The Periodiclly Forced Hrmonic Oscilltor S. F. Ellermeyer Kennesw Stte University July 15, 003 Abstrct We study the differentil eqution dt + pdy + qy = A cos (t θ) dt which models periodiclly forced hrmonic oscilltor. 1 Generl Solution of the Unforced Hrmonic Oscilltor Eqution The hrmonic oscilltor without externl forcing is modelled by the differentil eqution dt + pdy + qy =0 (1) dt where p = b/m, q = k/m, m is the mss of the bob, k is the spring constnt, nd b is the dmping coefficient. Due to the physicl interprettions of p nd q, we ssume tht p 0 nd tht q>0. Thechrcteristic equtionforthedifferentil eqution (1) is nd the eigenvlues re λ + pλ + q =0 λ 1 = p p p 4q λ = p + p p 4q 1
Due to the ssumptions p 0 nd q>0, no eigenvlue cn hve positive rel prt. There re thus four possibilities: 1. λ 1 < λ < 0 in which cse the hrmonic oscilltor is overdmped nd the generl solution of (1) is N (t) =c 1 e λ 1t + c e λ t.. λ = α + βi is n imginry eigenvlue with α < 0 nd β > 0 in which cse the hrmonic oscilltor is underdmped nd the generl solution of (1) is N (t) =c 1 e αt cos (βt)+c e αt sin (βt). 3. λ = βi is n imginry eigenvlue with β > 0 in which cse the hrmonic oscilltor is undmped nd the generl solution of (1) is N (t) =c 1 cos (βt)+c sin (βt). 4. λ < 0 is repeted eigenvlue in which cse the hrmonic oscilltor is criticlly dmped nd the generl solution of (1) is N (t) =c 1 e λt + c te λt. Note: We use the nottion N (t) to denote the generl solution of (1) becuse we re thinking of this solution s the nturl response when the hrmonic oscilltor experiences no externl forcing. Generl Solution of the Forced Hrmonic Oscilltor Eqution If we dd n externl forcing function, f (t), to the hrmonic oscilltor, we obtin the differentil eqution dt + pdy + qy = f (t). () dt The generl solution of the differentil eqution () is y (t) =N (t)+f (t)
where N is the nturl response (the generl solution of the corresponding unforced eqution) nd F is ny prticulr solution of the forced eqution (). We use the nottion F (t) to remind us tht this forced response. Once we hve found prticulr forced response, F (t), we will hve found the generl solution to (). 3 Some Useful Fcts From Trigonometry Lemm 1 If nd b re ny rel numbers with 6= 0,then b cos rctn = + b nd sin rctn b = b + b. Proof. We use the fct tht if x is ny rel number, then cos (rctn (x)) = 1 1+x nd sin (rctn (x)) = x 1+x. Setting x = b/, weobtin b cos rctn = q 1 1+ b = + b = + b. nd b sin rctn = q b 1+ b = b + b. Lemm If, b, nd re ny rel numbers, then for ll rel numbers, t, we hve cos (t)+b sin (t) =K cos (t φ) 3
where nd ( K = +b if 6= 0 b if =0 ½ rctn b φ = if 6= 0 π if =0. Proof. First suppose tht 6= 0nd let nd Then nd K = + b φ = rctn b. b K cos (φ) =K cos rctn = + b = + b b K sin (φ) =K sin rctn = + b = b b + b We conclude tht K cos (t φ) =K (cos (t)cos(φ)+sin(t)sin(φ)) = K cos (φ)cos(t)+ksin (φ)sin(t) = cos (t)+bsin (t) whichiswhtwewntedtoprove. 4
Now we consider the cse =0. In this cse, we define K = b nd φ = π nd observe tht K cos (φ) =0 K sin (φ) =b. Thus K cos (t φ) =K (cos (t)cos(φ)+sin(t)sin(φ)) = K cos (φ)cos(t)+ksin (φ)sin(t) =0cos(t)+bsin (t) = cos (t)+bsin (t). Remrk 3 It is useful to write function cos (t)+bsin (t) in the form K cos (t φ) becuse it llows us to tell the mplitude nd the phse shift of the oscilltions very esily. In fct, since K cos (t φ) =K cos t φ, we see tht the mplitude is K nd the phse shift is φ/. (Also,theperiod is π/ nd the frequency is / (π).) Exmple 4 A grph of the function is shown below. F (t) =cos(t) 5sin(t) 5
