Oscillator

An oscillator is a system capable of repeated motion or variation about a central state. A mass $m$ on an ideal spring with spring constant $k$ satisfiesThe restoring force accelerates the mass toward zero displacement; inertia carries it through that point, producing simple harmonic motion. The kinetic energy is largest at zero displacement, and the elastic potential energy is largest at the turning points.

Real oscillators can include damping, external driving and anharmonicity. A damped harmonic oscillator distinguishes the free transient from the forced periodic response, and a coupled oscillator adds exchange of energy with another dynamical coordinate. This makes an oscillator useful both as a physical device and as an approximation to small disturbances of much more complicated systems.

Coupled oscillators influence one another's motion through shared forces or other interactions. Two pendulums connected by a light spring provide a visible example: moving either bob changes the force on the other. For equal masses and lengths, the symmetric motion leaves the connecting spring unstretched, while the antisymmetric motion stretches it and has a higher frequency.

For a conservative linear system, the equations of motion take the form $M\ddot{\mathbf{q}}+K\mathbf{q}=0$, where $M$ is the mass matrix and $K$ the elastic stiffness matrix. A normal mode has a fixed displacement pattern $\mathbf{a}$ satisfying $K\mathbf{a}={\omega }^{2}M\mathbf{a}$. Expanding the initial motion in normal modes separates the evolution. Displacing just one pendulum generally excites several normal modes, producing beats and visible transfer of energy. A driving force excites only the modes onto which its spatial pattern has a nonzero projection.

For masses $m,2m$ connected in sequence to a moving support by springs $k,2k$, displacement coordinates give $m\mathrm{diag}(1,2)\ddot{\mathbf{x}}+k\left(\begin{array}{cc}3& -2\\ -2& 2\end{array}\right)\mathbf{x}=k(f,0{)}^{\mathsf{T}}$. The squared normal-mode frequencies are $(k/m)(2\pm \sqrt{3})$. For support motion $f=F\sin {\omega }_{0}t$ with ${\omega }_0^2=k/m$, the negligible-damping steady response is $(0,-F\sin {\omega }_{0}t/2)$. The first mass has a response zero between the two resonances. Finite damping generally makes that zero imperfect.

Two identical vertical springs connect a driven pivot to two equal masses in sequence. After subtracting the static gravitational equilibrium, the laboratory displacement equation is $m\ddot{\mathbf{x}}+k\left(\begin{array}{cc}2& -1\\ -1& 1\end{array}\right)\mathbf{x}=k(z,0{)}^{\mathsf{T}}$. For $z=z_0\cos {\omega }_{0}t$ and ${\omega }_0^2=k/m$, a particular solution is $\mathbf{x}=(0,-z_0\cos {\omega }_{0}t{)}^{\mathsf{T}}$. Thus the upper mass has zero driven response even though this frequency is not an eigenfrequency. The general solution also contains both free normal modes.

A free mass $M$ driven by $F_0\cos \omega t$ can be coupled by a single spring of stiffness $k$ to an auxiliary free mass $m$. With no wall attachment or damping, the periodic displacement amplitude of $M$ is $A=F_0(k-m{\omega }^2)/\{{\omega }^2[Mm{\omega }^2-k(M+m)]\}$. Tuning $k=m{\omega }^2$ gives antiresonance: $A=0$ while the auxiliary mass has signed amplitude $-F_0/k$, supplying an exactly cancelling spring force. A nearby pole occurs at $k=Mm{\omega }^2/(M+m)$. For relative tuning error $ϵ=(k-m{\omega }^2)/(m{\omega }^2)$ sufficiently small, the primary amplitude relative to its free response grows as $(M/m)|ϵ|$. A light absorber is therefore narrowband and may require large auxiliary travel. This cancellation dissipates no energy, and undamped free modes persist unless excluded by initial conditions.

In a finite-dimensional undamped linear oscillator, a bounded periodic particular solution can exist away from natural resonances. Adding any homogeneous normal mode gives another solution, and that free motion does not decay. Thus the late-time motion is not determined by the drive alone. At a resonant frequency with nonzero drive projection onto the corresponding mode, the particular solution grows secularly, so no bounded periodic steady response exists.

Place two equal masses $M$ on a massless mechanical string with spans $L,2L,L$ and uniform tension $T$. Fix one end and drive the other with displacement $h(t)$. Linear transverse force balance gives ${\ddot{y}}_1+{\omega }_0^2(3y_1-y_2)=0$ and ${\ddot{y}}_2+{\omega }_0^2(3y_2-y_1)=2{\omega }_0^{2}h$. The mass-normalized stiffness matrix has symmetric and antisymmetric normal modes with squared angular frequencies $2{\omega }_0^2$ and $4{\omega }_0^2$. The mode coordinates $(y_1+y_2)/2$ and $(y_1-y_2)/2$ separate the forced equations.

