115Lecture 13 Thursday March 5 , 2015Forced single-degree of freedom oscillatorsGeneric differential equation:meff {d2x/dt2}+ceff {dx/dt}+keff x = Feff(t)where F is an applied force, or maybe something equivalent…..examples: All of the above systems have Equations Of Motion (EOM) of the above generic form. The lasttwo are particular noteworthy because they are not driven by prescribed forces F(t) but rather byprescribed displacements y(t)…. yet they have the same form for their governing differentialequation.The first case is the most obvious. Effective quantities m, k, F, are the physical quantities.It is worth noting that if x(t) is to remain small and close to equilibrium, then the forcing F mustbe in some sense small also.The second case is a pendulum with ODE: mL2&&θ+ MgL sinθ= F(t)L cosθ which takes on ouradvertised general form when we replace sinθ with θ and cosθ with 1. Then Feff = LF(t)The fourth case has a force on the mass of k' { y-x } - k x acting towards the right. Thus theEOM (differential equation of motion) ismd2x/dt2 + {k+k'}x = k' y(t)which is of standard form. We identify k'y(t) as the effective forcing F(t). Note that the applieddisplacement y(t) is the cause of the right side of the equation, the "effective force" k'y(t). Alsonote the effective stiffness Keff is the sum k + k'.116A bit of sanity checking is in order ( because it is easy to make sign errors in these derivations)Notice that the effective stiffness k+k' is always positive (as it ought be for stability) Alsonotice that, if y > 0, the effective force is to the right, as it ought be. And if y is set to zero, thenwe recover a sensible unforced Simple Harmonic Oscillator.In the third case where m is on a moving table again we see that there is no explicit externalforce acting on the cart. The mass is forced inertially by the acceleration of the table d2y/dt2.The differential equation of motion is derivable by recognizing that the force on the mass is onlykx to the left - because x is the amount by which the spring is stretched. The absolutedisplacement of the mass is, however, not just x, but rather x + y. The acceleration with respectto the absolute is therefore the second derivative of x + y. Thusm d2 (x+y) / d t2 = - k x.which can be put into standard form m d2 x / d t2 + k x = - m d2 y / d t2. Again a sanity checkconfirms that keffective is positive, and that a table accelerating to the right corresponds to somekind of effective force - m d2 y / d t2 acting to the left. You will see a similar derivation of anODE for the case of an inertially excited cart as part of HW6B.2.These cases do not conserve energy, so our energy method wherein we write dE/dt = 0,doesn't work so well. Hence why I derived them by thinking in terms of force balance; F = ma.We'll see later that these systems can be derived using the energy method of LagrangeMechanics.In all the above cases, we get an ODE of the same general form. They are all "forcedlinear single-degree-of-freedom oscillators." Mathematically, these ODE's are classified aslinear 2nd order, constant coefficient, but inhomogeneous. The term on the right hand side is theinhomogeneous term because it is independent of the variable x being solved for. It is, ingeneral, an arbitrary function of time. We will not consider the (very hard and not so common)case in which it also has significant dependence on x.Solution of the Forced Problem, general commentsThe EOM can be written in the more abstract formL {x} = FWhere L is a "linear operator". Linear operators, of which "take a second derivative" is anexample, map functions of time into new functions of time, and do so in a linear manner. ( Inother courses the notion of a linear operator is more general, acting not necessarily on functions of time, butfunctions of space, or space and time, or matrices, or something more abstract… ) Linearity of an operatormeans it has the property that, for any functions f1(t) and f2(t) and constants a and b.L { a f1(t) + b f2(t) } = a L { f1(t) } + b L { f2(t) }117for example L operating on the sum of two functions gives the sum of L operating on eachfunction separately. This is the requirement for an operator to be linear. Clearly differentiationis a linear operator.The differential operator we have in on the left side of our ODEm{d2/dt2} + c {d/dt} + kis a linear operator, acting on our function of time x(t).Because of the property of linearity, we can establish that the differential equationL{x} = F(t) has a general solution of the form (for any constants c1 and c2)x(t) = c1h1(t) + c2h2(t)+ xp(t)This solution is written in terms of three functions h1(t) and h2(t) and xp defined as:h1(t) and h2(t) are any two linearly independent solutions of the homogeneous ODEwithout F(t). These solutions h1(t) and h2(t) were found by the looking for exp(λt), as discussedin previous lectures. Typically we will use h1 = cosωnt, h2 = sinωnt (if system is undamped), orh1 = exp(-ζωnt) cosωdt, h2 = exp(-ζωnt) sinωdt, (if system is damped)xp(t) is any particular solution of the full ODE L{x} = F(t).Thus if you have found one solution xp of the full ODE, and have the general solution of thehomogeneous case, c1h1(t) + c2h2(t) using the methods of Lectures 11 and 12, you have thegeneral solution of the inhomogeneous case. That this linear combination x(t) = c1h1(t) +c2h2(t)+ xp(t) solves the full ODE for any values of the constants c1 and c2 can be seen bysubstitution into L {x} = F. (Try it! )The combination c1h1(t) + c2h2(t) is sometimes called the "homogeneous" (because itsolves the homogeneous ODE) or "complementary" (because it complements the particular)solution. For reasons to be discussed later, the homogeneous part is sometimes also called the'transient' solution and the particular solution xp(t) is sometimes called the "steady statesolution." Using this prescription, we write the solution to our forced, damped, ODE in the form(using A and B instead of c1 and c2)x(t) = exp(-ct/2m) [ A cos ωdt + B sin ωdt ] + xp(t)or (if we prefer)x(t) = exp(-ct/2m) C cos (ωdt -ϕ) + xp(t)118The above forms each have two constants of integration (A and B, or C and ϕ) which may beadjusted to match the initial conditions. This is therefore not just a solution, it is the generalsolution. Any solution to the ODE, for any initial condition, may be put in either of