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Impulsive Methods • The big picture – Principle of Superposition – Overview of two methods. • Impulse superposition – Green’s function for underdamped oscillator • Exponential driving force – Green’s function for an undamped oscillator • Solution for constant force • Step function method – Why it works? – Undamped example • Purely time dependent force as Green’s integral • – Equivalence to double integral solution Non-Quiescent initial conditions Impulsive Methods 1 What will we do in this chapter? We develop the impulse (Green’s function) method for getting solutions for the harmonic oscillator with an arbitrary time dependent driving force. We do this using two techniques. In the first method we write the solution as a superposition of solutions with zero initial displacement but velocities given by the impulses acting on the oscillator due to the external force. An arbitrary driving force is written as a sum of impulses, The single impulse x(t) responses are added together in the form of a continuous integral. Another way to get this integral is based on adding the responses of the oscillator to a rectangular driving force by considering this as the superposition of a positive and negative step function. We next take the limit where the time base of the rectangle becomes infinitesimal and compute the response of the driving force to this “impulse” . Our solutions will be for quiescent initial conditions (zero initial displacement and velocity) We conclude by extending the treatment to different initial conditions. This basic Green’s function method is used in nearly all branches of physics. Impulsive Methods 2 The big picture F (t ) We are going to work out a general expression for the response of a damped mass-spring system to an arbitrary force as a function of time making some very clever uses of Superposition. We will view the force as a sum of rectangular infinitesimal impulses and add the x(t) solutions for each impulsive force. For an initially quiescent oscillator each impulse produces a solution equivalent to a free oscillator with initial velocity of v0 = FDt/m. The solution becomes a sum (integral) over such impulse responses. An alternative method of solution is to solve for the x(t) response of each impulse by viewing an impulse as the sum of a step function and an inverted step function. The difference of these step function responses is related to the derivative of the step function response. t f (t ) impulse t0 t0 t0 t0 (t t0 ) Impulsive Methods Step function t0 3 Impulse Superposition Method F (t ') D F (t ) F (ti ) t t Consider an underdamped harmonic oscillator at x 0 and at rest at t 0 subjected to an sharp impulse delivered at t t '. Immediately after the impulse the oscillator aquires a velocity of t ' D t ' D F (t ) F (t ') D v lim D a (t ) dt dt t' t' m m But as D we still have x(t'+D) 0 . At a time t t ' we just have the familiar solution for an F (t ')D underdamped oscillator with x o 0 and v 0 : m 1 F (t ') D x (t ) sin (t t ') exp t t ') m If we had a series of impulses of duration D at t i at a time t than max t i , superposition gives us: N 1 F (ti ) D x(t ) i 1 sin (t ti ) exp t ti ) m ti tN In the D limit the solution becomes the integral: sin (t t ') exp t t ') 1 t dt ' F ( t ') m 0 This result is called a Green's Integral for the undamped oscillator. This is a powerful result since it represents the solution to the differential equation for an arbitrary force mx 2m x kx F (t ). The only catch is that solution is for an initially quiescent oscillator x(0) x(0) 0. We sometimes write this as an integral over a "Green's" function x(t ) 1 dt ' F (t ') G (t , t ') where m 0 1 t t ' G (t , t ') e sin 1 t t ' (t t ') m1 x(t)= Impulsive Methods 4 Example: exponential driving force One then combines factors such as exp(-i1t ) F (t ) F0e t (t ) x(t)= Fo t t t ' t ' dt ' ( t ') e sin 1 t t ' m1 Fo e t t t ' i1 t t ' t ' i1 t t ' dt ' e e m1 2i 0 i1 t ' Fo e t e i1t e m1 2i i1 0 i1 t ' t F e e e o m1 2i i1 0 i1t e 1 Fo e e m1 2i i1 i1 t ' i1t t e e 1 Fo e m1 2i t ' i1t ' t i1t i1 t ' Reduction is tedious but straightforward Fo m 12 2 x(t)= e t e t cos t sin t 1 1 1 Sometimes difficult to work problems analytically but often easy to integrate Green’s functions on the computer! t t and exp(+i1t ) into sines and cosine. We show the response to an exponential with moderate and very high damping. At high damping the x(t) response is nearly the same as the driving force except for the clear x0 = 0 initial condition. 