Damped, Driven Oscillators and Resonance

Derivation

Damped, Driven Oscillators and Resonance

D-032 Home PU-101 Threads force · energy · waves Depends on Small Oscillations and Simple Harmonic Motion

Statement

For a mass on a linear spring subject to linear (viscous) damping and a sinusoidal driving force \(F(t)=F_0\cos\omega t\), the equation of motion \(m\ddot{x}+b\dot{x}+kx=F_0\cos\omega t\) has a long-time steady-state solution \(x(t)=A(\omega)\cos\!\big(\omega t-\delta(\omega)\big)\). We derive the amplitude \(A(\omega)\), the phase lag \(\delta(\omega)\), the driving frequency \(\omega_{\text{res}}\) that maximises \(A\), and show that the sharpness of the resonance peak is governed by the quality factor \(Q=\omega_0/\gamma\).

Why it matters

The driven damped oscillator is the canonical linear-response system: a single second-order ODE that recurs, essentially unchanged, in mechanical vibration, RLC circuits, atomic and molecular absorption lines, gravitational-wave bar detectors, and the Lorentz model of the dielectric constant. Its transfer function \(A(\omega)e^{-i\delta}\) is the prototype of every resonance curve in physics.

Resonance is where the payoff lives: driving near \(\omega_0\) lets a small periodic force build a large response, limited only by dissipation. The same mathematics tells an engineer how to avoid shaking a bridge apart and a spectroscopist how to read a linewidth as an inverse lifetime.

Assumptions

The restoring force is linear (Hooke's law), \(F_{\text{spring}}=-kx\).If the spring is nonlinear the response is no longer a single-frequency sinusoid; harmonics appear, the resonance curve bends over (Duffing oscillator) and can become multivalued and hysteretic.

Damping is linear in velocity, \(F_{\text{damp}}=-b\dot{x}\).Dry (Coulomb) friction or turbulent \(\propto\dot{x}^2\) drag makes the equation nonlinear; the steady state is no longer a pure cosine and the Lorentzian lineshape below is only approximate.

The drive is a single-frequency sinusoid \(F_0\cos\omega t\) of fixed amplitude and frequency.A broadband or amplitude-modulated drive requires superposing responses via the Fourier transform of \(F(t)\); a single \(A(\omega)\) no longer describes the motion.

We seek only the steady state, having let transients die.The homogeneous solution \(\propto e^{-\gamma t/2}\) is discarded. Dropped, the full solution contains a decaying transient at the natural frequency that beats against the drive early on; for times \(t\lesssim 1/\gamma\) our formula is incomplete.

Parameters \(m,b,k\) are constant and the system is one-dimensional.Time-varying stiffness gives a parametric oscillator (Mathieu equation) with its own instability tongues; coupled coordinates give normal modes, each an oscillator of this form.

Derivation

\[ m\ddot{x}+b\dot{x}+kx=F_0\cos\omega t \]

Newton's second law with the linear spring, viscous damper and applied drive summed as forces. A

\[ \ddot{x}+\gamma\dot{x}+\omega_0^2\,x=f_0\cos\omega t,\qquad \gamma\equiv\frac{b}{m},\ \ \omega_0^2\equiv\frac{k}{m},\ \ f_0\equiv\frac{F_0}{m} \]

Divide through by \(m\) and name the natural frequency \(\omega_0\), damping rate \(\gamma\) and force per unit mass \(f_0\). This isolates the two physical rates. A

\[ \ddot{z}+\gamma\dot{z}+\omega_0^2\,z=f_0\,e^{i\omega t},\qquad x=\operatorname{Re}z \]

Complexify: the drive \(\cos\omega t=\operatorname{Re}\,e^{i\omega t}\), and because the operator has real coefficients the real part of the complex solution solves the real equation. This turns trigonometry into algebra. B

\[ z(t)=\tilde{A}\,e^{i\omega t}\ \Rightarrow\ \big(-\omega^2+i\gamma\omega+\omega_0^2\big)\tilde{A}\,e^{i\omega t}=f_0\,e^{i\omega t} \]

