Damped and Forced Oscillations

Real oscillators never swing forever. Friction, air resistance and internal losses drain energy, so a freely oscillating system damps down; and if we want to keep it going we must drive it with a periodic force. This lesson develops both halves of that story rigorously. We solve the damped equation of motion with the auxiliary-equation method, classify the three damping regimes (under-, critical, over-), and then add a sinusoidal driver to derive the steady-state amplitude, the phase lag, the resonance condition and the quality factor. It is a beautiful application of the second-order linear ODE theory from the pure papers to a physical system.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M), following directly from SHM. The mathematics is the second-order constant-coefficient ODE — content you meet in the further-calculus compulsory material — now applied in a mechanics context. It is strongly AO2/AO3 (Paper 3 weights these at 25%/35%): you are expected to set up the model, solve it, classify the behaviour from the discriminant, and interpret resonance physically. Expect "show that", "hence classify", and "sketch" command words.

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2 Core theory — the damped equation and its three regimes

Add a resistive force proportional to velocity, $F_{\text{damp}} = -b\dot{x}$Fdamp​=−bx˙ with $b>0$b>0, to the spring model $m\ddot{x} = -kx$mx¨=−kx. The equation of motion is

$$m\ddot{x} + b\dot{x} + kx = 0 \quad\Longleftrightarrow\quad \ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0,$$mx¨+bx˙+kx=0⟺x¨+2γx˙+ω02​x=0,

where $\gamma = \dfrac{b}{2m}$γ=2mb​ is the damping coefficient and $\omega_0 = \sqrt{k/m}$ω0​=k/m​ the natural (undamped) angular frequency. This is a linear constant-coefficient ODE, so we try $x = e^{\lambda t}$x=eλt and obtain the auxiliary equation

$$\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0 \ \Longrightarrow\ \lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}.$$λ2+2γλ+ω02​=0 ⟹ λ=−γ±γ2−ω02​​.

The sign of the discriminant $\gamma^2 - \omega_0^2$γ2−ω02​ decides everything:

Discriminant	Roots	Name	General solution
$\gamma < \omega_0$γ<ω0​	complex conjugate	Under-damped	$x = A_0 e^{-\gamma t}\cos(\omega_d t + \phi)$x=A0​e−γtcos(ωd​t+ϕ)
$\gamma = \omega_0$γ=ω0​	real, repeated	Critically damped	$x = (C_1 + C_2 t)\,e^{-\gamma t}$x=(C1​+C2​t)e−γt
$\gamma > \omega_0$γ>ω0​	real, distinct	Over-damped	$x = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t}$x=C1​eλ1​t+C2​eλ2​t

Under-damped ($\gamma < \omega_0$γ<ω0​). The roots are $\lambda = -\gamma \pm i\omega_d$λ=−γ±iωd​ with the damped angular frequency

$$\omega_d = \sqrt{\omega_0^2 - \gamma^2} \ (< \omega_0),$$ωd​=ω02​−γ2​ (<ω0​),

so $x = e^{-\gamma t}\bigl(P\cos\omega_d t + Q\sin\omega_d t\bigr) = A_0 e^{-\gamma t}\cos(\omega_d t + \phi)$x=e−γt(Pcosωd​t+Qsinωd​t)=A0​e−γtcos(ωd​t+ϕ). This is an oscillation at the reduced frequency $\omega_d$ωd​, with amplitude decaying inside the exponential envelope $\pm A_0 e^{-\gamma t}$±A0​e−γt. The period $T_d = 2\pi/\omega_d$Td​=2π/ωd​ is longer than the undamped period.

Critically damped ($\gamma = \omega_0$γ=ω0​). The repeated root $\lambda = -\gamma$λ=−γ forces the $(C_1 + C_2 t)e^{-\gamma t}$(C1​+C2​t)e−γt form. This returns to equilibrium in the shortest possible time without overshooting — the design target for car suspensions and instrument needles.