1.4 1.2 0.21 exp( t ) x(t ) 1 0.8 0.6 0.4 0.2 0 1.2 0 10 20 30 0 10 20 30 1 0.8 40 50 540 50 1 0.6 0.4 0.2 0 Impulsive Methods A simpler example Consider solving the solving the undamped oscillator subject to a constant force x - 2 x Fo / m and a quiescent initial state using Green's function techniques. We can get build the Green's integral from the homogenous v solution: x(t ) x0 cos t 0 sin t for x 0 =0 and v0 F D / m . For x(0) x(0) 0, the response to an impulse delivered at t' is: v F (t ')D x(t ) 0 sin (t t ') ; v0 . Hence by m 1 t superposition x(t ) dt ' F (t ') sin (t t ') 0 m We can check this by turning the damping off of our damped general solution: 1 t t t ' x(t ) dt ' F ( t ') e sin 1 t t ' 0 m1 If =0, then 1 and 1 t x(t ) dt ' F (t ') sin t t ' as before m 0 For the case of a constant force F F0 (t ') Fo t sin (t-t') x(t ) dt ' m 0 Fo t F0 sin (t-t') t 't d t ' cos ( t t ') t ' 0 m 0 2 m 2 1 cos t Thus x(t ) F0 . m 2 Check by solution of the inhomogeneous DE with a constant particular sol. First solve: x 2 x F0 / m xH A sin t B cos t F /m F xP 0 2 x A sin t B cos t 0 2 m Solve for A and B using the initial conditions F x(0) x(0) 0 x(0) B 0 2 0 m F x(0) A 0 Thus x 0 2 1 cos t m This checks our Green's solution! Impulsive Methods 6 A simple non-oscillator example The Green's function approach works for other inhomogenous linear differential equations. An example is v v f (t ) . This describes a mass subjected to a viscous drag force of FDrag m v and a time dependent force of F (t ) m f (t ). We wish to find a Green's integral describing v(t ) for an abtrary f (t ). As before our technique extends only to the quiescent initial condition v(0) = 0 or release from rest. To obtain the Green's integral using the impulse method we need to find the the homogeneous solution for v(t) for an initial velocity v o that we will ultimately set to the impulse f D where D is the time duration of the impulse. For v v 0, the auxillary equation for v exp(r ) is just r 0 r - Hence v Ae- t ; v(0) A, thus v v0 e t Hence the response of a quiescent mass to a single impulse of duration D delivered at t' would be v(t ) f (t ')De- (t t ') or for an t arbitrary f (t ') : v(t ) dt ' f (t ')e - (t t ') 0 As an example, we apply this solution to a mass released form rest in gravity. Here f (t ') g . t t 0 0 v(t ) dt ' f (t ')e- (t t ') ge t dt ' e (t ') t exp( t ') g v(t ) ge t 1 exp t 0 This is indeed the solution we obtained for this problem in Physics 225. Recall the terminal velocity vterm g / Impulsive Methods 7 An alternative (step function) method It is also possible to obtain a Green’s integral through the response of the system to a unit step function. We will state the technique, give an example, and argue why it works. We start with a linear differential equation like A x Bx Cx f (t ) F (t ) m (1) We start by solving the differential equation A X BX CX 1 subjected to the initial condition X(0) 0 X (0) 0 . (This solution will be a sum of a homogeneous (transient) and particular solution. (2) Let X(t) be the solution to part (1) and let X(t) = dX dt (3) The Green's function is just G(t,t')=X(t-t') (t-t') t And the Green's integral is simply x(t ) dt ' f (t ') X (t - t ') - Impulsive Methods 8 An example of the step function method Consider the special case of an undamped F (t ) m We first solve: X 2 X 1 X H A sin t B cos t 1 1 X P 2 X A sin t B cos t 2 oscillator x 2 x Why does this work? We again write the force as