96Lecture 11 Tuesday Feb 24 , 2015 Begin single-degree-of-freedom Linear OscillatorsAll the systems pictured on the next page oscillate. But how can we quantitatively describe theirmotion? How are they similar or different? Let us explore some examples, starting with themathematically simplest: the massive cart (without internal stiffness) and the exactly linear (andmassless) spring, moving horizontally:We choose X(t) as our dynamical coordinate ( it is often important to be clear about the precisedefinition of whatever time-dependent variable we use to describe the motion) In this case we useX , defined as the instantaneous length of the spring, i.e. the distance to the right that the cart isfrom the wall. Later we will find it convenient to change our coordinate.One way to get the differential equation that governs X(t) is to invoke F= ma. The mass'sacceleration, in terms of X is just d2X/dt2 towards the right. The force on the mass is, (accordingto our understanding of linear springs) k * its stretch, or k (X(t)-L), towards the left. ( We havehad to introduce the natural length of the spring L. Interestingly, L will drop out of most of ourresults.) We confirm that our formula tells us the force is to the left if X > L; this assures us thatwe have the sign of our formula correct.We now write F= ma , or:m d2X/dt2 = -k (X(t) - L)This is our ODE. The Simple Harmonic Oscillator (SHO)Another way to get the ODE is to write an expression for the total energyE = T + PE = (1/2) m (dX/dt)2 + (1/2) k ( X(t) – L )2.where we have recalled the formula for the potential energy stored in a spring (1/2) k stretch2.We also note that there are no energy dissipating or generating elements in this model,so dE/dt =0. Thus0 = dE/dt = m (dX/dt) (d2 X/dt2) + k ( X(t) – L ) ( dX/dt)= (dX/dt) [ m (d2 X/dt2) + k ( X(t) – L ) ]97The ODE is essentially the same for all of theseoscillating systems. They are all of the form(simple harmonic Oscillator)Meff d2z(t)/dt2 + Keff z(t) = 0where z(t) is a measure of deviation away fromequilibrium. z is our "dynamical coordinate."Almost always for our applications thecoefficients Meff and Keff will be positive andindependent of t and z. If yours aren't, you mayhave made a mistake.For each physical system we must 1st choose themost useful coordinate z(t), and then do a bit ofwork to identify the values of Meff and Keff .98Formally solve ODE by seeking y(t) ~ exp(λt)Then λ2m exp(λt) + k exp(λt) = 0This is satisfied for all time if λ2 = −k/m.Ie., λ = ±i √k/m = ±i ωnThus y(t) = D1 exp(i ωnt) + D2 exp(-i ωnt) for some complex constants D1 and D2.If D2 = D1* then y is real, as it ought be.Euler's identity exp(i ωnt) =co ωnt +isinωnt thentells us y(t) = (D1 + D2 ) cos(ωnt) + i (D1 - D2 ) sin(ωnt)from which we can identifyA = (D1 + D2 ); B = i (D1 - D2 )If dE/dt is to be zero regardless of the speed of the mass, the part in square brackets must be zero.…. which gives us the ODE we have above.We will sometimes find that our ODE is best derived by invoking F=ma, or Torque = I α, orequivalent. On other occasions we will find it is easier to use an energy method, for examplesetting dE/dt = 0 if there is no dissipation.======How to solve this ODE?It is worth noting that the ODE indicates that there is an equilibrium value of X, such that F = 0and d2X/dt2 is zero. In the above example, that value is Xeq = L. Let us shift coordinates to aquantity that represents deviation away from equilibrium: y(t) = X(t) - Xeq. Then d2 X/dt2 = d2y/dt2 and X(t) – L = y. The ODE for y(t) is then:m d2y/dt2 = -k y(t)which is simpler than the ODE for X.This particular differential equation is verycommon, and can be expected to apply to abroad class of dynamical systems, especiallyif y is understood to represent smalldeviations away from equilibrium.We know the solution,y(t) = A cos ωnt + B sin ωntfor any values of the constants A and B, andfor ωnatural defined as √k/m.A and B are not specified by the ODE.They can be determined by using additional information, for example initial conditions. It is, forexample, easy to see that y (at t=0) is just A, and dy/dt at t = 0 is ωn B.A useful variation on the above form for y(t) isy(t) = C cos{ωnt -φ} C is the amplitude of the motion; φ is its phase.again with two apriori unspecified arbitrary constants C and φ. By direct substitution we can seethat this also satisfies the ODE for any values of the constants C and φ.99There are simple relations between A and B and C and φ. One way to get them is to expand Ccos{ωnt -φ} using the trig formula for the cosine of the sum of two angles:cos( a+b) = cosa cosb – sin a sinb.y(t) = C cos{ωnt -φ} = C cosφ cos ωnt + C sin φ sin ωntwhich permits us to identifyA = C cosφ and B = C sinφ. Also C = √A2 + B2 and tanφ = B/ A(but watch out for the + π ambiguity if you use tanφ = B/A to solve for φ.)It is easy to see thatC = the maximum value of y, andCωn = maximum value of dy/dt.Here is a plot of a typical y(t). It corresponds to C = 1.7, ωn= 2 and φ = 1 radian. It is also easyto see that it corresponds to A = about 0.92. B is a bit less obvious, but it can be worked out to be1.43 by either using amplitude equal to about 1.7 = sqrt(A^2+B^2) or by using dy/dt at t = 0(which is about 2.8 as could be seen by measurement off the graph) = ωnBThe period can be measured off the plot to be 3.14 – the time between successive maxima, or twicethe time between successive zeros.Let us write expressions for the PE using y= C cos{ωnt -φ}PE = (1/2) k y2= (1/2) k C2 cos2{ωnt -φ}which oscillates but is always positive. Its maximum value is k C2/2 corresponding to the PE thesystem has when y is at its maximum value C and dy/dt is zero.y =1.7cos(2t-1)= 0.92 cos2t +1.43sin2t100Similarly we may construct the Kinetic energyT = (1/2) m (dy/dt)2 = (1/2) m ωn2 C2 sin2{ωnt -φ}which oscillates but is always positive. Its maximum value is (1/2) m ωn2 C2 = k C2/2 . This isequal to the maximum PE but maximum KE occurs at a different time, when y = 0 and dy/dt ismaximum.The total energy is E = T + PE =E = (1/2) m ωn2 C2 sin2{ωnt -φ} + (1/2) k C2 cos2{ωnt -φ} = k C2/2 = PEmax=KEmaxThis is a constant, as it ought be.→ We note, the energy of the motion depends on the amplitude of the motion C→The period of the motion 2π/ωn is