Steady-state ansatz: the response oscillates at the drive frequency \(\omega\) with a complex amplitude \(\tilde{A}\) carrying magnitude and phase. Each \(\tfrac{d}{dt}\to i\omega\). Legal because the equation is linear with constant coefficients. B

\[ \tilde{A}=\frac{f_0}{(\omega_0^2-\omega^2)+i\gamma\omega} \]

Cancel the common \(e^{i\omega t}\) (never zero) and divide. The complex amplitude is fixed algebraically. A

\[ \tilde{A}=|\tilde{A}|\,e^{-i\delta},\qquad |\tilde{A}|=\frac{f_0}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad \tan\delta=\frac{\gamma\omega}{\omega_0^2-\omega^2} \]

Write the complex number in polar form. The modulus is \(f_0\) over the modulus of the denominator; the phase of \(\tilde{A}\) is minus the phase of the denominator, defining the lag \(\delta\ge 0\). B

\[ x(t)=\operatorname{Re}\!\big(|\tilde{A}|e^{-i\delta}e^{i\omega t}\big)=A(\omega)\cos(\omega t-\delta),\qquad A(\omega)=\frac{f_0}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}} \]

Take the real part to recover the physical displacement. The response lags the drive by \(\delta\). A

\[ \frac{d}{d\omega}\Big[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2\Big]=0\ \Rightarrow\ 2(\omega_0^2-\omega^2)(-2\omega)+2\gamma^2\omega=0 \]

The amplitude is maximal where its denominator is minimal, so differentiate the radicand and set it to zero. B

\[ \omega\big[-4(\omega_0^2-\omega^2)+2\gamma^2\big]=0\ \Rightarrow\ \omega_{\text{res}}=\sqrt{\omega_0^2-\tfrac{1}{2}\gamma^2} \]

Discard the trivial \(\omega=0\) root; solve the bracket. The amplitude resonance sits slightly below \(\omega_0\), pulled down by damping. Real only if \(\gamma^2<2\omega_0^2\). B

\[ A_{\max}=A(\omega_{\text{res}})=\frac{f_0}{\gamma\sqrt{\omega_0^2-\tfrac14\gamma^2}}\ \xrightarrow[\ \gamma\ll\omega_0\ ]{}\ \frac{f_0}{\gamma\omega_0},\qquad Q\equiv\frac{\omega_0}{\gamma} \]

Substitute \(\omega_{\text{res}}\) back into \(A\). For light damping the peak height is \(A_{\max}\approx f_0/(\gamma\omega_0)=Q\,f_0/\omega_0^2=Q\,x_{\text{stat}}\): the static deflection amplified by the quality factor \(Q\). C

Result

\[ A(\omega)=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad \tan\delta=\frac{\gamma\omega}{\omega_0^2-\omega^2},\qquad \omega_{\text{res}}=\sqrt{\omega_0^2-\tfrac12\gamma^2},\qquad Q=\frac{\omega_0}{\gamma} \]

Reading. The steady response oscillates at the drive frequency with amplitude \(A(\omega)\) and lags the force by \(\delta\). Far below resonance the mass moves in phase with the force (\(\delta\to 0\), spring-controlled); exactly at \(\omega=\omega_0\) the lag is \(\delta=\pi/2\) (damping-controlled, maximum power transfer); far above it the response is antiphase (\(\delta\to\pi\), inertia-controlled). The peak amplitude scales as \(Q\) times the static deflection, and a large \(Q\) means a tall, narrow resonance whose fractional full width at half power is \(\Delta\omega/\omega_0\approx 1/Q\).

Units check. \(f_0=F_0/m\) has units \(\mathrm{N\,kg^{-1}}=\mathrm{m\,s^{-2}}\). The denominator's terms are each frequency-squared-squared, \(\mathrm{s^{-4}}\), so its square root has units \(\mathrm{s^{-2}}\). Thus \(A\) carries \(\mathrm{(m\,s^{-2})/(s^{-2})=m}\), a length, as required. In \(\tan\delta\), numerator \(\gamma\omega\) and denominator \(\omega_0^2-\omega^2\) are both \(\mathrm{s^{-2}}\), so \(\delta\) is dimensionless. \(Q=\omega_0/\gamma\) is (rad s\(^{-1}\))/(s\(^{-1}\)), dimensionless.