4-4 - 0 t 4 - -4 This function hs period π/ =π nd frequency 1/π 0.3. The mplitude of the oscilltions is q K = +( 5) = 9 5.385 nd the phse shift is φ = rctn 5 1. = 0.6. In fct, we cn write the formul for the function F (exctly) s F (t) = 9 cos t rctn 5 or (pproximtely) s or s F (t) =5.385 cos (t +1.) F (t) =5.385 cos ( (t +0.6)) nd we see tht the grph of F is the grph of y =cos(t) mplified by fctor of 5.385 nd shifted to the left by bout 0.6 units. Exercise 5 AgrphofF (t) = 3cos 1 t +5sin 1 t is shown below. 6
6 4-10 -8-6 -4-0 4 t 6 8 10 - -4-6 Write the formul for F in the form F (t) =K cos (t φ) nd determine the period, frequency, mplitude, nd phse shift for this function. 4 Solving the Forced Eqution with Periodic Forcing We now consider the problem of finding prticulr solution of the forced eqution with periodic forcing: dt + pdy + qy =cos(t) (3) dt with given prmeters p 0, q>0, nd > 0. Notethtwererestricting our ttention to forcing function (cos (t)) tht hs mplitude 1 nd no phse shift. As will be seen, we will be ble to hndle more generl forcing functions (sines nd cosines with ny mplitude or phse shift) once we understnd how to find solution to eqution (3). We clim tht (in most circumstnces) eqution (3) hs solution of the form F (t) = cos (t)+bsin (t). We cn show tht this is correct once we find the right vlues of nd b. Keep the comment in most circumstnces in mind. In the clcultions tht we re bout to do, we will do some division opertions. As we know, division is oky s long s we re not dividing by zero. In the clcultions tht follow, we will ssume tht when we re never dividing by zero. Then, when we re done, we will go bck over our clcultions nd del with cses where there my hve been dnger of dividing by zero. 7
Defining the function F s F (t) = cos (t)+b sin (t), we observe tht df = sin (t)+bcos (t) dt nd d F dt = cos (t) b sin (t). If F is to be solution of eqution (3), then we must hve cos (t) b sin (t) + p ( sin (t)+bcos (t)) + q ( cos (t)+bsin (t)) =cos(t) for ll t. This gives us + bp + q cos (t)+ b p + bq sin (t) =cos(t) or, equivlently, q + pb cos (t)+ q b p sin (t) =cos(t). Sincetheboveequtionmustbetrueforllrelnumberst, itmustbetrue when t =0.Thisgivesus q + pb =1. Using t = π/ (), weobtin p + q b =0. Multiplying both sides of the first eqution by p, nd multiplying both sides of the second eqution by q gives p q +(p) b = p p q + q b =0. 8
Adding these two equtions gives us ³(p) + q b = p or p b = (p) +(q ). Hving found b, we cn now determine tht q = (p) +(q ). We thus clim tht solution of eqution (3) is q p F (t) = (p) +(q cos (t)+ ) (p) +(q sin (t). (4) ) Let us check tht this correct: Defining F s bove, we hve df dt = (q ) (p) +(q ) sin (t)+ p (p) +(q cos (t) ) nd d F dt = (q ) (p) +(q ) cos (t) p 3 (p) +(q sin (t) ) nd d F dt + pdf dt + qf = (q ) (p) +(q ) cos (t) p 3 (p) +(q sin (t) ) + p (q ) (p) +(q ) sin (t)+ p (p) +(q cos (t) ) q (q ) pq + (p) +(q cos (t)+ ) (p) +(q sin (t) ) = (q )+p + q (q ) (p) +(q ) cos (t) p 3 p (q )+pq + (p) +(q ) sin (t) =cos(t). 9
The bove computtion shows tht the function (4) is indeed solution of the differentil eqution (3). Observe tht the formul for F cn be written s 1 F (t) = q (p) +(q ) cos (t)+psin (t). (5) Assuming tht q 6=0, we cn use Lemm to conclude tht q cos (t)+psin (t) =K cos (t φ) (6) where q K = q (p) +(q ) q nd p φ = rctn. q By observing tht K =(p) + q, we cn conclude tht F (t) = 1 cos (t φ) K where K nd φ re s defined bove. Thus, the function F hs phse shift φ/ nd mplitude 1 K = 1. q(p) +(q ) Now let us consider the cse tht q =0. In this cse, we hve F (t) = 1 1 sin (t) = ³t p p cos π, showing tht F hs phse shift π/ () nd mplitude 1/ (p). All of the bove discussion relies on the ssumption tht either p 6= 0 or q 6=0. Notethtiftlestoneofp or q is not zero, then (p) +(q ) > 0 nd we re thus justified in dividing by this quntity nd in tking its squre root. In prticulr, if p>0 (mening tht the hrmonic oscilltor hs dmping present), then (p) +(q ) > 0 (becuse we re lso ssuming tht > 0). We hve therefore found complete solution to our problem in the cse tht dmping is present. We summrize s follows: 10
