TheHarmonicOscillator

Transcription

1 TheHarmonicOscillator S F Ellermeyer October 9, The differential equation describing the motion of a bob on a spring (a harmonic oscillator) is m d y dt + bdy + ky () dt In this equation, y denotes the position of the mass with y denoting the rest position, y> denoting a stretched position, y< denoting a compressed position The parameters in this equation are m mass of the bob k spring constant (coming from Hooke s Law) b damping constant (depending on the viscosity of the medium throughwhichthemassismoving) To be physically realistic, we assume that m>, k>, b The case b is not really physically realistic because it implies that there is no damping; ie, that the only force exerted on the bob is due to the spring Nonetheless, we will consider this case because it is mathematically interesting shows us what would happen if damping forces were absent (such as in a vacuum) The key to studying equation () is to rewrite the equation as a linear system of differential equations First, we divide both sides of the equation by m to obtain d y dt + pdy + qy dt

2 where p b/m q k/m Note that, because of our assumption that m>, k>, b, wehaveq> p Next, we define v dy/dt this gives us dv dt d y qy pdy qy pv dt dt We now have the linear system of differential equations dy dt v () dv qy pv dt At first, it might seem undesirable to convert the single differential equation () intoasystemofdifferential equations Our intuition is that studying one equation with one unknown should be easier than studying a system of two equations with two unknowns However, the second unknown (v) insystem () is physically meaningful It is the velocity of the bob If we want to determine the motion of a bob with initial position y () y initial velocity v () v, we must solve system () using initial conditions y () y (3) v () v Defining y Y v A q p we can write system () as dy AY (4) dt write the initial condition (3) as y Y () (5) v,

3 Eigenvalues To find the eigenvalues of the coefficient matrix A, q p we must solve the equation det (A λi) Setting λ q p λ gives us or λ ( p λ)+q λ + pλ + q Using the quadratic formula, we obtain The eigenvalues are thus λ p ± p p 4q λ p p p 4q λ p + p p 4q Whether the eigenvalues are real or imaginary depends on whether the quantity K p 4q is non negative or negative In particular, If K>, then K> we have two real distinct eigenvalues λ p K λ p + K 3

4 If K<, then K Ki we have two imaginary eigenvalues λ p K i λ p K + i (which, notice, are complex conjugates) 3 If K,then K we have repeated eigenvalues λ λ p We will consider cases,, 3 separately The Overdamped Case (K >) If K>, then we have two real distinct eigenvalues λ p K λ p K + for which (as our theory tells us), there must exist two corresponding linearly independent eigenvectors v v Also, the equilibrium point (, ) must be either a sink, a source, or a saddle In order to determine which of these it is, we need to take a closer look at the eigenvalues First, observe that λ < Hence, the origin cannot be a source This makes sense in terms of the harmonic oscillator because it should not be the case that the bob goes flying out of control in one direction However, such an unrealistic situation could also occur if it turned out that λ > (ie, if (, ) were to be a saddle) Hence, we suspect that (if our model is realistic) it must also be the case that λ < To prove that this is in fact true, note that if λ, thenwewouldhave p + K 4

5 which would give us K p since p (by assumption), we would then have K p Using the fact that K p 4q, we would then obtain which reduces to p 4q p q which is contrary to our assumption that q> We conclude that both eigenvalues, λ λ,mustbenegative hence that (, ) must be a sink Now, let us restate the condition K>in terms of the original model parameters m, k, b Since K p 4q, p b/m, q k/m, the condition K>becomes b k 4 > m m which simplifies to b > 4km In summary, if b > 4km then the bob must approach its rest position as t in eventually monotone fashion This is what we call the overdamped case We stretch the bob, let it go, it slowly moves back to its rest position Hence, the harmonic oscillator is not really an oscillator in the overdamped case Analytic Solution of the Overdamped Case To find an eigenvector for the eigenvalue λ p K, 5

6 we need to find a solution of the system (A λ I) v Writingthisequation in long form, we get à p K q p p K! x y which simplifies to p + K q p + x K y Upon inspection, we see that an eigenvector is v p + K To find an eigenvector for the eigenvalue λ p + K, we need to find a solution of the system (A λ I) v Writingthisequation in long form, we get à p+ K q p p+ K which simplifies to p K q p K! x y x y Upon inspection, we see that an eigenvector is v p K The general solution of system () is thus Y k e λt v + k e λt v k e λ t p + + k K e λ t à k e λt k e λ t k ³p + ³ K e λt + k p K p K e λ t! 6

7 Example Supposethatwehaveabob springsystemwithbobofmass m, spring constant k, damping constant b 3 Then, since b > 4km, we are in the overdamped case Now, suppose that we stretch the spring a distance of 4 units from the rest position let it go (with initial velocity v ) What will the motion of the bob be? First, we note that q k m p b m 3 K p 4q λ p K λ p + K Using the general solution obtained above, we have k e Y t k e t 4k e t +k e t In order to satisfy the initial condition 4 Y (), we must solve k k 4 4k +k The solution of this system is k, k 4 Wethushave 4e Y t +8e t 8e t 8e t In particular, the position of the bob at time t is its velocity at time t is y 4e t +8e t v 8e t 8e t 7

8 t Graph of y 4e t +8e t 3 The Underdamped Undamped Cases (K <) If K<, wehaveanimaginaryeigenvalue λ p K + i Therealpartofthiseigenvalueisα p/ the imaginary part is β K/ Sinceweareassumingthatp, we can see that α Furthermore, if p>, thenα < if p,thenα Hence, if p>, then the equilibrium point (, ) is a spiral sink if p, then the equilibrium point (, ) is a center Furthermore, all solutions spiral around (, ) with period π β π K 4π K Since the condition K<is equivalent to the condition b < 4km, wecan summarize our findings in terms of the original model parameters as follows: If b < 4km, then the bob oscillates about its rest position If p>, then the oscillations are damped meaning that they decay in amplitude to amplitude zero as t However, ifp (meaning that there actually is no damping), then the oscillations are sustained (meaning that their amplitude does not decay) The case p> ( K>) is called the underdamped case The case p ( K > ) iscalledtheundamped case 8

