Some Basic Information about M-S-D Systems
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- Cornelius Goodman
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1 Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order, consan coefficien sysems. Recall ha he homogeneous differenial equaion ha we are dealing wih may be wrien in he form which has characerisic equaion m d d ) x + γ d d x + ω ox =, or where γ = c/m and ω o m λ + c λ + k =, λ + γλ + ω o =, = k/m. Using hese values of he consans, he roos are λ ± = γ ± 4 γ 4 ωo = γ ± γ ωo ± ωo = ±iω o, γ = = ( ) ωo γ 1 ± 1, γ. γ The Harmonic Oscillaor The firs alernaive leads o he Harmonic Oscillaor which has general soluion We make he folowing remarks: x() = C 1 cos (ω o ) + C sin (ω o ). (a) ω o is called he naural frequency. This is he number of waves passing a fixed observer in uni ime. 1
2 (b) The quaniy π ω o is called he wave lengh. I is usually called λ and is he disance a which he profile repeas iself. (c) The ime aken for one complee wave o pass any paricular poin is called he period, T, of he wave. Since he cosine and sine funcions are periodic of period π, he period is given by T = λ/π = 1/ω o. (d) x() = A cos (ω o δ) wih A = C 1 + C and cos (δ) = C 1 /A, sin (δ) = C /A or an (δ) = C /C 1 is called he phase-ampliude form of he soluion; A is he ampliude and δ is he phase shif. Example.1 : Le m = 1 and k = 36. Then ω o = 6 and he soluion of he iniial value problem ẍ + 36 x = wih iniial condiions x() = 1, ẋ() = 1 can be deermined from he general form of he soluion x() = C 1 cos (6 ) + C sin (6 ) ẋ() = 6C 1 sin (6 ) + 6C cos (6 ) Then x() = 1/ = C 1 and ẋ() = 1 C or C = 1/6. Hence he required soluion is x() = 1 cos (6 ) sin (6 ). (1 ) The AMPLITUDE of he vibraion is A = + ( 1 5 6) =.57. The PHASE 18 SHIFT is compued from he definiion of δ aking care ha he correc branch of he inverse angen is chosen. Here C 1 = 1/ and C = 1/6 so sin (δ) > and cos (δ) > which implies ha δ lies beween and π/. ( 1 ) an (δ) = ( 1 6) = 6 = 3, and we may conclude ha, approximaely, δ = 1.49 radians or, again approximaely, degrees. The quesion of deermining he spring consan is a maer of experimenal measuremen. Consider he following example:
3 Example. : A mass weighing 6 lbs. sreches a spring 6 in. Deermine he funcion x() ha describes he displacemen of he mass if he mass is released from res 18 in. below he equilibrium posiion. From his daa, we need o deermine he spring consan or he raio of he spring consan o he mass. Cauion: You mus be sure ha he unis are consisen! Hooke s Law says ha he spring force is linear and of he form F = k x. Since we have ha he force applied o he spring is 6 lbs., i follows ha he equaion 6 = k.5 so ha k = 1lbs./f.. Nex, he mass mus be deermined from he equaion F = m a. Since F = 6 and g = 3f./sec., we have ha he mass m = 15/8 slugs. Now, ω o = k/m = (1 8)/15 = 64 sec. Wih hese values of he parameers he equaion of moion becomes ẍ + 64 x =, and since he iniial posiion x() = 1.5f. (18 in. is 1.5 f.) and he mass is released from res, ẋ() =, we can solve he iniial value problem o see ha x() = cos (6 ). Now ry he following exercise: Exercise.3 : A mass of 1 gm. sreches a spring 5 cm. If he mass is se in moion from is equilibrium posiion wih a downward velociy if 1 cm./sec., and if here is no damping, deermine he posiion x of he mass a any ime. When does he mass firs reurn o he equilibrium posiion? 3 The Damped Oscillaor If damping is presen, hen γ and here are he usual hree cases which depend on he sign of he descriminan which, in urn, depends on he relaive sizes of he parameers. This relaive dependence is bes seen in he form of he descriminan ha we derived above. Spefically, remembering ha ω o = g/m and γ = c/m, he hree cases are: 3
