4.10g - Damped oscillations | OCR A-Level Further Mathematics A

Model a restoring force together with resistance proportional to velocity. You will solve the resulting second order equation and distinguish underdamping, critical damping and overdamping by what their solutions predict.

Building a damping model

Does adding resistance simply make an oscillator swing more slowly, or can it stop repeated swinging altogether? The answer depends on how the resistance compares with the restoring force and the mass.

Suppose displacement $x$ is measured from equilibrium. A restoring force $-kx$ , with $k>0$ , points towards equilibrium. A resistive force $-c\dot x$ , with $c>0$ , opposes velocity. When a particle moves left, $\dot x<0$ and this resistive force points right: resistance changes direction with the motion.

You already know

Newton's second law gives $m\ddot x=F$ . Without resistance, $m\ddot x=-kx$ gives SHM. For a second order homogeneous equation, the auxiliary roots determine whether the solution contains real exponentials, a repeated-root factor, or sine and cosine.

Take a $1\,\mathrm{kg}$ particle with restoring force $-5x\,\mathrm{N}$ and resistance $-2\dot x\,\mathrm{N}$ , with $x$ in metres and time in seconds. At $x=0.2$ , $\dot x=-0.5$ , the restoring force is $-1\,\mathrm{N}$ and the resistance is $+1\,\mathrm{N}$ . The acceleration is momentarily zero, although neither force vanishes.

Linear damping model

Damping removes energy from an oscillating system through a resistive effect. For a particle of constant mass $m>0$ , restoring coefficient $k>0$ and velocity-resistance coefficient $c>0$ , the unforced linear model is

m\ddot x+c\dot x+kx=0

It assumes the two force laws remain proportional over the motion and no external driving force acts. Resistance $-c\dot x$ is not a restoring force: it depends on velocity, not position.

All coefficients here are constants. The mass has units $\mathrm{kg}$ , $c$ has units $\mathrm{kg\,s^{-1}}$ , and $k$ has units $\mathrm{kg\,s^{-2}}$ , so all three terms are forces. This model equation is not in the formula booklet; learn how to construct it from the stated forces.

Worked example 1: forces with different jobs

Question. A $1\,\mathrm{kg}$ particle has restoring force $-5x\,\mathrm{N}$ and resistive force $-2\dot x\,\mathrm{N}$ . Form its equation of motion and find its acceleration when $x=0.2\,\mathrm{m}$ , $\dot x=-0.5\,\mathrm{m\,s^{-1}}$ .

Plan. Add the signed forces, then substitute the position and velocity into the acceleration equation.

\begin{aligned} \ddot x&=-5x-2\dot x && F=ma,\ m=1\\ \ddot x+2\dot x+5x&=0 && \text{standard form}\\ \ddot x&=-5(0.2)-2(-0.5)=0 && \text{substitute both signed values} \end{aligned}

Why retain the sign of velocity? The resistive force must reverse when the motion reverses.

Answer. $\ddot x+2\dot x+5x=0$ ; the acceleration at that instant is $0\,\mathrm{m\,s^{-2}}$ .

Check. The two forces are $-1$ and $+1\,\mathrm{N}$ , giving zero resultant.

Change only the resistance coefficient from $2$ to $4$ . The same negative velocity now produces a larger positive resistive force.

Worked example 2: stronger resistance

Question. The same $1\,\mathrm{kg}$ particle now has forces $-5x\,\mathrm{N}$ and $-4\dot x\,\mathrm{N}$ . Form its equation and find the acceleration at $x=0.2$ , $\dot x=-0.5$ , in SI units.

Plan. Replace the damping coefficient while keeping the restoring force and mass unchanged.

\begin{aligned} \ddot x&=-5x-4\dot x && \text{resultant force divided by mass}\\ \ddot x+4\dot x+5x&=0 && \text{standard form}\\ \ddot x&=-5(0.2)-4(-0.5)=1 && \text{substitute} \end{aligned}

Why can the acceleration point away from equilibrium? Resistance opposes the present velocity; at this instant it exceeds the restoring force. That does not make resistance a source of energy.