Theorem 6 Consider the differentil eqution dt + pdy + qy =cos(t) dt where p 0, q>0, nd > 0. Also, ssume tht either p>0or q 6=0. Under the bove ssumptions, solution of this differentil eqution is q p F (t) = (p) +(q cos (t)+ ) (p) +(q sin (t). ) Furthermore, F hs mplitude 1 q (p) +(q ) nd phse shift φ/ where ( ³ rctn p if q φ = 6=0 q π if q =0. Theorem 6 shows us how to find prticulr solution of the periodiclly forced hrmonic oscilltor differentil eqution if p>0 (mening tht dmping is present) or if q 6= (even if dmping is not present). We must still consider the cse in which p =0(mening tht dmping is not present) nd q =. This is the specil cse in which we get the phenomenon known s resonnce. Before proceeding to study the cse of resonnce, let us mke some observtions bout the conclusions of Theorem 6: First, suppose tht p>0 (mening tht dmping is present). In this cse, the forced response hs mplitude 1. q(p) +(q ) It cn be seen from this formul for the mplitude tht if p is very lrge or is very lrge, then the mplitude of the forced response is very smll. Next, suppose tht p =0(mening tht no dmping is present) nd tht q 6= : In this cse, the forced response hs mplitude 1 q. 11
nd we see tht if q is very close to, then the mplitude of the forced response is very lrge. This is wht we cll the ner resonnt cse. Exmple 7 Consider the undmped, forced hrmonic oscilltor differentil eqution +4y =cos(t). dt A prticulr solution of this eqution is The generl solution is F (t) = 1 cos (t). 3 y (t) =c 1 cos (t)+c sin (t)+ 1 cos (t). 3 Suppose tht we wish to find the prticulr solution tht stisfies initil conditions y (0) = y 0 (0) = 0. Since we must solve This gives us y 0 (t) = c 1 sin (t)+c cos (t) 1 sin (t), 3 c 1 + 1 3 =0 c =0. nd the prticulr solution whose grph is shown below. c 1 = 1 3 c =0 y (t) = 1 3 cos (t)+1 cos (t), 3 1
0. 0 10 0 t 30 40-0. -0.4-0.6 Exmple 8 In the previous exmple, we hd q =4nd =1so q =3 is firly lrge, mening tht the forced response hs firly smll mplitude (1/3). Let us see wht hppens if we choose such tht is closer to q. For exmple consider +4y =cos(1.9t). dt In this cse, q =4 (1.9) =0.39 nd the forced response is F (t) = 1 cos (1.9t). 0.39 This forced response hs mplitude 1/0.39.5641. The generl solution of the bove differentil eqution is y = c 1 cos (t)+c sin (t)+ 1 cos (1.9t). 0.39 Suppose tht we wish to find the prticulr solution tht stisfies initil conditions y (0) = y 0 (0) = 0. Since y 0 (t) = c 1 sin (t)+c cos (t) 1.9 sin (1.9t), 0.39 we must solve c 1 + 1 0.39 =0 c =0. 13
This gives us c 1 = 1 0.39 c =0 nd the prticulr solution y (t) = 1 1 cos (t)+ cos (1.9t), 0.39 0.39 whose grph is shown below. 4 0-10 0 30 40 50 60 70 80 t -4 Forced hrmonic oscilltors re fscinting, re they not? 5 The Cse of Resonnce We now consider the differentil eqution dt + pdy + qy =cos(t) dt where p =0nd q =.Thus,thedifferentil eqution tht we re considering is dt + y =cos(t) with prmeter > 0. We clim tht this differentil eqution hs solution of the form F (t) =t ( cos (t)+b sin (t)). 14
As before, we determine wht nd b must be in order to mke this be true: Letting F be s defined bove, we obtin df dt = t ( sin (t)+b cos (t)) + ( cos (t)+b sin (t)) =( + tb)cos(t)+( t + b)sin(t) nd d F dt = + tb sin (t)+b cos (t) + t + b cos (t) sin (t) = t +b cos (t)+ tb sin (t). If F is to be solution, we must hve t +b cos (t)+ tb sin (t) + (t cos (t)+tb sin (t)) =cos(t) for ll rel numbers t. Simplifiction of the bove eqution gives us b cos (t) sin (t) =cos(t). Setting t =0yields nd setting t = π/ () yields b =1 =0. We conclude tht =0nd b =1/ (). Let us check tht F (t) = 1 t sin (t) is solution of our differentil eqution: First note tht df dt = 1 (t cos (t)+sin(t)) 15
