1. Free Undamped MotionMOTION IN A STABILITY REGION (PART I) 4 MOTION IN A STABILITY REGION (PART I) When motion is confined to one independent degree-of-freedom, the linearized equation that governs the motion is of the form (4 – 1) fkxxcxm=++&&& In this section, we analyze Eq. (4 – 1) in more detail than in Chapter 1. The system’s free motion (f = 0) is analyzed first and then its forced motion. The analysis performed for the free motion is called a transient analysis. The transient analysis is independent of the non-homogeneous term f that appears on the right side of the differential equation. The non-homogeneous term is often a force but it can also arise as a result of prescribing the displacement at a point in the system. The right side of the differential equation is generally called the excitation. It is shown how the time dependence of an excitation affects a system’s time response. We start with the constant excitation. Static loads, weight forces, and prescribed displacements are the most frequent examples. The constant excitation change’s a MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) system’s equilibrium position. Next, we consider the harmonic excitation. Systems that contain unbalanced rotating elements, like a washing machine, milling machines, and rotating shafts, are systems that are acted on by harmonic excitations. The harmonic excitation causes a system to undergo harmonic motion. 1. Free Undamped Motion Figure 4 – 1: The mass-spring-damper system First consider the free undamped system (See Fig. 4 – 1). Letting f = 0 and c = 0 in Eq. (4 – 1) yields (4 – 2) 0=+ kxxm&& Equation (4 – 2) is a homogeneous constant-coefficient linear differential equation. As with any constant-coefficient linear differential equation, the solution is a combination of complex exponential functions. We start by looking at the single complex exponential function (4 – 3) stex= MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) where s is a complex number that needs to be determined. Substitute Eq. (4 – 3) and its second time derivative into Eq. (4 – 2) to get 0)(2=+ststkeesmDividing by ste (4 – 3) 02=+kms The values of s for which x = satisfies the differential equation are ste (4 – 4a, b) mkisnn=±=ωω, where .1−=i The two solutions are (4 – 5) titinnexexωω−==21 These two solutions may seem a bit odd; after all they’re complex. The complex solutions are actually just building blocks from which the real solution is constructed. Recall that and in Eq. (4 – 5) are complex harmonic functions of the form (See Fig. 4 – 2) tineωtineω− (4 – 6) titetitenntinntinnωωωωωωsincossincos−=+=− MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) Figue 4 – 2: est for pure imaginary s The real solution is a linear combination of the two complex solutions. From Eq. (4 – 5) and Eq. (4 – 6) (4 – 7) )sin(cos)sin(cos212211titAtitAxAxAxnnnnωωωω−++=+= where A1 and A2 are constants. By rearranging terms, the real solution is written as (4 – 8) tCtBxnnωωsincos += in which B = A1 + A2 and C = i(A1 – A2). MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) Equation (4 – 8) is called the transient solution or the homogeneous solution of Eq. (4 – 1). The constants B and C depend on whether the system was initially displaced, was initial moving, or both. The constants B and C can also be written as (4 – 9) φφsincosACAB== Substituting Eq. (4 – 9) into Eq. (4 – 8) (4 – 10) )cos(φω−= tAxn Figure 4 – 3 using the trigonometric identity ).cos(sinsincoscosφωωφωφ−=+ tttnnn We see in Eq. (4 – 10) that the transient solution of a free undamped system is a harmonic function. It’s amplitude is A, its natural frequency is ωn, and MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) its phase angle is φ. (See Fig. 4 – 3). The natural frequency is found from Eq. (4 – 4b). Its standard units are rad/s. The natural frequency is also sometimes expressed in terms of cycles per seconds which is the same as a Hertz, abbreviated Hz. Since 1 cycle is 2π radians it follows that rad/s2Hz1π= and so 1 Hz is about six times larger than 1 rad/s. The system’s natural period is (4 – 11) kmTnnπωπ22== 2. Free Damped Motion Now consider the free damped system (See again Fig. 4 – 1). Letting f = 0 in Eq. (4 – 1) yields (4 – 12) 0=++ kxxcxm&&& The procedure for solving Eq. (4 – 12) is the same as the procedure followed for solving Eq. (4 – 2). Start by assuming a solution in the form of a complex exponential function. Substitute Eq. (4 – 3) into Eq. (4 – 12) to get 0)()(2=++stststkesecesm Divide by to yield the quadratic equation ste (4 – 13) 02=++kcsms Its roots are (4 – 14) []mkmcmcmkccms −⎟⎠⎞⎜⎝⎛±−=−±−=2222421 MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) Let’s now distinguish between three levels of damping depending on whether the quantity under the square root is positive, zero, or negative. The roots are (4 – 15) ⎪⎪⎩⎪⎪⎨⎧<−±−=−>−=±−=overdampeddampedcriticallydunderdampeisnnnnnddαωωαααωααωαωωωα2222 in which mcmkn2==αω Figure 4 – 4 est at time t MAE 461: DYNAMICS AND CONTROLSMOTION IN A STABILITY REGION (PART I) The three levels of damping are referred to as under-damped, critically damped, and over-damped. The function stex= is complex in under-damped systems and real otherwise. Figure 4 – 4 shows stex= in the complex plane for under-damped systems. A. Under-damped In under-damped motion the two solutions to Eq. (4 – 12) are (4 – 16) )sin(cos)sin(cos)(2)(1titeeeextiteeeexddttittiddttittiddddωωωωαωαωααωαωα−===+===−−−−−−−+− The general solution is a linear combination of these complex solutions. )sin(cos)sin(cos212211titeAtiteAxAxAxddtddtωωωωαα−++=+=−−Regrouping terms, (4 – 17) )sincos( tCtBexddtωωα+=− in which B and C are constants. We see in Eq. (4 – 17) that the transient solution of a free under-damped system is a damped harmonic function. Its natural damping rate is α and its damped natural frequency is ωd (See Fig. 4 – 5). The