a sum of square impulses. F (t ) t Each square impulse can be written as a Solve for A and B using the initial conditions superposition of an upright and inverted step 1 function: (t-t0) X (0) X (0) 0 X (0) B X (0) A 0 Thus X= and X 1 2 2 0 f (t ) 1 cos t d 1 sin t 1 cos t Thus: 2 dt t0 t x(t ) dt ' f (t ') X (t - t ') t0 t0 (t t0 ) - 1 t sin (t-t') x(t ) dt ' F (t ') as before! m - Impulsive Methods t0 t0 unit step function 9 Why did our simple prescription work? f (ti ) D We write the force as a sum of upright and inverted unit step functions multiplied by the force in the center t t1 t2 t3 ti f (t ) LimD0 f (ti ) (t ti ) (t ti D ) i Let X(t) be the solution to A X BX CX (t ) (or the solution to A X BX CX 1 for t 0) By superposition: x(t) LimD0 f (ti ) X (t ti ) X (t ti D ) ti t dX LimD0 X (t ti ) X (t ti D ) D dt t t ' X (t ti )D Thus x(t ) f (ti ) X (t ti ) D dt ' f (t ') X (t - t ') t ti t - We write the response as a sum of responses for each rectangular impulse. These response differences are just the time derivatives in the infinitesimal limit. The sum in the limit of small D becomes an integral but only forces ahead of time considered are included. Impulsive Methods 10 f (t ) F (t ) m Another Non-Oscillator Example These two forms do not look alike but we can integrate the Green solution by parts: t0 t t0 t0 Consider a free particle subjected to an external x(t ) 0 dt ' (t - t ') f (t ') 0 udv force satisfying the Diff Eq: x F (t ) / m f (t ). Let dv f (t ') dt ' and u -(t '- t ) t' Assume x(0) x(0) 0 and f (t t0 ) 0. We v(t ) f (t '') dt '' ; du dt ' 0 already know the solution to this problem is a t ' t0 t0 t' t0 double integral of the form: 0 udv (t - t ')0 f (t '')dt '' 0 vdu t t' 0 0 x(t ) dt ' dt '' f (t '') . Since f (t t0 ) 0, we can write this solution as t0 t ' 0 t0 t' 0 0 0 x(t ) (t - t0 ) f (t '')dt '' dt ' f (t '')dt '' Hence we essentially can write the single t0 t0 t' Green's integral as the familiar double integral x(t ) (t t0 ) dt '' f (t ') dt ' dt '' f (t '') 0 0 0 solution for a purely time dependent force. Let us try to solve this using Green's functions. We could have also obtained the Green's integral from impulse method. In absence of external force We solve X 1 for X (0) X (0) 0. The F (t ')D x(t ) v (t - t ') (t - t ') f (t ') D (t - t ') solution is X (t ) t 2 / 2. The Green function m t is then G (t - t ') X (t - t ') (t - t ')(t t ') By superposition : x(t ) dt '(t - t ') f (t ') 0 t0 and x(t ) dt ' (t - t ') f (t ') 0 Impulsive Methods 11 Non-quiescent initial conditions This is clearly the correct solution when f (t ') 0. Lets consider solving x 2 x f (t ) with We can also check the solution for the below case: x(0) 0 ; x(0) v0 and the driving acceleration f (t ) f0 f(t) first becomes non-zero at t > 0. We cannot D 0 t write solution in the form x(t)= dt' G(t-t') f(t') 0 D since if f(t)=0 this form gives x(t)=0 when in v reality x(t) = 0 sin t. The solution is to t t0 consider t 0 dt' G(t-t') f(t') as the "particular" solution and the complete solution as t v0 x(t ) sin t dt ' G (t - t ') f (t ') 0 where G(t-t') is the Green's function for the initial condition x(0)=x(0)=0. Hence x(t ) v0 sin t 1 t dt ' 0 f (t ') sin (t t ') x(t ) v0 sin t x(t t0 ) v0 1 t0 D t0 sin t dt ' f 0 sin (t t ') f0D sin (t t0 ) We can now try to check this answer by starting our clock at t t - t0 . At t 0 the mass is at v x(t 0) 0 sin t0 and x(t 0) v0 cos t0 After the impulse is applied x(t D) and x(t D) f 0 D v0 cos t0 since F t f0D v m v0 12 sin t0 Checking our solution in impulsive limit We can think of x(t D) and x(t D) as a new set of "initial conditions" that starts the oscillator at t t0 or t 0 The oscillator moves according to: v0 f 0 D v0 cos t0 x(t ) sin t0 cos t sin t where you can easily verify the "initial" conditions. We can re-arrange the formula as: v0 f0D x sin t0 cos t cos t0 sin t sin t x x v0 v0 sin t t0 sin t f0D f0D sin t . Using t t - t0 we have sin t - t0 which equals our Green's solution. Impulsive Methods 13