Answer. $\ddot x+4\dot x+5x=0$ , with acceleration $+1\,\mathrm{m\,s^{-2}}$ .

Check. The resultant is $-1+2=1\,\mathrm{N}$ , agreeing with $m\ddot x$ .

The sign of each force follows its own physical role, rather than a memorised direction on the page.

Watch out

Wrong: write resistance as $-c|\dot x|$ , always pointing left. Right: the signed component is $-c\dot x$ , so it always opposes motion.

Use a new resistance coefficient in the same force model.

Underdamping

With modest resistance, the particle still crosses equilibrium repeatedly, but its oscillations shrink.

For $\ddot x+2\dot x+5x=0$ , the roots are $-1\pm2i$ , so the solution is a sine–cosine oscillation multiplied by $e^{-t}$ . Its oscillation has angular frequency $2$ , whereas the undamped model $\ddot x+5x=0$ has angular frequency $\sqrt5$ : damping affects timing as well as size.

Underdamped solution

For $m\ddot x+c\dot x+kx=0$ , with $m,k,c>0$ and $c^2<4mk$ , the auxiliary roots are

-\frac{c}{2m}\pm i\beta,\qquad \beta=\frac{\sqrt{4mk-c^2}}{2m}

The solution has the form

x=e^{-ct/(2m)}(A\cos\beta t+B\sin\beta t)

This is underdamping: a non-zero solution oscillates with a decreasing envelope. It repeats its phase every $2\pi/\beta$ , but its displacement is not periodic because its size decreases.
Not in the formula booklet: learn it.

You can find the roots directly instead of memorising their expanded formula. The sine–cosine bracket can be written $R\sin(\beta t+\phi)$ ; hence $|x|\le Re^{-ct/(2m)}$ . The bounding curves need not meet the actual local maxima, because the exponential is changing while the particle moves.

Worked example 3: release from rest with damping

Question. Solve $\ddot x+2\dot x+5x=0$ , given $x(0)=1$ , $\dot x(0)=0$ . Describe the motion as $t$ increases.

Plan. Use the complex auxiliary roots and differentiate the full product before applying the velocity condition.

\begin{aligned} m^2+2m+5=0&\Rightarrow m=-1\pm2i && \text{auxiliary equation}\\ x&=e^{-t}(A\cos2t+B\sin2t) && \text{complex roots}\\ x(0)=A&=1 && \text{displacement condition}\\ \dot x&=e^{-t}[(-A+2B)\cos2t+(-B-2A)\sin2t] && \text{product rule}\\ \dot x(0)=-1+2B=0&\Rightarrow B=\frac12 && \text{velocity condition} \end{aligned}

Why differentiate the exponential too? It contributes $-A$ to the initial velocity, so omitting it gives the wrong sine coefficient.

Answer. $x=e^{-t}(\cos2t+\tfrac12\sin2t)$ . The motion is underdamped: it crosses equilibrium repeatedly and tends to equilibrium with shrinking oscillations.

Check. $x(0)=1$ , $\dot x(0)=-1+1=0$ . The bound $|x|\le\tfrac{\sqrt5}{2}e^{-t}$ tends to zero.

Change only the initial velocity to $-1$ . The equation and decay rate stay fixed, but the initial phase changes.

Worked example 4: a moving start

Question. Solve $\ddot x+2\dot x+5x=0$ , given $x(0)=1$ , $\dot x(0)=-1$ .

Plan. Keep the same general solution and use the changed velocity condition to find $B$ .

\begin{aligned} x&=e^{-t}(A\cos2t+B\sin2t) && m=-1\pm2i\\ A&=1 && x(0)=1\\ -A+2B&=-1 && \dot x(0)=-1\\ B&=0 && \text{solve} \end{aligned}

Why is a cosine-only bracket compatible with non-zero initial velocity? The decaying multiplier itself is changing at $t=0$ .

Answer. $x=e^{-t}\cos2t$ .

Check. $\dot x=e^{-t}(-\cos2t-2\sin2t)$ , so $\dot x(0)=-1$ ; the graph lies between $\pm e^{-t}$ .

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4.10g - Damped oscillations

Building a damping model

Underdamping

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