Lecture 19. Mechanical Vibrations1. An Application!2. Free, Undamped MotionDifferential EquationsLECTURE 19Mechanical Vibrations1. An Application!We’re finally at a point where we can stop discussing solutions and properties of second orderlinear equations. Now, we’ll turn our attention to an application: mechanical vibrations. Theparticular application we’ll be considering is an object of given mass m hanging from a spring ofnatural length l, but there are a number of applications in various branches of engineering thatonly differ from our setup by some specifics and notation.Our convention is always that downward displacements and forces are positive, while upwarddisplacements and forces are negative. It’s important to be consistent. We also measure all dis-placements of the object from its equilibrium position. Thus, if our displacement function is u(y),u = 0 corresponds to the center of gravity as it hangs at rest from the spring.The first task is to develop a differential equation to model the displacement u of the object.First, recall Newton’s Second Lawma = F,where m is the mass of the object. We want our equation to be for displacement, so we’ll replacea by u00, and Newton’s Second Law becomesmu00= F (t, u, u0).What are the various forces acting on the object? We will consider four different forces, some ofwhich may not act on the system in every situation.(1) Gravity, FgThe gravitational force always acts on the object. It’s given byFg= mg,where g is the acceleration due to gravity. For simpler computations, we’ll take g = 10m/s.Notice that this force is positive, since it always acts downward.(2) Spring, FsSince we’re attaching the object to a spring, the spring will itself always exert a forceon the object. We’ll assume Hooke’s Law governs this force. Hooke’s Law says that thespring force is proportional to the displacement of the spring from its natural length. Whatis the displacement of the spring? When we attach an object to a spring, the spring getsstretched. Let’s denote the length of the stretched spring by L. Then its displacementfrom its natural length is L + u.So, the spring force is given byFs= −k(L + u),where k > 0 is the spring constant. Why is that negative there? It ensures that the forcehas the correct direction. If u > −L, i.e., the spring has been stretched beyond its naturallength, then u+L > 0 and so Fs< 0, which is good because we would expect the spring topull upward on the object in this situation. If u < −L, so that the spring is compressed,the spring force would push the object back downwards, and so we expect to find Fs> 0.Surely enough, that’s what we get.1Differential Equations Lecture 19: Mechanical Vibrations(3) Damping, FdWe will consider some situations where a damper is attached to the system. They willnot always be present, but we’ll always note when there is a damper involved. Damperswork to counteract motion (a good example of one is the shocks on your car), so that theywill always oppose the direction of the object’s velocity.In other words, if the object has downward velocity u0> 0, we’d want the dampingforce to be acting in the upwards direction, so that Fd< 0. Similarly, if u0< 0, we wantFd> 0. We’ll further assume that all of our dampers are linear. So we end up withFd= −γu0,where γ > 0 is the damping coefficient.(4) External Forces, F (t)This is more of a catchall than a particular force. Sometimes we’ll want our spring-mass system to have some external forces acting on it (for example, we might hook oursystem up to a generator which will exert some additional force on it0. We call F (t)the forcing function, and it’s just the sum of any of these external forces we have in aparticular situation.It’s important to reiterate here that in a given problem, all of these forces may not necessarilyact on the spring-mass system. So our force function will change depending on the particularsituation. When actually writing down our differential equation for a particular situation, we’llneed to be aware of what forces are and are not present. But let’s go ahead and write down themost general form of our equation, and we’ll discuss how it changes in particular cases.We have F (t, u, u0) = Fg+ Fs+ Fd+ F (t), so that Newton’s Second Law becomesmu00= mg − k(L + u) −γu0+ F (t),or, upon reordering this a little,mu00+ γu0+ ku = mg − kL + F (t).Let’s now think about what happens when the object is at rest. When the object is at equilib-rium u = 0, there are only two forces acting on the object: gravity and the spring force. Since theobject is at rest, these two forces must be balancing each other out: Fg+ Fs= 0. In other words,mg = kL.So our equation simplifies to(19.1) mu00+ γu0+ ku = F (t),and this is the most general form of our equation, with all forces present. Along with this differentialequation, we’ll have initial conditions of the formu(0) = u0Initial displacement from the equilibrium positionu0(0) = u00Initial velocity.It’s important to keep our sign conventions in mind when writing these down.Before we start talking about particular cases, we need to discuss how we might go aboutfiguring out these constants k and γ if they’re not explicitly given to us in a problem. Let’s startwith the spring constant k. We know that if the spring is attached to some object with mass m,the object stretches the spring by some length L when it’s at rest. We observed above that atequilibrium, mg = kL. Thus, if we know how much some object with a known mass stretches thespring when it’s at rest, we can computek =mgL.2Differential Equations Lecture 19: Mechanical VibrationsThis may not necessarily be the same object as the one in the spring-mass system, but that doesn’tmatter.How do we compute γ? If we don’t explicitly know the damping coefficient from the beginning,we’ll know how much force the damper exerts to oppose motion of a given speed. Then we can set|Fd| = γ|u0|, where |Fd| is the magnitude of the damping force and |u0| is the speed of motion. Sowe have γ =Fdu. We’ll see examples of this computation when we consider damped motion.Let’s start looking at specific cases. These will be defined by which forces are acting on ourspring-mass system.2. Free, Undamped MotionWe’ll begin by assuming that there are no dampers or external forces acting on our system.This is the simplest situation, and as we have no damping, we can take γ = 0. Our
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