I8, I11 - Damped oscillations | AQA A-Level Further Mathematics

Resistance changes a spring's ideal oscillation into a return towards equilibrium with a shrinking amplitude or no repeated crossings. I11 models a damping force proportional to velocity, and I8 connects the differential equation to light, critical and heavy damping. You will retain force signs, identify each case and interpret what its solution predicts.

Velocity-proportional resistance

Can adding more resistance make a displaced particle take longer to return close to equilibrium? Resistance removes motion, but excessive resistance can also prevent a spring from bringing the particle back quickly. We need a signed force law before comparing the outcomes.

Choose a positive direction and let $x$ be displacement from equilibrium. The velocity is $\dot x$ , which can be positive or negative. A force that opposes motion must reverse direction when velocity reverses.

Linear damping force

A velocity-proportional damping force has signed component

F_{\mathrm d}=-c\dot x, \qquad c>0

Its magnitude is $c|\dot x|$ ; its direction is opposite to the velocity. The constant $c$ has units $\mathrm N\,\mathrm s\,\mathrm m^{-1}$ . This model is not a formula supplied in the examination booklet.

The force is zero at an instant when the particle is at rest. That does not mean the resultant force is zero: a displaced spring can still accelerate the particle. During motion the damping force removes mechanical energy, because it acts against the direction of travel. The model assumes linear resistance; it is not a universal law for every fluid or every speed.

Worked example 1: motion in the positive direction

Question. A damper has $c=3\,\mathrm N\,\mathrm s\,\mathrm m^{-1}$ . Find its signed force when velocity is $+2\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Plan. Substitute the signed velocity, retaining the minus sign in the force law.

F_{\mathrm d}=-c\dot x=-3(2)=-6\,\mathrm N

Answer. The force is $-6\,\mathrm N$ , meaning $6\,\mathrm N$ in the negative direction.

Check. The force points opposite to the positive velocity. Its magnitude is $c|\dot x|=3(2)=6\,\mathrm N$ .

Reverse only the velocity. The force magnitude is unchanged, but its sign must reverse.

Worked example 2: motion in the negative direction

Question. Find the force from the same damper when velocity is $-2\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Plan. Use the negative velocity in the same signed law.

F_{\mathrm d}=-3(-2)=+6\,\mathrm N

Answer. The force is $+6\,\mathrm N$ , opposing the negative motion.

Check. The product of the force and velocity is $(6)(-2)<0$ , consistent with resistance acting against motion. Using $-c|\dot x|$ would incorrectly give a negative force in both examples.

The restoring force and damping force do different jobs. The spring opposes displacement; the damper opposes velocity. Their directions need not agree.

Watch out

Wrong: always drawing damping upwards or always assigning it a negative component. Use $-c\dot x$ with a signed velocity. Its sign changes whenever the direction of motion changes.

Calculate a fresh signed force.

Forming the damped-motion equation

For an ideal spring with restoring force $-kx$ , combine spring and damper forces before applying Newton's second law. With no external driving force,

\begin{aligned} m\ddot x&=-kx-c\dot x\\ m\ddot x+c\dot x+kx&=0 \end{aligned}

Here $m>0$ , $k>0$ , and $c>0$ for passive damping. For a vertical spring, the same equation holds with displacement measured from equilibrium: $mg-k(e_0+x)-c\dot x=-kx-c\dot x$ , since $ke_0=mg$ . Weight is included before that cancellation.

Unforced damped spring model

Dividing by $m$ , write

\ddot x+2\beta\dot x+\omega_0^2x=0, \qquad \beta=\frac{c}{2m}, \qquad \omega_0=\sqrt{\frac{k}{m}}

The positive constant $\omega_0$ is the natural angular frequency of the corresponding undamped spring. The damped oscillation frequency, when oscillations occur, will generally differ. These model and solution formulae are not supplied in the formula booklet: learn them.

Each term of the unnormalised equation has force units. The two terms $c\dot x$ and $kx$ have plus signs on the left because the corresponding forces on the right oppose velocity and displacement respectively.

Worked example 3: moving towards equilibrium

Question. A $2\,\mathrm{kg}$ particle has spring stiffness $10\,\mathrm N\,\mathrm m^{-1}$ and damping coefficient $4\,\mathrm N\,\mathrm s\,\mathrm m^{-1}$ . Form its equation and find its acceleration when $x=0.3\,\mathrm m$ , $\dot x=-0.2\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Plan. Add the signed spring and damping forces, then divide the resultant by mass.

\begin{aligned} 2\ddot x&=-10x-4\dot x\\ \ddot x+2\dot x+5x&=0\\ 2\ddot x&=-10(0.3)-4(-0.2)\\ &=-3+0.8=-2.2 \end{aligned}

Answer. The equation is $2\ddot x+4\dot x+10x=0$ ; acceleration at the stated instant is $-1.1\,\mathrm{m}\,\mathrm{s}^{-2}$ .

Check. The spring pulls in the negative direction and the damper opposes the negative velocity, so it pulls positively. Their net force remains negative, with smaller magnitude than the spring force alone. The acceleration has the corresponding sign.

Reverse only the velocity at the same displacement. The damping force now assists the restoring force rather than opposing it.

Worked example 4: moving away from equilibrium

Question. For the same particle, find its acceleration when $x=0.3\,\mathrm m$ , $\dot x=+0.2\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Plan. Use the same equation with the changed signed velocity.

\begin{aligned} 2\ddot x&=-10(0.3)-4(0.2)\\ &=-3-0.8=-3.8\\ \ddot x&=-1.9 \end{aligned}

Answer. Acceleration is $-1.9\,\mathrm{m}\,\mathrm{s}^{-2}$ .

Check. Both forces now point negatively: both resist the particle's motion away from equilibrium. The acceleration magnitude is larger than in Worked example 3, even though the displacement and speed are unchanged.

At release from rest the damping force is initially zero. Two different dampers can therefore give the same initial acceleration for the same displacement, mass and spring, yet produce different subsequent curves.

Watch out

Wrong: $m\ddot x=-kx+c\dot x$ for a passive damper with $c>0$ . The damping force is $-c\dot x$ ; a positive velocity term on the force side would feed motion rather than oppose it.

Form a new equation before substituting the state.

Light, critical and heavy damping

The auxiliary equation of $m\ddot x+c\dot x+kx=0$ is $m\lambda^2+c\lambda+k=0$ . Its discriminant compares damping with the restoring system:

\Delta=c^2-4mk

For $m,k,c>0$ , the three cases are as follows.

Condition	Damping	General solution and behaviour
$0<c<2\sqrt{mk}$ , so $\Delta<0$	Light	$x=e^{-\beta t}(A\cos\nu t+B\sin\nu t)$ , $\nu=\sqrt{\omega_0^2-\beta^2}>0$ ; shrinking oscillations
$c=2\sqrt{mk}$ , so $\Delta=0$	Critical	$x=(A+Bt)e^{-\beta t}$ ; repeated negative root, no repeated oscillation
$c>2\sqrt{mk}$ , so $\Delta>0$	Heavy	$x=Ae^{\lambda_1t}+Be^{\lambda_2t}$ , with distinct negative real roots; no repeated oscillation

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Mechanics

Statistics

Discrete Mathematics

I8, I11 - Damped oscillations

Velocity-proportional resistance

Forming the damped-motion equation

Light, critical and heavy damping

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