For a cosine drive of amplitude $h_0$, the periodic response of an end-driven two-bead string has signed amplitudes $A_1=2h_0/[(2-s)(4-s)]$ and $A_2=(3-s)A_1$, where $s={\omega }^2/{\omega }_0^2$. At $s=3$, $A_2=0$ while $A_1=-2h_0$. Equal and opposite transverse forces from the driven end and moving first bead cancel at the second bead. This antiresonance differs from a natural resonance: it is a zero of one measured response between the two natural frequencies. Free normal modes remain unless initial conditions exclude them.

For equal masses $m>0$, positive end spring constants $k_1>k_2$ and coupling spring constant $\kappa >0$, the stiffness matrix is $K=((k_1+\kappa ,-\kappa ),(-\kappa ,k_2+\kappa ))$. Its eigenvalues divided by $m$ are the squared normal mode angular frequencies ${\omega }_{\pm }^2=(k_1+k_2+2\kappa \pm D)/(2m)$, where $D=\sqrt{(k_1-k_2{)}^2+4{\kappa }^2}$. Holding the opposite mass fixed gives ${\omega }_j^2=(k_j+\kappa )/m$. Since $D>k_1-k_2$, subtraction proves the title ordering. Holding a mass fixed retains the coupling stiffness; removing the coupling spring produces a different system.

Consider identical harmonic oscillators with unit mass, natural angular frequency ${\omega }_0>0$, and positive coupling $\kappa \ll {\omega }_0^2$:An initial impulse on the first gives $x_1(0)=x_2(0)={\dot{x}}_2(0)=0$ and ${\dot{x}}_1(0)=v$. The normal mode coordinates $u=x_1+x_2$ and $w=x_1-x_2$ have angular frequencies ${\omega }_0$ and ${\omega }_a=\sqrt{{\omega }_0^2+2\kappa }$. Solving their initial value problems provesLet $\Delta ={\omega }_a-{\omega }_0\sim \kappa /{\omega }_0$. To leading order in $\kappa /{\omega }_0^2$, sum-to-product identities give the displacement amplitude envelopes $(|v|/{\omega }_0)|\cos (\Delta t/2)|$ and $(|v|/{\omega }_0)|\sin (\Delta t/2)|$. Thus the first nearly complete energy transfer occurs at $\pi /\Delta$, and localization returns at $2\pi /\Delta$. More precisely, at $t=\pi /\Delta$, the first oscillator has exactly zero velocity and residual displacement $\frac{v}{2}(1/{\omega }_0-1/{\omega }_a)\sin ({\omega }_{0}t)$, of order $|v|\kappa /{\omega }_0^3$.

Here transferred energy means kinetic energy plus the uncoupled quadratic potential energy; the coupling potential energy is relatively small. The fast phase need not make the receiving oscillator pass through zero displacement at the transfer time. Expanding $\Delta t$ as small is invalid on the transfer scale $t=O(1/\kappa )$. This lemma complements absolute and signed beat envelopes by keeping that slow phase.

For equal lengths, end masses $m$, middle mass $M$, and equal springs between adjacent bobs, reflection symmetry gives mode vectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2m/M,1)$. Their squared frequencies are displayed above. These vectors are orthogonal in the inner product weighted by $\mathrm{diag}(m,M,m)$. Equal end masses, rather than equality of all three masses, suffice for this symmetry reduction.

Consider three coupled pendulums with equal end masses, with end masses $m$, middle mass $M$, equal lengths $l$ and connecting spring constants $k$. Give only the first bob a small horizontal displacement $a$, then release all three from rest. Write ${\omega }_0^2=g/l$, ${\omega }_1^2=g/l+k/m$ and ${\omega }_2^2=g/l+k/m+2k/M$. Decomposing $(a,0,0)$ into the three normal mode patterns givesEach cosine enforces zero initial velocity. At $t=0$, the three terms sum to the prescribed end displacement and cancel at the other bobs, which checks the coefficients independently of the time evolution.

The middle bob has no contribution from the antisymmetric normal mode $(1,0,-1)$, so a trace of that bob alone misses one frequency. This is a simple example of how a sensor's position determines which normal modes it can detect. The expression assumes the small-angle regime and horizontal coupling at the bobs.

For $x_2=(a/2)[\cos {\omega }_{1}t-\cos {\omega }_{2}t]$, the slowly varying signed factor is $a\sin [({\omega }_2-{\omega }_1)t/2]$. The absolute amplitude envelope repeats in $2\pi /|{\omega }_2-{\omega }_1|$, half the period of that signed factor.

Two equal small-angle pendulums coupled by a spring have displacement patterns $(1,1)$ and $(1,-1)$, with squared angular frequencies $g/l$ and $g/l+2k/m$. An initial displacement of only the first pendulum excites both patterns equally.