1. Impulsive Methods1 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

2. Impulsive Methods2 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.

3. Impulsive Methods3 The big picture 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 v 0 = F  t  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. Step function impulse

4. Impulsive Methods4 Impulse Superposition Method

5. Impulsive Methods5 Example: exponential driving force 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 x 0 = 0 initial condition. Sometimes difficult to work problems analytically but often easy to integrate Green’s functions on the computer!

6. Impulsive Methods6 A simpler example

7. Impulsive Methods7 A simple non-oscillator example

8. Impulsive Methods8 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.

9. Impulsive Methods9 An example of the step function method unit step function Why does this work? We again write the force as a sum of square impulses. Each square impulse can be written as a superposition of an upright and inverted step function:  (t-t 0 )

10. Impulsive Methods10 Why did our simple prescription work? We write the force as a sum of upright and inverted unit step functions multiplied by the force in the center 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  becomes an integral but only forces ahead of time considered are included.

11. Impulsive Methods11 Another Non-Oscillator Example

12. 12 Non-quiescent initial conditions

13. Impulsive Methods13 Checking our solution in impulsive limit