4 Case Type 1 ( ωo γ ) > overdamping (c > 4mk) 1 ( ωo γ ) = criical damping (c = 4mk) 1 ( ωo γ ) = underdamping (c < 4km) Example 3.1 : Criical Damping An 8 lb. weigh is aached o a spring of lengh 4 fee. A equilibrium, he spring has lengh 6 f. Deermine he displacemen from equilibrium as a funcion of ime, x() if he drag coefficien is and (a) he mass is released from is equilibrium posiion wih a downward iniial velociy of 1 f./sec.; (b) he mass is released 6 in. above he equilibrium wih and iniial velociy of 5 f./sec. in he upward direcion. As before, one can compue boh he mass and he spring consan which are respecively, 1/4 slug and 4. So he differenial equaion is ẍ + 8 ẋ + 16 x =, and is o be solved wih wo ses of iniial condiions: (a) x() =, ẋ() = 1 and (b) x() =.5, ẋ() = 5. For his equaion, we are in he criically damped case since he characerisic equaion is λ +8λ+16 = or (λ+4) =. Hence he general soluion of he homogeneous equaion is x() = C 1 e 4 + C e 4. The firs se of iniial condiions lead o a soluion x a () = e 4. The graph of he soluion is given in he nex figure. For he second se of iniial condiions, he soluion is x b () =.5 e e 4. Again we can plo he soluion Noe ha in his las case here is an overshoo of he equilibrium ha occurs for his paricular choice of iniial condiion. 4
5 x() Figure 1: The Soluion x a x() Figure : The Soluion x b 5
6 Exercise 3. : Assume ha he sysem described by he equaion mẍ + cẋ + kx = is eiher criically damped or overdamped. Show ha he mass can pass hrough he equilibrium posiion a mos once, regardless of he iniial condiions. Example 3.3 : The Overdamped Case A 3 lb. weigh sreches a spring 8 fee. If he resisive force due o damping is 5 ẋ, deermine he displacemen as a funcion of ime if he mass is released from 1 f. below he equilibrium posiion wih (a) an upward velociy of 1 f./sec.; (b) an upward velociy of 6 f./sec. The corresponding differenial equaion is ẍ + 5 ẋ + 4 x =, which has characerisic roos λ 1 = 1 and λ = 4. The general soluion is hen This is he case of overdamping. x() = C 1 e + C e 4. Now, for he firs se of iniial condiions x() = 1, ẋ() = 1 (noe ha he sign indicaes ha he iniial velociy is upward) he corresponding soluion is x a () = e. 1.8 x() Figure 3: The Soluion x a For he second se of iniial condiions, x() = 1, ẋ() = 6, he corresponding soluion is x b () = 5 3 e 4 3 e. 6
7 1.8 x() Figure 4: The Soluion x b We now urn o he case of underdamping. Consider he following example: Example 3.4 : The Underdamped Case A 16 lb. weigh sreches a spring fee. Deermine he displacemen as a funcion of ime if he resisive force due o damping is.5 ẋ and he mass is released from he equilibrium posiion wih a downward velociy of 1 f./sec. In his case he spring consan k = 8 lb./f. while he mass is.5 slugs. The iniial condiions are x() =, ẋ() = 1 and he governing equaion is whose characerisic equaion has roos ẍ + ẋ + 16 x =, λ ± = 1 ± i 3 7, so ha, using Euler s formula, he general soluion of he equaion is x() = e [ C 1 cos C sin 3 ] 7. Using he iniial condiions, we find ha he soluion is x() = 7 1 e sin.
