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Physics LibreTexts

14.2: Two Coupled Linear Oscillators

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Consider the two-coupled linear oscillator, shown in Figure 14.2.1, which comprises two identical masses each connected to fixed locations by identical springs having a force constant κ. A spring with force constant κ couples the two oscillators. The equilibrium lengths of the outer two springs are l while that of the coupling spring is l. The problem is simplified by restricting the motion to be along the line connecting the masses and assuming fixed endpoints. The small displacements of m1 and m2 are taken to be x1 and x2 with respect to the equilibrium positions l and l+l respectively. The restoring force on m1 is κx1κ(x1x2) while the restoring force on m2 is κx2κ(x2x1). This coupled double-oscillator system exhibits basic features of coupled linear oscillator systems.

12.2.1.PNG
Figure 14.2.1: Two coupled linear oscillators. The equilibrium spring-lengths are l for the outer springs and l for the coupling spring. The displacement from the stable locations are given by x1 and x2. The separation between the two masses is r and the location of the center-of-mass is Rcm.

Assuming m1=m2=m, then the equations of motion are

m¨x1+(κ+κ)x1κx2=0m¨x2+(κ+κ)x2κx1=0

Assume that the motion for these coupled equations is oscillatory with a solution of the form

x1=B1eiωtx2=B2eiωt

where the constants B may be complex to take into account both the magnitude and phase. Substituting these possible solutions into the equations of motion gives

mω2B1eiωt+(κ+κ)B1eiωtκB2eiωt=0mω2B2eiωt+(κ+κ)B2eiωtκB1eiωt=0

Collecting terms, and cancelling the common exponential factor, gives

(κ+κmω2)B1κB2=0(κ+κmω2)B2κB1=0

The existence of a non-trivial solution of these two simultaneous equations requires that the determinant of the coefficients of B1 and B2 must vanish, that is

|κ+κmω2κκκ+κmω2|=0

The expansion of this secular determinant yields

(κ+κmω2)2κ2=0

Solving for ω gives

ω=κ+κ±κm

That is, there are two characteristic frequencies (or eigenfrequencies) for the system

ω1=κ+2κm

ω2=κm

Since superposition applies for these linear equations, then the general solution can be written as a sum of the terms that account for the two possible values of ω.

12.2.2.PNG

Figure 14.2.2: Displacement of each of two coupled linear harmonic oscillators with κ=4 and κ=1 in relative units.

Figure 14.2.2 shows the solutions for a case where κ=4 and κ=1, in arbitrary units, with the initial condition that x2=D, and x1=˙x1=˙x2=0. The two characteristic frequencies are ω1=6m and ω2=4m. The characteristic beats phenomenon is exhibited where the envelope over one complete cycle of the low frequency encompasses several higher frequency oscillations. That is, the solution is

x2(t)=D4[eiω1t+eiω1t+eiω2t+eiω2t]=Dcos[(ω1+ω22)t]cos[(ω1ω22)t]

while

x1(t)=D4[eiω1t+eiω1teiω2teiω2t]=Dsin[(ω1+ω22)t]sin[(ω1ω22)t]

The energy in the two-coupled oscillators flows back and forth between the coupled oscillators as illustrated in Figure 14.2.2.

A better understanding of the energy flow occurring between the two coupled oscillators is given by using a (x1,x2) configuration-space plot, shown in Figure 14.3.1. The flow of energy occurring between the two coupled oscillators can be represented by choosing normal-mode coordinates η1 and η2 that are rotated by 45 with respect to the spatial coordinates (x1,x2). These normal-mode coordinates (η1,η2) correspond to the two normal modes of the coupled double-oscillator system.


This page titled 14.2: Two Coupled Linear Oscillators is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform.

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