Mass-Spring-Damper Systems The Theory

1 Mass-Spring-Damper Systems : The , Bournemouth University 2001 Page 1 of 9 Mass-Spring-Damper SystemsThe TheoryThe Unforced Mass- spring SystemThe diagram shows a mass, M, suspended from a spring of naturallength l and modulus of elasticity . If the elastic limit of the springis not exceeded and the mass hangs in equilibrium, the spring willextend by an amount, e, such that by Hooke s Law the tension in thespring, T, will be given by Tel= For system equilibrium, this will be balanced by the weightso leTMg == (1)If the spring is pulled down a further distance, y, (with y positive downwards) therestoring force will now be the new tension in the spring , T, given by () =+Teyl ,and so the net force acting DOWNWARDS is MgT ()= += MgeylMgelyl.

2 TBut, from equation (1), Mgel= ,so the net force downwards = yl(2)From Newton s 2nd Law, Force = mass x acceleration 22dtydM=(3)so, combining (2) and(3)The above analysis hasresulted in a second-order differentialequation with dependent variable y (displacement) and independent variable t (time)and system parameters M, and l. (See box on next page for discussion onparameters and variables)For the Mass-Spring-Damper s 2nd order differential equation, TWO initial conditionsare given, usually the mass s initial displacement from some datum and its the system above is unforced, any motion of the mass will be due to the initialconditions ONLY. Typical initial conditions could be ()y02= and ()40+=y&.

3 Withdownward as the positive direction, y measured in centimetres and t in seconds, theseinitial conditions say that at t = 0 the mass is instantaneously 2 cm above the datumand is travelling with a velocity of 4 cm/s in the downward += TMgMgMass- spring -Damper Systems : The , Bournemouth University 2001 Page 2 of 9 The Unforced Mass-Spring-Damper SystemThe above system is unrealistic since it does not take into account theresistance to motion due to friction in the spring or air resistance. Oncethe mass is set in motion, that system will continue moving can be introduced into the system physically, schematicallyand mathematically by incorporating all resistances into a dashpot (seediagram).It can be shown experimentally that in such cases the resistance tomotion is directly proportional to the velocity of the mass and,naturally, opposes the motion.

4 This is not unreasonable - the faster themass moves, the greater the resistance is exerted upon it (compare howmuch more difficult it is running, rather than walking, through water).So the damping force, DRdydt= . (R > 0)Here, R is the constant of proportionality and is called the damping inclusion of the damping modifies the equations of the previous case thus:This time, the net downward force will be MgT - D Mg T D()= + = MgeylRdydtylRdydt . And, again using Newton s 2nd Law, this results inor,where k = / lDigression on Variables and Parameters In this system y(t) is the output of thesystem once the mass has been initially displaced and released. It is a time (t) is called the dependent variable and t is the independent variable since the valueof displacement y depends on (is a function of) time, t.

5 Note that in any such system,the displacement y will vary (unless it is a constant) as time, t, varies. However, in anygiven system M, and l will always take just the one value for all time. It is possibleto change them - but if they are changed, this results in a different Mass-Spring-Damper system and hence a completely different differential equation to that remain constant like this within any system (such as M, and l) areparameters of the that the system above has no input it is unforced. Nothing forces the system tomove; any movement is a consequence only of an initial displacement or an ++= MdydtRdydtky220++= Mass-Spring-Damper Systems : The , Bournemouth University 2001 Page 3 of 9 This is, once again, a second-order differential equation, but this time with parametersM, R and k.

6 Parameter k is in terms of parameters and l, and parameter R isdependent upon the viscosity of the fluid in the dashpot, for Forced Mass-Spring-Damper SystemConsider now the case of the mass being subjected to a force, f(t), in thedirection of (t)This time, the net downward force will be MgT - D + f(t) Mg T D f(t)()()()= + += +MgeylRdydtftylRdydtft . Again using Newton s 2nd Law, this results inor,Note how the only difference here is that the input to the system, f(t), the forcingterm, appears on the right hand side of the differential equation rather than the zerowhen the system was unforced ( zero input).Forcing TermsThe forcing term, the input to the system, given by f(t) can take various forms and canbe modelled readily by standard mathematical functions.

7 An unforced system, modelled by using f(t) = 0 A constant applied force uses f(t) = c (where c is a constant) A constantly changing force (ramp input), f(t) = mt + c (m, c constants) A quadratically changing force, f(t) = at2 + bt + c (a, b and c constants) An oscillating force (sinusoidal input), f(t) = tbta cossin+( here a, b & are constants where is the angular frequency of the applied oscillations) An exponentially changing input, f(t) = aebt (a, b constants)Solving the Mass-Spring-Damper Second-Order Differential EquationObtaining the solution of second order differential equations is outside of the remit ofthis Theory sheet. You should be learning these methods on your course - methodssuch as the classical Complementary Function and Particular Integral method, or the Laplace Transforms method.

8 Here the emphasis is on using the accompanyingapplet and tutorial worksheet to interpret (and even anticipate) the types of of Solution of Mass-Spring-Damper Systems and their InterpretationThe solution of Mass-Spring-Damper differential equations comes as the sum of twoparts: the complementary function (which arises solely due to the system itself), and the particular integral (which arises solely due to the applied forcing term).()tflydtdyRdtydM=++ 22()MdydtRdydtkyf t22++=---(#) used laterMass- spring -Damper Systems : The , Bournemouth University 2001 Page 4 of 9 The particular integral is the easier part of the solution to consider. The Mass-Spring-Damper differential equation is of a special type; it is a linear second-order differentialequation.

9 In mathematical terms, linearity means that y, dy/dt and d2y/dt2 only occur tothe power 1 (no y2 or (d2y/dt2)3 terms, for example). In real-world terms, linearitymeans What goes in, comes out ! If you apply an oscillating force to such a system,oscillations will result. A constant applied force (input) will produce a constantdeflection, y (output). As you can imagine, if you hold a Mass-Spring-Damper systemwith a constant force, it will maintain a constant deflection from its datum is the steady state part of the it gets to the steady state solution is governed by the system itself (is it light andspringy or perhaps heavy and slow?) and hence dependent on the values of M, R andk. The way in which the mass reaches its steady-state solution, called the transient, isreflected in the complementary function, which itself is dependent on the relativesizes of R2 and linear second order differential equation is related to a second order algebraicequation, kydtdyRdtydM++22 is related directly to cbxax++2.

10 For a secondorder algebraic equation the discriminant b2 4ac plays an important part in decidingthe type of solution to the equation cbxax++2= 0. Similarly the discriminant R2 4Mk determines the type of solution to the differential equation kydtdyRdtydM++22 =0, the system with the forcing term taken out it is this equation (with f(t) = 0)that produces the transient 4Mk > 0 (or R2 > 4Mk) produces a complementary function (transient) of theformtmtmBeAey21+= with A, B, m1 and m2 all constant with m1 and m2 both produces an exponential decaying transient. How long the transient takes to dieaway will depend upon the time constants of the two exponential decay terms (seenext section for discussion on time constants).