Over-damped ($\gamma > \omega_0$γ>ω0​). Two distinct negative real roots $\lambda_{1,2} = -\gamma \pm \sqrt{\gamma^2-\omega_0^2}$λ1,2​=−γ±γ2−ω02​​; the motion is a sum of decaying exponentials with no oscillation, crawling back to equilibrium more slowly than the critical case. Both roots are negative (since $\sqrt{\gamma^2-\omega_0^2} < \gamma$γ2−ω02​​<γ), so every term decays — a damped system, whatever the regime, always returns to rest.

It is worth dwelling on why the repeated-root case needs the extra factor of $t$t. When the auxiliary equation has a single root $\lambda = -\gamma$λ=−γ, the function $e^{-\gamma t}$e−γt supplies only one independent solution, but a second-order ODE needs two to match two initial conditions (position and velocity). The standard theory supplies the missing one as $t\,e^{-\gamma t}$te−γt; you can verify by substitution that $t e^{-\gamma t}$te−γt solves $\ddot{x} + 2\gamma\dot{x} + \gamma^2 x = 0$x¨+2γx˙+γ2x=0. This is precisely why a critically damped system can still move away from and then back to equilibrium rather than simply collapsing — the linear-in-$t$t prefactor lets it carry an initial velocity.

Energy decay and the logarithmic decrement

For light damping the amplitude follows $A(t) = A_0 e^{-\gamma t}$A(t)=A0​e−γt; since energy scales as amplitude squared,

$$E(t) = E_0 e^{-2\gamma t}$$E(t)=E0​e−2γt

— energy decays at twice the rate of the amplitude. A convenient experimental measure of damping is the logarithmic decrement, the natural log of the ratio of successive peaks one period apart:

$$\delta = \ln\!\left(\frac{A_n}{A_{n+1}}\right) = \gamma T_d = \frac{2\pi\gamma}{\omega_d}.$$δ=ln(An+1​An​​)=γTd​=ωd​2πγ​.

It is constant for linear (velocity-proportional) damping, which makes a log-amplitude-against-cycle-number graph a straight line of gradient $-\delta$−δ. This is the practical experimental route: measure several successive peaks, plot $\ln A_n$lnAn​ against $n$n, and the gradient gives $\delta$δ, from which $\gamma = \delta/T_d$γ=δ/Td​ and ultimately the damping constant $b = 2m\gamma$b=2mγ follow. The related time constant $\tau = 1/\gamma$τ=1/γ is the time for the amplitude to fall to $1/e\approx37\%$1/e≈37% of its value; the energy time constant is $1/(2\gamma) = \tau/2$1/(2γ)=τ/2 because energy decays twice as fast. A lightly damped system (small $\gamma$γ) has a long $\tau$τ and rings for many cycles; a heavily damped one barely completes one swing.

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3 Forced oscillations and the steady state

Now apply a sinusoidal driving force $F_0\cos\Omega t$F0​cosΩt:

$$m\ddot{x} + b\dot{x} + kx = F_0\cos\Omega t \quad\Longleftrightarrow\quad \ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = \frac{F_0}{m}\cos\Omega t,$$mx¨+bx˙+kx=F0​cosΩt⟺x¨+2γx˙+ω02​x=mF0​​cosΩt,

where $\Omega$Ω is the driving angular frequency. The general solution is the complementary function (one of the three decaying solutions above — the transient) plus a particular integral — the steady state. After the transient dies away the system oscillates at the driving frequency:

$$x_{\text{ss}} = A(\Omega)\cos(\Omega t - \delta),$$xss​=A(Ω)cos(Ωt−δ), $$\boxed{A(\Omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \Omega^2)^2 + (2\gamma\Omega)^2}}}, \qquad \tan\delta = \frac{2\gamma\Omega}{\omega_0^2 - \Omega^2}.$$A(Ω)=(ω02​−Ω2)2+(2γΩ)2​F0​/m​​,tanδ=ω02​−Ω22γΩ​.

