Damped harmonic oscillator

Ringdown is the freely decaying motion of a resonator after an applied drive is removed. For an underdamped damped harmonic oscillator, the amplitude decays as $e^{-\gamma t}$ while the angular frequency is $\nu =\sqrt{{\Omega }^2-{\gamma }^2}$. In the weak-damping limit, the cycle-averaged stored energy decays as $e^{-2\gamma t}$, giving the quality factor from ringdown $Q\simeq \Omega /(2\gamma )$.

For a damped harmonic oscillator $\ddot{y}+\gamma \dot{y}+{\omega }_0^{2}y=0$, differentiation of the mechanical energy $E=m({\dot{y}}^2+{\omega }_0^2{y}^2)/2$ gives $\dot{E}=-m\gamma {\dot{y}}^2$. This identity is exact. In the weak-damping regime $\gamma \ll {\omega }_0$, the leading period average has equal kinetic energy and potential energy, so $\overline{{\dot{y}}^2}\simeq \overline{E}/m$ and $\dot{\overline{E}}\simeq -\gamma \overline{E}$. Thus the averaged energy has exponential decay, while the instantaneous energy has a small oscillatory correction. Its derivative vanishes at every turning point, ruling out an exact pointwise nonconstant exponential. The quality factor is $Q\simeq {\omega }_0/\gamma$ in this approximation.

A spring's support displacement supplies force $k{x}_1$, while a damper attached to a fixed wall supplies $-b\dot{x}$. Measuring from static equilibrium gives $m\ddot{x}+b\dot{x}+kx=k{x}_1$. The sinusoidal displacement transfer function is the displayed ratio, with low-frequency limit1 and high-frequency limit $-k/(m{\omega }^2)$. Moving both the spring and damper supports would introduce a different numerator. A forced particular response must be supplemented by a transient unless initial conditions already select steady motion.

For fixed spring constant $k$ and linear damping coefficient $b$, replacing mass $m$ by $\lambda m$ changes the exponential envelope rate from $\gamma =b/(2m)$ to $\gamma /\lambda$. The damped angular frequency becomes ${\omega }_{d,\lambda }^2=({\omega }_d^2+{\gamma }^2)/\lambda -{\gamma }^2/{\lambda }^2$. Thus envelope decay times scale exactly as mass, while the familiar inverse-square-root frequency scaling is a weak-damping approximation. A constant gravitational force shifts equilibrium without changing either rate.

For distinct roots $r_{\pm }=(-\gamma \pm \sqrt{{\gamma }^2-4mk})/(2m)$, the deterministic response is $X=(v_0-r_{-}{x}_0)e^{r_{+}t}/(r_{+}-r_{-})+(r_{+}{x}_0-v_0)e^{r_{-}t}/(r_{+}-r_{-})$. It satisfies $X(0)=x_0$ and $\dot{X}(0)=v_0$. With zero-mean future noise and prescribed initial data it is also the conditional ensemble mean, not a full stochastic trajectory.

For $ϵ=km/{\gamma }^2\ll 1$, momentum relaxes in ${\tau }_m=m/\gamma$ while the trap relaxes in ${\tau }_k=\gamma /k$. After the fast transient, the slow amplitude is $x_0+{\tau }_m{v}_0+O(ϵ)$. Neglecting the initial-velocity contribution additionally requires $|{\tau }_m{v}_0|\ll |x_0|$; small Reynolds number alone does not prove this condition.

For $L=(D_t+a)(D_t+b)$ with $a>b>0$, $H$ denotes the Heaviside step function. This Green function vanishes at negative time and satisfies $LG=\delta (t)$. Its positive-time part solves the homogeneous ordinary differential equation, is zero at the origin, and has right derivative one. Thus it is continuous with a unit derivative jump, which gives precisely the Dirac delta function in its second distributional derivative.

With the negative-exponential Fourier transform convention,Both poles lie in the upper half-plane. Contour integration closes above for positive time and below for negative time, giving the title expression. When the integral converges, convolution $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$ supplies the response with no freely added homogeneous transient growing toward negative infinity. The continuous limit $a\to b$ is $G(t)=H(t)t{e}^{-bt}$.

For $a>b>0$ with $a,b\ne 1$, the solution of $(D_t+a)(D_t+b)y=e^{-|t|}$ that decays at both infinities isIts Fourier transform is $2/[(1+k^2)(a+ik)(b+ik)]$. The lower half-plane contains only the pole $-i$, while the upper half-plane contains $i,ia,ib$. Inverting by the residue theorem gives the two expressions. Direct substitution verifies the equation separately on each side, and $y,y^{\prime }$ match at zero, so there is no spurious impulsive source. The negative-time response is nonzero because the forcing already acts before zero, even though the Green function is causal. Coinciding rates $a=1$ or $b=1$ require limits with polynomial times exponential terms.

For $\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0$, an initial velocity $v_0$ at zero displacement givesAll three cases begin with slope $v_0$. Only underdamping produces repeated crossings through equilibrium; critical damping and overdamping remain on the initial side of equilibrium for this particular initial condition. The overdamped long-time tail decays at rate $\gamma -s$, which becomes slow for strong damping.