8 This is clearly an oscillaory soluion, bu he ampliude of he oscillaions decreases exponenially..3 x() Figure 5: The Underdamped Soluion Noice ha he soluion x() as. The oscillaion is NOT periodic. Moreover, from our physical inuiion, we would expec ha in his case, he moion would evenually sop in he equilibrium sae. This is NOT he case here; his is only an approximae model for he acual physical siuaion, much of he approximaion being made in our assumpion of he damping forces acing on he sysem. One can see he exponenial decay more clearly in he following figure where he solid lines are he graphs of he funcions ± 7 1 e. Exercise 3.5 : Here is a problem where you can use MAPLE o illusrae he dependence of he soluion of he damped oscillaor on he damping facor. Given he iniial value problem ẍ + c ẋ + 6 x = x() =, ẋ() = 1 find he soluion of he equaion for each of he values c = 6,c = 4 6, and c = 6. Idenify, for each value of he parameer c which of he cases you are in and plo he hree soluions on a single display. 8
9 .3. x() Forced Oscillaions Figure 6: The Exponenial Decay of Oscillaions We now urn o he problem of deermining he behavior of a forced damped oscillaor. The general equaion we will wrie as m ẍ + c ẋ + k x = F(), where F() is he forcing funcion. This problem can be solved for reasonable forcing funcions by he Mehod of Undeermined Coefficiens. We will give some simple examples, bu even hese will reveal some basic behavior. As usual, we begin wih he undamped case, or he Forced Harmonic Oscillaor. m ẍ + k x = F(). In his case, le us ry he simple case in which F() = e. In order o find a paricular soluion of he nonhomogeneous equaion, we ake a rial funcion in he form x() = A e. NOTE: we know ha he general soluion of he harmonic oscillaor is a linear combinaion of cosine and sine funcions so his rial funcion is NOT a soluion of he homogeneous equaion. Differeniaing he rial funcion wice yields ẍ() = 4 A e and subsiuing his resul ino he forced oscillaor equaion yields or m ( 4 A e ) + k A e = A e, 9
10 4 m A + k A = 1, and so, rearranging (4 m + k) A = 1. Hence A = 1/(4m + k). From his calculaion we can conclude ha he general soluion has he form: ( ) 1 x() = C 1 cos (ω o ) + C sin (ω o ) + e. m (4 + ωo) Noice ha, whaever he iniial condiions are, he las erm on he righ-hand side decays rapidly wih ime. Wha we expec o see in any paricular soluion is ha he iniial porion of he soluion will differ from he soluion of an unforced harmonic oscillaor, bu ha as ime increases, he soluion seles down o he pure oscillaions. If we ake a paricular se of iniial condiions, say x() = 1, ẋ() = 1, i.e., we sar from 1 uni below he equilibrium wih velociy 1 downward, and if we choose paricular values for he parameers, say m = 1 and k = 1/, hen he soluion is: x() = 1 ( ) 3 cos ( ) sin + 4 ) 3 e(, From wha is said above, if we ake he homogeneous equaion wih iniial condiions x() = 1/3, ẋ() = 5/3, he soluion of his iniial value problem is jus x() = 1 ( ) 3 cos ( ) sin and we expec he oscillaion of he forced equaion o approach he his oscillaory soluion for large. This phenomenon is illusraed in Figures 7-9 where he doed curve in he las figure represens he ransien soluion. We will see he occurance of he ransien/seady sae soluions again. Now we wan o look a he resul of imposing a sinusoidal forcing erm. Again, we look a he harmonic oscillaor wih ( m ) = 1 and( k = 1/. The general soluion of he homogeneous problem is x() = C 1 cos + C sin ), so ha we see he naural frequency is ω o = 1/. We sar wih he forcing frequency he same, i.e. ( F() = cos ). Our experience wih he Mehod of Undeermined ( Coefficiens ( leads us o expec ha we should find a soluion of he form [A cos )+B sin )] and ha he soluion o he full problem would grow in ampliude as ges large. Le us illusrae by doing he compuaions here. 1
11 Figure 7: Soluion for Forced Oscillaor Figure 8: The Transien Soluion Figure 9: The Seady Sae vs. Full Soluion 11