(The cleanest derivation writes the driver as $\operatorname{Re}(F_0 e^{i\Omega t})$Re(F0​eiΩt), tries $x = \operatorname{Re}(Ce^{i\Omega t})$x=Re(CeiΩt), and reads off $C = \dfrac{F_0/m}{\omega_0^2 - \Omega^2 + 2i\gamma\Omega}$C=ω02​−Ω2+2iγΩF0​/m​; the modulus gives $A$A and the argument gives $-\delta$−δ. This is De Moivre / complex-exponential technique straight from the pure papers.)

The word transient deserves emphasis. The complementary function always contains a decaying factor $e^{-\gamma t}$e−γt, so whatever the initial conditions, it dies away after a few time-constants $1/\gamma$1/γ and only the steady state survives. This is why the long-term behaviour is independent of how the system was started: the driver "forgets" the initial conditions. In the short term, though, the transient and steady-state oscillations beat against each other, which is why a struck bell or a plucked-then-bowed string sounds different at the start of a note than once it has settled. When a question imposes initial conditions on a driven system, you must therefore write $x = x_{\text{transient}} + x_{\text{ss}}$x=xtransient​+xss​ and fix the transient's constants so the sum matches the data at $t=0$t=0 — a frequent source of lost marks.

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4 Resonance and the quality factor

Resonance is the peak of $A(\Omega)$A(Ω). Since $A$A is largest when its denominator is smallest, write $D(u) = (\omega_0^2 - u)^2 + 4\gamma^2 u$D(u)=(ω02​−u)2+4γ2u with $u = \Omega^2$u=Ω2 and minimise:

$$\frac{\mathrm{d}D}{\mathrm{d}u} = -2(\omega_0^2 - u) + 4\gamma^2 = 0 \ \Longrightarrow\ \omega_0^2 - u = 2\gamma^2 \ \Longrightarrow\ u = \omega_0^2 - 2\gamma^2 .$$dudD​=−2(ω02​−u)+4γ2=0 ⟹ ω02​−u=2γ2 ⟹ u=ω02​−2γ2.

Hence the amplitude-resonance frequency is

$$\Omega_{\text{res}} = \sqrt{\omega_0^2 - 2\gamma^2} \qquad (\text{provided } \omega_0^2 > 2\gamma^2),$$Ωres​=ω02​−2γ2​(provided ω02​>2γ2),

slightly below $\omega_0$ω0​. The second derivative $\mathrm{d}^2D/\mathrm{d}u^2 = 2 > 0$d2D/du2=2>0 confirms this is a minimum of $D$D, hence a maximum of $A$A. For light damping $(\gamma \ll \omega_0)$(γ≪ω0​), $\Omega_{\text{res}} \approx \omega_0$Ωres​≈ω0​, and the peak amplitude is

$$A_{\max} \approx \frac{F_0}{2m\gamma\omega_0} = \frac{F_0}{b\omega_0},$$Amax​≈2mγω0​F0​​=bω0​F0​​,

inversely proportional to the damping: light damping gives a tall, sharp peak; heavy damping a low, broad one (and if $\omega_0^2 \le 2\gamma^2$ω02​≤2γ2 there is no peak at all — the amplitude just falls from its static value). The sharpness is captured by the quality factor

$$\boxed{Q = \frac{\omega_0}{2\gamma}} = 2\pi\times\frac{\text{energy stored}}{\text{energy lost per cycle}}.$$Q=2γω0​​​=2π×energy lost per cycleenergy stored​.

A high $Q$Q means a sharp resonance and slow energy loss (a tuning fork has $Q\sim10^3$Q∼103; a car suspension is deliberately low-$Q$Q).