For an isotropic harmonic oscillator with mass $m$, restoring constant $k$ and linear drag coefficient $b$, put $\omega =\sqrt{k/m}$ and $\gamma =b/(2m)$. Choose initial position ${\mathbf{r}}_0=(c,c)$ and tangential velocity ${\mathbf{v}}_0=\omega (-c,c)$, which would give anticlockwise circular motion without damping. For underdamping, $\nu =\sqrt{{\omega }^2-{\gamma }^2}$ givesTaking the squared norm and multiplying by $k/2$ gives the exact potential energyOnly to leading order in $\gamma /\omega \ll 1$ does this reduce to $k{c}^2{e}^{-2\gamma t}$. The exact expression has $\dot{U}(0)=0$, consistent with the initially tangential velocity, while the leading approximation does not reproduce this short-time derivative. Angular momentum obeys $L_z(t)=L_z(0)e^{-2\gamma t}$ by angular momentum decay under uniform linear drag.

For the initially circular damped planar oscillator with overdamping $\gamma >\omega$, put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}$. The exact solution replaces $\cos \nu t,\sin \nu t/\nu$ by $\cosh \alpha t,\sinh \alpha t/\alpha$. Resolving along the starting radius and its anticlockwise tangent yieldsIts limiting angle is the title value. Both coordinates remain positive when $c>0$, so the trajectory returns to the origin without repeated revolutions. It initially moves tangentially and bends toward its slow decaying eigenmode, rather than following a radial straight line from the start.

The fixed-time regular perturbation is $(1-ϵt)\sin t+O({ϵ}^2)$. Retaining the exponential decay and the first accumulated phase correction gives $e^{-ϵt}[\sin t-{ϵ}^{2}t\cos t/2]+O({ϵ}^2)$ on $ϵt=O(1)$. When ${ϵ}^{2}t=O(1)$ its phase series is unordered. For positive damping the absolute error is nevertheless exponentially small on that later scale, so phase failure and failure in absolute displacement must be distinguished.

For $L=D_t^2+2k{D}_t+k^2+{\omega }^2$, with $k>0$ and $\omega >0$, the causal Green function is $G(t)=H(t)e^{-kt}\sin (\omega t)/\omega$, where $H$ is the Heaviside step function. It satisfies $LG=\delta (t)$: it is continuous at zero, with a unit jump in its first derivative. Its smooth positive-time kernel $g$ obeys $g(0)=0,g^{\prime }(0)=1$. The zero-frequency limit is $H(t)t{e}^{-kt}$. Shifting $G$ gives the zero-initial-data response to a delayed Dirac delta function.

With the Fourier transform convention $\hat{f}(\xi )=\int f(t)e^{-i\xi t}dt$, the transformed equation gives $\hat{G}(\xi )=[k^2+{\omega }^2-{\xi }^2+2ik\xi {]}^{-1}$. Its poles $\xi =ik\pm \omega$ lie in the upper half of the complex plane. Inverting by contour integration therefore gives zero for negative time and an exponentially damped oscillation for positive time.

For $\beta >0$, $0<\gamma <{\omega }_0$ and zero prehistory, the equation $y^{\prime \prime }+2\gamma y^{\prime }+{\omega }_0^{2}y=F{e}^{-\beta t}H(t)$ hasOne obtains it by convolution with the causal Green function of a damped oscillator. Alternatively, a particular exponential has amplitude $F/[(\beta -\gamma {)}^2+{\Omega }^2]$; the two homogeneous terms enforce $y(0)=y^{\prime }(0)=0$. These continuity conditions avoid adding impulsive forcing at switch-on. The forcing may have either sign; its Fourier transform is $-iF/(\omega -i\beta )$ with the negative-exponential transform convention.

For $x^{\prime \prime }+2\gamma x^{\prime }+({\gamma }^2+{\mu }^2)x=g(t)$ with $\gamma ,\mu >0$, zero initial data give $x(t)={\mu }^{-1}{\int }_0^t{e}^{-\gamma (t-s)}\sin (\mu (t-s))g(s)ds$. If $|g(s)|\le C{e}^{-as}$ for $a>0$, the causal Green function of a damped oscillator givesBoth bounds tend to zero. A decaying homogeneous transient can be added for arbitrary finite initial data. The coincident decay rates require the extra polynomial factor and cannot be handled by dividing by zero in the first expression.

Integrating the smooth positive-time impulse kernel gives the zero-data response to the Heaviside step function. For nonzero $\omega$,At zero frequency and nonzero $k$ the limit is $[1-e^{-kt}(1+kt)]/k^2$; when both parameters vanish it is $t^2/2$. The integral representation avoids ambiguous divisions in these limiting cases.

Adding a load $M$ to a spring-supported table shifts its static downward position by $\Delta =Mg/k$. If the load is deposited with negligible impulse and moves with the table, the moving mass is $m+M$. Starting at the former equilibrium with zero velocity, an underdamped response isIts transient envelope decays over $1/\beta$. Critical damping $b=2\sqrt{k(m+M)}$ gives $x=\Delta [1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. A nonzero impact impulse changes the initial velocity and therefore the transient.

For a step-response kernel with $K(0)=0$, differentiating the start time produces no delta term from the response itself, because its coefficient is zero. The negative shift derivative is the impulse response. More generally an additional $\delta (t-b)K(0)$ term appears if the kernel has a nonzero initial value. The identity holds distributionally as well as for the ordinary smooth pieces.

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.