Limiting cases

Low frequency \(\omega\ll\omega_0\): \(A\to f_0/\omega_0^2=F_0/k\), the static (DC) deflection; \(\delta\to 0\), motion follows the force. Spring-dominated regime.
Resonance \(\omega=\omega_0\): denominator \(=\gamma\omega_0\), so \(A=f_0/(\gamma\omega_0)=Q\,F_0/k\); \(\delta=\pi/2\) exactly. Velocity is in phase with the force, so the drive delivers maximum average power.
High frequency \(\omega\gg\omega_0\): \(A\to f_0/\omega^2\to 0\); \(\delta\to\pi\). The mass cannot keep up; inertia-dominated, motion antiphase to force.
Undamped \(\gamma\to 0\): \(A\to f_0/|\omega_0^2-\omega^2|\) diverges at \(\omega_0\); \(Q\to\infty\), \(\omega_{\text{res}}\to\omega_0\). Resonance catastrophe of the ideal oscillator.
Heavy damping \(\gamma\ge\sqrt{2}\,\omega_0\): \(\omega_{\text{res}}\) becomes imaginary; \(A(\omega)\) has no interior peak and falls monotonically from its DC value. No resonance.

Breaks when

Amplitudes grow large enough to leave the linear regime. A real spring stiffens or softens (\(kx\to kx+\alpha x^3\)); the resonance curve leans (Duffing), becomes multivalued, and jumps discontinuously as \(\omega\) is swept. The single-valued Lorentzian \(A(\omega)\) is then wrong.
The transient has not yet decayed. For \(t\lesssim 1/\gamma\) (high-\(Q\) systems can mean many cycles) the discarded homogeneous solution \(e^{-\gamma t/2}\cos(\omega_1 t)\) beats against the drive; the observed amplitude rings up rather than sitting at \(A(\omega)\).
Damping is not linear in velocity. Coulomb friction (constant magnitude) or quadratic drag makes the equation nonlinear; the steady state acquires harmonics and the phase relation \(\tan\delta=\gamma\omega/(\omega_0^2-\omega^2)\) fails.
The drive frequency is swept quickly through resonance. A time-dependent \(\omega(t)\) violates the fixed-frequency ansatz; the response chirps and rings up asymmetrically (relevant to swept-sine testing and gravitational-wave chirps).

Failure modes

Confusing the three resonant frequencies. Amplitude resonance is at \(\sqrt{\omega_0^2-\tfrac12\gamma^2}\), velocity/power resonance is exactly at \(\omega_0\), and the free damped oscillation is at \(\omega_1=\sqrt{\omega_0^2-\tfrac14\gamma^2}\). Students quote "\(\omega_0\)" for all three.
Using \(\gamma=b/m\) versus \(\gamma=b/2m\) inconsistently. Many texts write \(\ddot{x}+2\beta\dot{x}+\omega_0^2 x=\cdots\) with \(\beta=b/2m\). Mixing conventions puts stray factors of 2 in \(Q\), \(\omega_{\text{res}}\) and the linewidth.
Forgetting that the drive amplitude can itself depend on \(\omega\). If the physical forcing is \(m\) times a base acceleration, or an eccentric of fixed displacement, then \(F_0\propto\omega^2\); the peak shifts and the low-frequency limit changes.
Sign/branch error in \(\delta\). Above \(\omega_0\), \(\omega_0^2-\omega^2<0\), so \(\arctan\) must be taken in the second quadrant to give \(\delta\in(\pi/2,\pi)\); a naive calculator value returns a spurious negative lag.
Reading \(A_{\max}=f_0/(\gamma\omega_0)\) as exact. It is the light-damping approximation; the exact peak is \(f_0/(\gamma\sqrt{\omega_0^2-\gamma^2/4})\).
Setting the transient to zero without justification. The steady state is only reached after transients decay; quoting \(A(\omega)\) for an experiment that ran a single cycle is wrong.

