$x = ⅓(4e^(−t) − e^(−4t)). Both start at 1 at rest and neither crosses the center, but the critical curve is lower all the way: 0.41 against 0.48 at t = 1, and 0.09 against 0.18 at t = 2.

The three cases

The discriminant $k^2-4{\omega }^2$k² − 4ω² names the motion. Negative is light damping, $x=e^{-\frac{kt}{2}}\left(A\cos qt+B\sin qt\right)$x = e^(−kt/2)(A cos qt + B sin qt), which swings with a shrinking amplitude. Zero is critical damping, $x=\left(A+Bt\right)e^{-\frac{kt}{2}}$x = (A + B t)e^(−kt/2). Positive is heavy damping, $x=A{e}^{m_{1}t}+B{e}^{m_{2}t}$x = A e^(m₁t) + B e^(m₂t), with no swing.

For $\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+5x=0$d²x/dt² + 2 dx/dt + 5x = 0, $k^2-4{\omega }^2=4-20=-16$k² − 4ω² = 4 − 20 = −16, so the damping is light. For $\frac{d^{2}x}{d{t}^2}+6\frac{dx}{dt}+9x=0$d²x/dt² + 6 dx/dt + 9x = 0, it is 36 − 36 = 0, so the damping is critical.

Here the equation is $\frac{d^{2}y}{d{t}^2}+b\frac{dy}{dt}+4y=0$d²y/dt² + b dy/dt + 4y = 0, with y in place of x and b in place of k. On the left are the roots of $m^2+bm+4=0$m² + bm + 4 = 0; on the right is y from y = 1 at rest. At b = 2 the discriminant is −12 and the roots are $-1\pm 1.73i$−1 ± 1.73i, so y swings inside a dashed envelope. As b rises the roots move round the circle of radius 2, meet at −2 when b = 4, and then split along the real axis, and the swing is gone.

The usual mistakes

Comparing k with $\omega$ω instead of $k^2$k² with $4{\omega }^2$4ω². For $\frac{d^{2}x}{d{t}^2}+4\frac{dx}{dt}+5x=0$d²x/dt² + 4 dx/dt + 5x = 0, k = 4 is larger than $\omega =\sqrt{5}$ω = √5, but $k^2=16$k² = 16 is less than $4{\omega }^2=20$4ω² = 20, so the damping is light.

Taking $k=\omega$k = ω as the critical value. Critical damping is $k=2\omega$k = 2ω: for ${\omega }^2=4$ω² = 4 it is k = 4.

Taking $\omega$ω as the rate of the swing under light damping. The swing goes at $q=\sqrt{{\omega }^2-\frac{k^2}{4}}$q = √(ω² − k²/4), which is slower: for k = 2 and $\omega =2$ω = 2 it is $\sqrt{3}$√3, and each swing takes $\frac{2\pi }{\sqrt{3}}\approx 3.63$2π/√3 ≈ 3.63 rather than $\pi$π.

Calling heavy damping the fastest return because it has the most resistance. The extra resistance slows the return, which is why the heavy curve is above the critical one.

A door on a hydraulic closer

In the application below, a door obeys $\frac{d^{2}x}{d{s}^2}+5\frac{dx}{ds}+4x=0$d²x/ds² + 5 dx/ds + 4x = 0, with time s in seconds: k = 5 and ${\omega }^2=4$ω² = 4, so $k^2=25$k² = 25 is more than $4{\omega }^2=16$4ω² = 16 and the damping is heavy. It is the heavy case above, released from 1.2 radians instead of 1, so $x=1.6e^{-s}-0.4e^{-4s}$x = 1.6e^(−s) − 0.4e^(−4s). The application finds −1 and −4 as eigenvalues of a matrix, and they are the roots of the auxiliary equation.