9 3 Analytic Solution of the Underdamped Undamped Cases To find a complex eigenvector corresponding to the complex eigenvalue λ p K + i, we must solve (A λi) v which is written in long form as à p! K i x q p K y i or equivalently as p Ki q p Ki Upon inspection, we see that v x y p Ki is a (complex) eigenvector This means that Y e λt v is a complex solution to system (4) We now use Euler s formula to break this solution into its real imaginary parts: Y e λt v e (αt+βti) p Ki (cos(βt)+i sin (βt)) p Ki (cos (βt)+isin (βt)) e αt e αt cos(βt) i sin (βt) p cos (βt)+ K sin (βt)+ p sin (βt) K cos (βt) i e αt cos (βt) pe αt cos (βt)+ Ke αt + sin (βt) e αt sin (βt) pe αt sin (βt) Ke αt cos (βt) Note that, as guaranteed by out theory, the real imaginary parts of this complex solution, e Y Re αt cos (βt) pe αt cos (βt)+ Ke αt sin (βt) 9 i

10 Y Im e αt sin (βt) pe αt sin (βt) Ke αt cos (βt) are themselves solutions they are linearly independent solution of system (4) is thus The general Y k Y Re + k Y Im e k αt cos (βt) pe αt cos (βt)+ e Ke αt + k αt sin (βt) sin (βt) pe αt sin (βt) Ke αt cos (βt) k e αt cos (βt) k e αt sin (βt) k pe αt cos (βt)+k Ke αt sin (βt)+k pe αt sin (βt) k Ke αt cos (βt) This means that the position of the bob at time t is y k e αt cos (βt) k e αt sin (βt) the velocity of the bob at time t is ³ v ³k p k K e αt cos (βt)+ k K + k p e αt sin (βt) Example Supposethatwehaveabob springsystemwithbobofmass m, spring constant k, damping constant b Then, since b < 4km ( b>), we are in the underdamped case Now, suppose that we stretch the spring a distance of 4 units from the rest position let it go (with initial velocity v ) What will the motion of the bob be? First, we note that q k m p b m K p 4q 4 λ p K + i +i Using the general solution given above, we obtain y k e t cos t k e t sin t

11 v (k k ) e t cos t +(k +k ) e t sin t To find the solution satisfying the initial condition y () 4 v (), we must solve k 4 k k The solution of this system is k, k Thus, we have y 4e t cos t +4e t sin t v 8e t sin t t Graph of y 4e t cos t +4e t sin t The oscillations get very small after time t 4or so The graph below shows a magnified view of y 4e t cos t +4e t sin t

12 8 6 y t Example 3 Supposethatwehaveabob springsystemwithbobofmass m, spring constant k, damping constant b Then, since b < 4km ( b ), we are in the undamped case Now, suppose that we stretch the spring a distance of 4 units from the rest position let it go (with initial velocity v ) What will the motion of the bob be? First, we note that q k m p b m K p 4q 4 λ p K + i i Using the general solution given above, we obtain y k cos t k sin t v k cos t +k sin t To find the solution satisfying the initial condition y () 4 v (),

13 we must solve k 4 k The solution of this system is k, k Thus, we have y 4cost v 4sint 4 y t - -4 Graph of y 4cost 4 The Critically Damped Case (K ) The case K p 4q is called the critically damped case This is the case where we have a real repeated eigenvalue λ p/ Inthis case, the equilibrium point (, ) is a sink However, note that a critically damped system is very sensitive to perturbations Even a very slight change in the spring constant or the damping constant will cause K to become either positive or negative which will result in either an overdamped or an underdamped spring bob system 4 Analytic Solution of the Critically Damped Case To find eigenvectors for λ, wesolve p q p x p y 3

14 which is equivalent to p q p x y This means we need to solve the system px +y qx py To see that this system is redundant (as expected), we multiply the first equation in the system by p multiply the second equation in the system by to obtain the equivalent system p x +py 4qx +py Since p 4q, the system is redundant Furthermore note that since we are assuming that q>thenitmustbethecaseherethatp 6 Aneigenvector for λ is thus v p it is not possible that every vector is an eigenvector for λ (as could happen if we were to remove our assumption that q>) In this case, our theory tells us that every solution of system (4) is of the form Y k e λt v + k e λt (tv + v ) where v is a vector satisfying (A λi) v v To find v we must thus solve the system p x q p y p which is equivalent to p q p x y 4 p This system has solution v 4

15 We conclude that the general solution of equation (4) is Y k e λt + k p e λt t + p e λt (k + k t) e λt (p (k + k t) k ) The position of the bob at time t is thus y e λt (k + k t) the velocity of the bob at time t is v e λt (p (k + k t) k ) Example 4 Supposethatwehaveabob springsystemwithbobofmass m, spring constant k, damping constant b Then, since b 4km, we are in the critically damped case Now, suppose that we stretch the spring a distance of 4 units from the rest position let it go (with initial velocity v ) What will the motion of the bob be? First, we note that p b/m λ p/ Using the general solution given above, we obtain y e t (k + k t) v e t ( (k + k t) k ) To find the solution satisfying the initial condition we must solve y () 4 v (), k 4 k k The solution of this system is k, k Thus, we have y 4e t ( + t) v 4te t 5

16 4 3 y t Graph of y 4e t ( + t) 6