Discussion

The phase lag \(\delta\) is the clearest way into resonance. The three terms in \(m\ddot{x}+b\dot{x}+kx\) are the inertial, dissipative and elastic forces, with phases \(\omega t+\pi\), \(\omega t+\pi/2\) and \(\omega t\) relative to the displacement. At low \(\omega\) the spring term dominates the balance against the drive and the mass moves with the force; at high \(\omega\) inertia dominates and the mass moves against it. Exactly at \(\omega_0\) the spring and inertial terms cancel identically (\(-m\omega_0^2+k=0\)), leaving the damping force alone to balance the drive. There the response is in quadrature with the force, velocity is in phase with the force, and the time-averaged power input \(\langle F\dot{x}\rangle\) is maximised. Power resonance therefore sits precisely at \(\omega_0\), even though the amplitude peak is pulled slightly lower.

The quality factor \(Q=\omega_0/\gamma\) has three equivalent readings every physicist should carry: it is the amplification of the static deflection at resonance (\(A_{\max}\approx Q\,F_0/k\)); it is \(2\pi\) times the ratio of stored energy to energy dissipated per cycle (\(Q=2\pi\,E/\Delta E_{\text{cycle}}\)); and it is the inverse fractional linewidth, \(Q=\omega_0/\Delta\omega\), where \(\Delta\omega\) is the full width between the half-power points. A tuning fork has \(Q\sim10^3\), a quartz crystal \(\sim10^5\), an atomic-clock transition \(\sim10^{9}\) or more, and LIGO's test-mass suspensions reach \(\sim10^7\). Large \(Q\) means both a sharp filter and a long ring-down, \(\tau=2/\gamma=2Q/\omega_0\); the two are the same fact viewed in frequency and in time.

The complex amplitude \(\tilde{A}(\omega)=f_0/[(\omega_0^2-\omega^2)+i\gamma\omega]\) is the frequency-domain Green's function of the oscillator. Its two poles sit in the lower half of the complex-\(\omega\) plane at \(\omega=\pm\omega_1-i\gamma/2\), with \(\omega_1=\sqrt{\omega_0^2-\gamma^2/4}\); the imaginary part is the decay rate of the free oscillation and the real part its ringing frequency. Because the poles lie in the lower half-plane the time response is causal (zero for \(t<0\)), and the real and imaginary parts of \(\tilde{A}(\omega)\) obey the Kramers–Kronig relations. This is the microscopic origin of the Lorentzian absorption line: \(\operatorname{Im}\tilde{A}\propto\gamma\omega/[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2]\) is the power absorbed, and near resonance it reduces to a Lorentzian of half-width \(\gamma/2\). The same pole structure underlies the electric susceptibility in the Lorentz oscillator model of dielectrics and the natural linewidth of a radiating atom, where \(\gamma\) is the spontaneous-emission rate.

Common misconceptions. Resonance does not require zero damping — it requires the drive to match a natural frequency; damping only limits how tall and how narrow the peak becomes. "Resonance frequency" is ambiguous: state whether you mean amplitude (\(\sqrt{\omega_0^2-\gamma^2/2}\)) or power (\(\omega_0\)). And the steady-state amplitude does not depend on initial conditions at all — those affect only the transient, which decays away.

Worked examples

A mass \(m=0.50\ \mathrm{kg}\) on a spring \(k=200\ \mathrm{N\,m^{-1}}\) has damping coefficient \(b=1.0\ \mathrm{kg\,s^{-1}}\), driven by \(F_0=2.0\ \mathrm{N}\). Find \(\omega_0\), \(Q\), the resonant amplitude and the phase lag when driven at \(\omega=\omega_0\).

Set up the reduced parameters symbolically first. A

\[ \omega_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.50}}=\sqrt{400}=20\ \mathrm{rad\,s^{-1}},\qquad \gamma=\frac{b}{m}=\frac{1.0}{0.50}=2.0\ \mathrm{s^{-1}} \]

Natural frequency and damping rate from their definitions. A

\[ Q=\frac{\omega_0}{\gamma}=\frac{20}{2.0}=10,\qquad f_0=\frac{F_0}{m}=\frac{2.0}{0.50}=4.0\ \mathrm{m\,s^{-2}} \]

Quality factor and force per unit mass. A

\[ A(\omega_0)=\frac{f_0}{\gamma\omega_0}=\frac{4.0}{(2.0)(20)}=0.10\ \mathrm{m},\qquad \delta=\frac{\pi}{2}\ (90^\circ) \]