The phase lag $\delta$δ tells the rest of the story:

Regime	Phase lag $\delta$δ	Interpretation
$\Omega \ll \omega_0$Ω≪ω0​	$\approx 0$≈0	response follows the driver
$\Omega = \omega_0$Ω=ω0​	exactly $\tfrac{\pi}{2}$2π​	driver in phase with velocity → maximum power transfer
$\Omega \gg \omega_0$Ω≫ω0​	$\approx \pi$≈π	response in antiphase with the driver

The $\delta = \pi/2$δ=π/2 result is the deep one: at resonance the force is in step with the velocity, not the displacement, so it does positive work on every part of the cycle — that is why the amplitude builds up.

Why resonance matters

The same amplitude curve governs a remarkable range of physical systems, and the engineering lesson is always the same: either tune to resonance to amplify a wanted response, or avoid it to prevent a catastrophic one.

System	Role of resonance
Wind instruments, organ pipes, strings	Driven at a natural frequency, the air column or string resonates strongly — this is how a note is sustained
Radio / LCR tuning circuit	The receiver is tuned so its natural frequency matches the carrier, picking out one station
Bridges and tall buildings	Engineers ensure natural frequencies lie away from wind, footfall or seismic driving frequencies; dampers are added to lower $Q$Q
Vehicle suspension	Deliberately near-critical (low $Q$Q) so bumps do not set up a sustained bounce
MRI scanners, atomic clocks	Exploit extremely high-$Q$Q resonances for precision

A famous cautionary tale is a suspension bridge driven into large oscillations by a steady wind — a vivid reminder that an undamped structure near resonance has, in our idealised model, an unbounded amplitude. Real structures survive only because of damping (and because true resonance is rarely sustained), which is exactly the $A_{\max}\propto 1/b$Amax​∝1/b relationship made physical.

The quality factor also fixes how selective the resonance is. The peak spans a band of driving frequencies of width roughly $\Delta\Omega \approx 2\gamma$ΔΩ≈2γ between the two points where the power has fallen to half its maximum, so $Q = \omega_0/\Delta\Omega$Q=ω0​/ΔΩ: a radio circuit with $Q = 100$Q=100 tuned to $\omega_0$ω0​ responds strongly only within about $1\%$1% of that frequency, which is what lets it separate adjacent stations. High $Q$Q therefore means both a tall peak and a narrow one — sharp selectivity bought at the price of slow energy loss. This dual role of $Q$Q, controlling peak height, peak width and ringing time all at once, is why it is the single most quoted number when an engineer specifies an oscillator.

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5 Worked examples with mark scheme

Example 1 — classifying the damping

A damped oscillator has $m = 0.5\,\text{kg}$m=0.5kg, $k = 50\,\text{N m}^{-1}$k=50N m−1, $b = 2\,\text{N s m}^{-1}$b=2N s m−1. Find (a) the natural frequency, (b) the damped frequency, (c) the type of damping.

(a) $\omega_0 = \sqrt{k/m} = \sqrt{50/0.5} = \sqrt{100} = 10\ \text{rad s}^{-1}.$ω0​=k/m​=50/0.5​=100​=10 rad s−1. (M1 A1)

(b) $\gamma = \dfrac{b}{2m} = \dfrac{2}{1} = 2\ \text{s}^{-1}$γ=2mb​=12​=2 s−1, so

$$\omega_d = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{100 - 4} = \sqrt{96} = 4\sqrt{6} \approx 9.80\ \text{rad s}^{-1}. \quad (\textbf{M1 A1})$$ωd​=ω02​−γ2​=100−4​=96​=46​≈9.80 rad s−1.(M1 A1)

(c) Since $\gamma = 2 < \omega_0 = 10$γ=2<ω0​=10, the discriminant $\gamma^2 - \omega_0^2 < 0$γ2−ω02​<0: the system is under-damped. (A1)

(M1 $\omega_0=\sqrt{k/m}$ω0​=k/m​; A1; M1 $\omega_d=\sqrt{\omega_0^2-\gamma^2}$ωd​=ω02​−γ2​; A1 $4\sqrt6$46​; A1 "under-damped" with reason.)

Example 2 — steady-state amplitude on and off resonance