Oscillation

An oscillation is a repeated variation about an equilibrium or central value. A normal mode describes a single frequency of oscillation.

Equality of repeated time intervals; for an oscillator, independence of its oscillation period from amplitude in the stated regime. Ideal simple harmonic motion is isochronous. A simple pendulum has this property only approximately at small amplitude.

Coupled oscillators exchange energy through interactions. Linear equations can be transformed into normal mode coordinates; the spatial pattern of an applied force determines which normal modes are driven.

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.

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.

Anharmonicity is departure from the motion of a harmonic oscillator, often caused by nonquadratic terms in a potential energy. It can make the oscillation frequency depend on amplitude and generate higher harmonics. A quartic oscillator has one such anharmonic potential energy.