At \(\omega=\omega_0\) the denominator reduces to \(\gamma\omega_0\) and \(\tan\delta\to\infty\). The static deflection is \(F_0/k=0.010\ \mathrm{m}\), so the amplitude is \(Q=10\) times larger, as expected. B

\[ \omega_0=20\ \mathrm{rad\,s^{-1}},\quad Q=10,\quad A(\omega_0)=0.10\ \mathrm{m},\quad \delta=90^\circ \]

Reading. A modest \(2\ \mathrm{N}\) drive produces a \(10\ \mathrm{cm}\) swing because resonance amplifies the \(1\ \mathrm{cm}\) static deflection tenfold. Units check. \(f_0/(\gamma\omega_0)=(\mathrm{m\,s^{-2}})/(\mathrm{s^{-1}\cdot s^{-1}})=\mathrm{m}\). Consistent.

For the same system, at what driving frequency is the amplitude actually maximal, and what is the fractional bandwidth \(\Delta\omega/\omega_0\) between the half-power points?

Use the amplitude-resonance and linewidth formulas symbolically. B

\[ \omega_{\text{res}}=\sqrt{\omega_0^2-\tfrac12\gamma^2}=\sqrt{400-\tfrac12(2.0)^2}=\sqrt{398}=19.95\ \mathrm{rad\,s^{-1}} \]

Amplitude peak, pulled just below \(\omega_0\) by the small damping. B

\[ A_{\max}=\frac{f_0}{\gamma\sqrt{\omega_0^2-\tfrac14\gamma^2}}=\frac{4.0}{2.0\sqrt{400-1}}=\frac{4.0}{2.0\times19.975}=0.1001\ \mathrm{m} \]

Exact peak height; it exceeds \(A(\omega_0)=0.100\ \mathrm{m}\) only in the fourth decimal, confirming that for \(Q=10\) the distinction is tiny. B

\[ \frac{\Delta\omega}{\omega_0}\approx\frac{1}{Q}=\frac{1}{10}=0.10\ \Rightarrow\ \Delta\omega\approx\gamma=2.0\ \mathrm{rad\,s^{-1}} \]

The full width between half-power points is \(\Delta\omega\approx\gamma\) for high \(Q\); here a 10% bandwidth. C

\[ \omega_{\text{res}}\approx19.95\ \mathrm{rad\,s^{-1}},\quad A_{\max}\approx0.100\ \mathrm{m},\quad \Delta\omega\approx2.0\ \mathrm{rad\,s^{-1}}\ (10\%) \]

Reading. At \(Q=10\) the amplitude peak lies only \(0.05\ \mathrm{rad\,s^{-1}}\) below \(\omega_0\) and is barely taller than the value at \(\omega_0\); the resonance is 10% wide. Units check. \(\Delta\omega\) has units \(\mathrm{s^{-1}}\); the fractional width is dimensionless.

