Forced harmonic oscillator

A forced harmonic oscillator responds at both its natural and forcing frequencies.

A nonrotating frame whose origin travels on a circle carries a spatially uniform fictitious force. A harmonic trap fixed to that origin therefore drives its relative coordinate by the displayed equation. Away from resonance its circular particular response has radius $R{\Omega }^2/({\omega }^2-{\Omega }^2)$. At ${\omega }^2={\Omega }^2\ne 0$, the response grows linearly in time. Rotating the coordinate axes adds Coriolis and centrifugal terms but does not change this physical resonance condition.

For $\ddot{x}+\gamma \dot{x}+{\omega }_0^2(x-x_0)=0$, the particular path $x=x_0$ is a solution precisely when ${\ddot{x}}_0+\gamma {\dot{x}}_0=0$. An exponentially moving equilibrium $x_0=d{e}^{-\gamma t}$ satisfies this condition. If the initial position and velocity match it, uniqueness removes all homogeneous oscillations even when the natural system is underdamped. The motion is sustained dynamically by linear drag, with zero instantaneous spring extension.

For a forced harmonic oscillator with constant viscous damping coefficient $m\gamma >0$, the average power divided by its maximum is ${\gamma }^2{\omega }^2/[({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2]$. It peaks at ${\omega }_0=\sqrt{k/m}$. Its two positive half-power angular frequencies are ${\omega }_{1,2}=(\sqrt{4{\omega }_0^2+{\gamma }^2}\mp \gamma )/2$. Hence their separation is exactly $\gamma$, and their product is ${\omega }_0^2$. The narrow-resonance approximation ${\omega }_{1,2}\simeq {\omega }_0\mp \gamma /2$ is needed for symmetry about the peak, not for the exact bandwidth in angular-frequency units.

For the forced harmonic oscillator $m\ddot{z}+m\gamma \dot{z}+kz=F_0\cos (\omega t+{\varphi }_0)$, a response $z=A\cos \omega t$ requires $A(k-m{\omega }^2)=F_0\cos {\varphi }_0$ and $Am\gamma \omega =F_0\sin {\varphi }_0$. Squaring and adding gives the amplitude $A=F_0/\sqrt{(k-m{\omega }^2{)}^2+(m\gamma \omega {)}^2}$ for positive $A,F_0$. The two coefficient equations determine the full quadrant of the forcing phase, while their ratio alone can lose that information. In particular the amplitude cannot generally be independent of damping: at $m=k=F_0=\omega =1$ and ${\varphi }_0=\pi /2$, it is $A=1/\gamma$.

For the convention $x=\mathrm{Im}(RF{e}^{i\omega t})$, viscous damping $b$ gives $R=[m({\omega }_0^2-{\omega }^2)+ib\omega {]}^{-1}$. Put $\delta =\omega -{\omega }_0$ and $\gamma =b/(2m)$. If both are small relative to ${\omega }_0$, the displayed approximation follows. Its squared modulus has half-width $\gamma$, and the displacement lags the force by $\pi /2$ at $\omega ={\omega }_0$.

Let $y^{\prime \prime }+{\omega }^{2}y=F(t)$ with $\omega >0$, zero initial displacement and velocity, and real-valued forcing with ${\int }_0^{\infty }|F(t)|dt<\infty$. The causal Green function gives $y(t)={\omega }^{-1}{\int }_0^{t}F(s)\sin [\omega (t-s)]ds$. Expanding the sine difference, the two coefficients tend to finite limits by absolute convergence. Thus $y(t)$ approaches a sinusoid whose amplitude is the displayed modulus of the one-sided Fourier transform, divided by $\omega$. The displacement error is at most ${\omega }^{-1}{\int }_t^{\infty }|F(s)|ds$, which tends to zero. This describes the surviving free oscillation after a transient force; an arbitrary nonzero initial oscillation must be added before taking the final amplitude.

For a supported mass $m$, total spring constant $k$, and viscous coefficient $b$, a harmonic applied force has response $|x|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. Far above the natural angular frequency, light damping gives $|x|\simeq F_0/(m{\omega }^2)$. For a prescribed displacement bound $x_{\max }$, this estimates $m\gtrsim F_0/({\omega }^2{x}_{\max })$, but the finite spring term can make the estimated equality slightly violate the bound. If $F_0/x_{\max }>b\omega$, the high-mass exact threshold isThis model describes a specified external force. Motion transmitted through a moving support instead has a base-displacement forcing term and a different dependence on damping.

For a lightly damped forced harmonic oscillator with known force amplitude $F_0$, measure the static displacement $x_s$, the resonance angular frequency ${\omega }_r$ and the peak displacement $x_p$. The response is $|x_0|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. To leading order at large quality factor,The exact displacement resonance is ${\omega }_r^2=k/m-b^2/(2m^2)$, whereas the undamped natural angular frequency is ${\omega }_0=\sqrt{k/m}$. A lightly damped half-power bandwidth obeys $\Delta \omega \simeq b/m$ and $Q\simeq {\omega }_0/\Delta \omega$. Half power corresponds to displacement $x_p/\sqrt{2}$, not $x_p/2$. Graph readings should be reported as estimates rather than extra significant digits from the model inversion.

This damped harmonic oscillator receives a constant force only while moving toward increasing $x$. For $0<k<\omega$, a positive maximum $X$ evolves without forcing to a minimum $-rX$, then with forcing to the next maximum $r^{2}X+(1+r)a/{\omega }^2$, where $r=e^{-k\pi /\sqrt{{\omega }^2-k^2}}$. Its unique periodic maximum is $a/[{\omega }^2(1-r)]$ and its period is $2\pi /\sqrt{{\omega }^2-k^2}$. Position and velocity are continuous at switches, but acceleration has jumps. The slope $r^2<1$ of this turning-point return map shows attraction of nearby turning-point amplitudes.