nd d F dt = 1 t sin (t)+cos (t)+cos (t) = 1 t sin (t)+cos (t). Thus d F dt + F = 1 t sin (t)+cos (t) 1 + t sin (t) =cos(t) showing tht F is indeed solution. We remrk tht since ³ sin (t) = cos t π, we cn lso write the formul for F s F (t) = 1 ³t t cos π. This forced response hs phse shift of π/ () ndnmplitudethtgrows linerly s time goes on. If is very lrge, then the mplitude grows slowly; wheres, if is smll, then the mplitude grows quickly. Exmple 9 Consider the undmped, forced hrmonic oscilltor differentil eqution +4y =cos(t). dt A prticulr solution of this eqution (whose grph is shown below) is F (t) = 1 t sin (t). 4 16
10 8 6 4 0 - -4-6 -8-10 10 0 30 40 t The generl solution is y (t) =c 1 cos (t)+c sin (t)+ 1 t sin (t). 4 Suppose tht we wish to find the prticulr solution tht stisfies initil conditions y (0) = y 0 (0) = 0. Since y 0 (t) = c 1 sin (t)+c cos (t)+ 1 t cos (t)+1 sin (t), 4 we must solve c 1 =0 c =0. This gives us c 1 =0 c =0 nd the prticulr solution is whose grph is shown bove. y (t) = 1 t sin (t) 4 17
6 More Generl Periodic Forcing Functions Now tht we know how to solve nd study solutions of the forced hrmonic oscilltor eqution dt + pdy + qy =cos(t), (7) dt we would like to be ble to solve hrmonic oscilltor equtions with more generl sinusoidl forcing functions. In prticulr, we would like to be ble to solve dt + pdy + qy = A cos (t θ) (8) dt where A,, ndθ re given constnts. Fortuntely, it is esy to find solution of eqution (8) once we hve found solution of eqution (7). In prticulr, suppose tht F is solution of eqution (7) nd define G (t) =AF t θ. Then G 00 (t)+pg 0 (t)+qg(t) =AF 00 t θ + paf 0 t θ + qaf t θ = A F 00 t θ + pf 0 t θ + qf t θ = A cos t θ = A cos (t θ) which shows tht G is solution of eqution (8). solution of the forced eqution In summry, if F is dt + pdy + qy =cos(t), dt then G (t) =AF t θ is solution of the forced eqution dt + pdy + qy = A cos (t θ). dt 18
Exmple 10 Solve the forced eqution dt +5dy +6y =3cos(t 5). dt Solution 11 First, we note tht the generl solution of the unforced eqution dt +5dy +6y =0 dt is N (t) =c 1 e 3t + c e t. Next, we consider the forced eqution dt +5dy +6y =cos(t). dt Since p =5,q=6, =1, q =5,nd(p) +(q ) =50,wesee by eqution (4) tht prticulr solution of this forced eqution is F (t) = 1 10 cos (t)+ 1 sin (t). 10 Finlly, we consider the forced eqution dt +5dy +6y =3cos(t 5). dt A prticulr solution of this eqution is G (t) =AF t θ 1 =3 10 cos (t 5) + 1 sin (t 5). 10 In conclusion, the generl solution of the forced eqution is dt +5dy +6y =3cos(t 5) dt y (t) =c 1 e 3t + c e t + 3 (cos (t 5) + sin (t 5)). 10 19
Exmple 1 Solve the forced eqution dt +5dy +6y =sin(4t π). dt Solution 13 First we use the trigonometric identity ³ sin (α) =cos α π to obtin ³ sin(4t π) =cos 4t π π. Thus, we cn rewrite our problem s dt +5dy +6y =cos dt 4t 3π Thus, A =, =4,ndθ =3π/. The generl solution of the unforced eqution, is dt +5dy +6y =0 dt N (t) =c 1 e 3t + c e t. Also, since p =5, q =6, q = 10, nd(p) +(q ) =500,wesee solution of the forced eqution. dt +5dy +6y =cos(4t) dt is F (t) = 1 50 cos (4t)+ 1 sin (4t). 5 This mens tht solution of the forced eqution dt +5dy +6y =cos 4t 3π dt 0
is G (t) =AF t θ 1 = 50 cos 4 t 3π + 15 8 sin 4 t 3π 8 = 1 5 cos 4t 3π + 5 sin 4t 3π. In conclusion, the generl solution of the forced eqution dt +5dy +6y =cos 4t 3π dt is y (t) =c 1 e 3t + c e t + 1 5 sin 4t 3π cos 4t 3π. 1