Problems

An oscillator has \(\omega_0=50\ \mathrm{rad\,s^{-1}}\) and \(\gamma=5.0\ \mathrm{s^{-1}}\). Compute \(Q\) and the number of oscillations for the free amplitude to fall to \(1/e\).
Solution
\(Q=\omega_0/\gamma=50/5.0=10\). The free amplitude decays as \(e^{-\gamma t/2}\), so the \(1/e\) time is \(\tau=2/\gamma=0.40\ \mathrm{s}\). The oscillation period is \(T\approx 2\pi/\omega_0=0.1257\ \mathrm{s}\), giving \(\tau/T=0.40/0.1257\approx3.2\) oscillations. (Equivalently, the amplitude \(1/e\) time in cycles is \(Q/\pi\approx3.2\).)
A driven oscillator with \(m=2.0\ \mathrm{kg}\), \(k=800\ \mathrm{N\,m^{-1}}\), \(b=8.0\ \mathrm{kg\,s^{-1}}\) is driven at \(\omega=15\ \mathrm{rad\,s^{-1}}\) with \(F_0=10\ \mathrm{N}\). Find the steady-state amplitude and phase lag.
Solution
\(\omega_0=\sqrt{800/2.0}=20\ \mathrm{rad\,s^{-1}}\), \(\gamma=8.0/2.0=4.0\ \mathrm{s^{-1}}\), \(f_0=10/2.0=5.0\ \mathrm{m\,s^{-2}}\). Denominator terms: \(\omega_0^2-\omega^2=400-225=175\ \mathrm{s^{-2}}\); \(\gamma\omega=4.0\times15=60\ \mathrm{s^{-2}}\). \(\sqrt{175^2+60^2}=\sqrt{30625+3600}=\sqrt{34225}=185.0\ \mathrm{s^{-2}}\). \(A=5.0/185.0=0.0270\ \mathrm{m}=2.70\ \mathrm{cm}\). \(\tan\delta=60/175=0.343\Rightarrow\delta=18.9^\circ\) (drive below resonance, so \(\delta<90^\circ\)).
Show that the two frequencies at which the average power absorbed falls to half its resonant maximum satisfy \(\omega_0^2-\omega^2=\pm\gamma\omega\), and hence that the full width at half maximum in power is \(\Delta\omega\approx\gamma\) for light damping.
Solution
The average power absorbed is \(\langle P\rangle=\tfrac12 b\omega^2 A^2\propto \omega^2/[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2]\). Near resonance the numerator varies slowly, so \(\langle P\rangle\) is maximal where the denominator is minimal, i.e. at \(\omega\approx\omega_0\), value \(\propto1/(\gamma^2\omega_0^2)\). Half power requires the denominator to double: \((\omega_0^2-\omega^2)^2+\gamma^2\omega^2=2\gamma^2\omega^2\), i.e. \((\omega_0^2-\omega^2)^2=\gamma^2\omega^2\), so \(\omega_0^2-\omega^2=\pm\gamma\omega\). Writing \(\omega=\omega_0\pm\tfrac12\Delta\omega\) and using \(\omega_0^2-\omega^2\approx-2\omega_0(\omega-\omega_0)\) gives \(|2\omega_0(\omega-\omega_0)|=\gamma\omega_0\), so each half-width is \(\gamma/2\) and the FWHM is \(\Delta\omega\approx\gamma\). Hence \(Q=\omega_0/\Delta\omega\).
A tuning fork rings at \(f_0=440\ \mathrm{Hz}\) and its sound amplitude decays to \(1/e\) in \(2.0\ \mathrm{s}\). Estimate its quality factor \(Q\) and the resonance linewidth \(\Delta f\).
Solution
\(\omega_0=2\pi f_0=2\pi(440)=2765\ \mathrm{rad\,s^{-1}}\). Amplitude decays as \(e^{-\gamma t/2}\) with \(1/e\) time \(2/\gamma=2.0\ \mathrm{s}\Rightarrow\gamma=1.0\ \mathrm{s^{-1}}\). \(Q=\omega_0/\gamma=2765/1.0\approx2.8\times10^3\). Linewidth \(\Delta f=f_0/Q=440/2765\approx0.16\ \mathrm{Hz}\) (equivalently \(\Delta\omega=\gamma=1.0\ \mathrm{rad\,s^{-1}}\Rightarrow\Delta f=\gamma/2\pi\approx0.16\ \mathrm{Hz}\)).
An oscillator with \(\omega_0=100\ \mathrm{rad\,s^{-1}}\) is driven exactly at resonance and reaches a steady amplitude of \(5.0\ \mathrm{cm}\); the static (DC) deflection under the same peak force is \(0.20\ \mathrm{cm}\). Find \(Q\), \(\gamma\), and the amplitude-resonance frequency \(\omega_{\text{res}}\).
Solution
At \(\omega_0\), \(A(\omega_0)=Q\times(F_0/k)=Q\,x_{\text{stat}}\), so \(Q=A(\omega_0)/x_{\text{stat}}=5.0/0.20=25\). Then \(\gamma=\omega_0/Q=100/25=4.0\ \mathrm{s^{-1}}\). Amplitude resonance: \(\omega_{\text{res}}=\sqrt{\omega_0^2-\tfrac12\gamma^2}=\sqrt{10000-8}=\sqrt{9992}=99.96\ \mathrm{rad\,s^{-1}}\), essentially \(\omega_0\) for this high-\(Q\) system.