A Single Spring & Mass OscillatorDescription of the ModelFinding the General SolutionAnalysis of ResultsCoupled OscillatorsDescription of the ModelFinding the General SolutionBehavior of the SystemNormal modesConclusionA Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 1 of 21Go BackFull ScreenCloseQuitCoupled OscillatorsAlex Gagen & Sean LarsonMay 18, 2000AbstractWe know that a single spring and mass system obeys simple harmonicmotion (SHM). But how does a system of two spring and mass oscillators,coupled by a third spring behave? The analysis of this system of coupledoscillators is a great introduction to the concept of normal coordinates,and normal modes of vibration. For the normal modes of vibration, themotion is SHM. However, the general behavior of the two mass system isnot SHM. It is a much more interesting motion in which the energy of thesystem oscillates back and forth between the two masses . The analysishere shows the general solution for the two mass system, in the undamped,undriven case.A Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 2 of 21Go BackFull ScreenCloseQuitkx=0mFigure 1: A Single Spring and Mass System1. A Single Spring & Mass Oscillator1.1. Description of the ModelBefore we discuss the coupled oscillators, we will review the simple problem of asingle spring and mass oscillator. This system consists of a single mass attachedto a spring, and allowed to slide on a horizontal surface. A picture of this canbe seen in Figure 1. For simplicity, we will ignore any damping forces acting onthe m ass , such as air resistance, or friction. With these assumptions, we needonly consider the force that the spring exerts on the mass. We will assume thatthe spring obeys Hooke’s law, and thus exerts a linear re storing force on themass. This means that when the mass is displaced by a distance ∆x from itsA Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 3 of 21Go BackFull ScreenCloseQuitequilibrium position, the spring exerts a force Fs= −k∆x on the mass. We willrestrict our system to one-dimensional motion along the x-axis.1.2. Finding the General SolutionWe can describe the state of our system at any time in terms of the mass’sdisplacement from equilibrium x(t), and velocity v(t) =dxdt. Given initial con-ditions for the system, x(0) and v(0), we would like to be able to solve for themasses position as a function of time.Newton’s Second LawNewton’s second law is an important tool in solving this problem as well as themore complex problem of the coupled oscillators. Newton’s second law says thatthe sum of the forces acting on a body is equal to the product of the object’smass and its acceleration. Mathematically this is stated as, F = ma. In ourcase the displacement of the mass is x(t), so the acceleration isd2xdt2. So we stateNewton’s second law as:F = md2xdt2And we know that the net force acting on the mas s is simply the force due tothe spring, which we know to be:Fs= −kxSo we write Newton’s Second Law for our spring and mass system as:Fs= −kx = md2xdt2A Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 4 of 21Go BackFull ScreenCloseQuitWhich is rearranged:mx00+ kx = 0This equation is traditionally rewritten in the form:x00+kmx = 0 (1)Now we make a simple substitution to write the equation in a s impler form.ω =rkm(2)which is known as the natural frequency of the mass. The reas on for this willbe revealed later. Now we have the second order differential equation:x00+ ω2x = 0 (3)This is the final form of our differential equation describing the motion of themass.The Method of the Lucky GuessWe will solve this equation in a way that at first seems to be a bit of a swindle,but in fact is a common technique for solving differential equations. We guess asolution. Though this is not a completely random guess, but rather, an educatedguess. Our guess is:x = eλtA Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 5 of 21Go BackFull ScreenCloseQuitwhere λ must be found. Taking the first and second derivatives of our guess, weget:x0= λeλtandx00= λ2eλtwhich we substitute back into Equation 2 and simplify, giving us the character-istic polynomial p(λ).p(λ) = λ2+ ω2= 0 (4)which we solve for λλ = ±iωand obtain our solution.x = eiωt(5)But this is not the form we want it in. So we expand it using Euler’s formula.x = cos ωt + i sin ωt (6)Now we use a theorem from differential equations that states that for complexvalued solutions of a linear differential equation, both the real and imaginaryparts of the complex solution, are solutions. And we form our general solutionby taking a linear combination of these two solutions.x(t) = C1cos ωt + C2sin ωt (7)Using trigonometry, we can rewrite our general solution in a more useful form:x = A cos(ωt − φ) (8)A Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 6 of 21Go BackFull ScreenCloseQuit0 5 10 15−1−0.500.51txFigure 2: Simple Harmonic Motion: x(t) = A cos(ωt − φ), A = 1, ω = 1, φ = 01.3. Analysis of ResultsFrom our solution, we now know that the masses motion has an amplitude of A,an angular frequency of ω, and a phase shift of φ radians. Now we can see whythe substitution for ω in Equation 2 was made. As we can see in Equation 8,the natural frequency ω is the angular frequency of the masses motion. A graphof the mass’s motion where A = 1, ω = 1, and φ = 0 can be seen in Figure 2.A Single Spring & . . .Coupled OscillatorsConclusionHome PageTitle PageContentsJJ IIJ IPage 7 of 21Go BackFull ScreenCloseQuit12Ckkkx=0x=0mmFigure 3: Coupled Oscillators2. Coupled Oscillators2.1. Description of the ModelNow that we have reviewed the concepts of simple harmonic motion, we areready to discuss the problem of the coupled oscillators. As seen in Figure 3, wehave two spring and mass oscillators with stiffness constants k, and mass es mcoupled together by a third spring of stiffness kc. Also it should be noted thatkc<< k. So we can say that our system is weakly coupled. As we did withthe single spring and mass system, we will ignore any damping forces, such asfriction or air-resistance. And the motion of the masses is again restricted to onedimension. We will make the same assumption that the springs obey Hooke’slaw and thus exert a