12 The Mehod of Undeermined Coefficiens ells us ha we should ake a rial soluion of he form x() = [A cos (/sqr) + B sin (/sqr)]. Then differeniaing wo imes yields: ẋ() = A cos(/ ) + B sin(/ ) + [ A sin(/ ) + B cos(/ ] ) ẍ() = [ A sin(/ ) + B cos(/ ] [ ) + A sin(/ ) + B cos(/ ] ) + [ A cos(/ ) + B ] sin(/ ) = [ A sin(/ ) + B cos(/ ] [ ) + A cos(/ ) B ] sin(/ ). Subsiuing back ino he differenial equaion yields: ẍ() + 1 x() = [ A cos(/ ) B sin(/ ] ) + [ A sin(/ ) + B cos(/ ] ) + [ A cos(/ ) + B sin(/ ] ) = [ A sin(/ ) + B cos(/ ] ) = cos (/ ). Equaing coefficiens of like erms, we find ha A = and / B = 1 or B = /, from which we conclude ha a paricular soluion of he non-homogeneous equaion is x nh () = ( ) ( sin ). Now we can wrie down he general soluion of he non-homogeneous problem ( ) ( ) x() = C 1 cos + C sin + ( ) ( sin ). The consans are, as usual, deermined by applying he iniial condiions o his full soluion and NOT o jus he soluion of he homogeneous equaion. In he case ha we ake iniial condiions x() = 1, ẋ() = 1, he soluion is 1
13 ( ) x() = cos + ( ) sin + ( ) ( sin ). Noe he las erm corresponding o x nh has a facor of and we expec he phenomenon of resonance. Here is a graph of he soluion Figure 1: The Phenomenon of Resonance If we now increase he frequency of he forcing funcion, we see anoher phenomenon. Le us consider he equaion ẍ + 1 x = cos ( 3 4 ). This ime, since he forcing funcion does no have he same frequency as he naural frequency of he sysem, he es funcion should be aken in he form A cos (3/4) + B sin (3/4). The usual calculaions lead o a soluion of he non-homogeneous equaion x nh () = 16 cos (3/4) so ha, aking iniial condiions x() =, ẋ() =, i.e. saring from he equilibrium posiion, he soluion of he iniial value problem is ( ) x() = 16 cos 16 cos ( ) 3. 4 The graph of his soluion in Figure 11 illusraes he phenomenon of beas. Exercise 4.1 Use he Mehod of Undeermined Coefficiens o solve he differenial equaion and hen use he general form of he soluion o deermine he unknown coefficiens. In his way, verify ha he given soluion is correc. 13
14 Figure 11: The Phenomenon of Beas If we increase he frequency of he forcing funcion o, say, 1.5, hen he soluion of he same iniial value problem as in he previous case leads o a soluion: x() = 4 ( ) 7 cos 4 ( ) 3 7 cos. Exercise 4. From he general form of he soluion x() = c 1 cos (/ )+sin (/ ) (4/7) cos (3/) find he values of c 1, c for which he soluion saisfies he iniial condiions x() =, ẋ() =. The nex figure shows he soluion. 5 Forced Damped Oscillaions We urn now o he case ha he forced mass-spring sysem is aached o a dashpo. Tha means ha we are dealing wih a sysem wih fricion and he governing equaion is given by m ẍ + γ ẋ + k x = F o cos (ω ). We know ha, wih damping, all soluions of he unforced sysem decay exponenially. We have discussed behavior of he responses of over-damped, criically damped, and underdamped sysems. Thus, in all cases, he soluion of he forced equaion can be wrien 14
15 Figure 1: The Response for Larger Frequency as x() = x h () + x p () where x h is he general soluion of he homogeneous equaion and represens a ransien response while he paricular soluion, x p, of he nonhomogeneous problem persiss wih ime and is called he seady sae response. In order o find he soluion of he nonhomogeneous problem, we again use he Mehod of Undeermined Coefficiens, looking for a rial soluion in he form which has derivaives x p () = A cos (ω ) + B sin (ω ) ẋ() = A ω sin (ω ) + B ω cos (ω ), and ẍ() = A ω cos (ω ) B ω sin (ω ). Subsiuing hese relaions ino he differenial equaion and combining like erms in he usual way, we find ha [(k m ω ) A + γ ω B] cos (ω ) + [(k m ω ) B γ ω A] sin (ω ) = F o cos (ω ). Equaing he coefficiens on boh sides of his equaion we ge he sysem (k m ω ) A + γ ω B = F o γ ω A + (k m ω ) B =. 15
16 Solving his sysem for he undeermined coefficiens A and B, we find ha A = F o (k m ω ) k m ω ) + γ ω and B = F o γ ω) k m ω ) + γ ω. I follows ha he seady sae soluion has he form x p () = F o (k m ω ) + γ ω {(k m ω ) cos (ω ) + γ ω sin (ω )}. Noice ha his soluion involves a facor muliplying a bounded oscillaion and he magniude of his facor is a muliple of he ampliude of he original forcing funcion, namely F o. This muliplicaive facor is 1 ((k m ω ) + γ ω and is referred o as he frequency gain or simply he gain of he sysem since i represens he chance in he magniude of he sinusoidal inpu. Noice ha he frequency gain has he unis of lengh/force. The magniude of he seady sae oscillaion depends, of course, on he forcing frequency and is given by A(ω) = F o M(ω) where M(ω) = 1 ((k m ω ) + γ ω = (1/m) ( k ω) ( m + γ ). ω m 16
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