Section 4.3 UNDAMPED FORCING AND RESONANCE

A trip down memory lane… Remember section 1.8, Linear Equations? Example: Solve the first-order linear nonhomogeneous equation: The general solution is a sum of The general solution to the associated homogeneous equation A particular solution to the given nonhomogeneous equation

The solution The general solution to is Explain where each term in the solution comes from. What methods were used to obtain the solution?

An unexpected wrinkle… Here’s a new equation: How is it different from the previous equation? How is it the same? What’s the name of this kind of DE? Guess. What methods can we use? Guess again!

Method Example: Solve the second-order linear nonhomogeneous equation: The general solution is a sum of The general solution to the associated homogeneous equation A particular solution to the given nonhomogeneous equation

Homogeneous solution Solve which is the DE for the _________________________. The general solution is undamped harmonic oscillator

One solution to nonhomogeneous DE Remember that we just need one solution to Guess: Plug in to DE: So  = 4/3 and

General solution to original DE Putting it all together, the general solution to is The solution to the IVP y(0) = 0 and y’(0) = 0 is

Exercise p. 419, #11

Why these pictures? Use BeatsAndResonance and set the parameters a and  to see the graphs of the solution Experiment with changing the values of a and . You should see that (except in very special cases) the solutions look like the product of trig functions. Here’s a handy-dandy trig formula: Now set A = 3t/4 and B = t/4. So

The solution The solution to the IVP is This formula is not in the book (they make you sweat it out with complex exponentials). What happens when a   ? When a and  are not close? When a =  ? Experiment with the applet.

Some answers Again, the solution to the IVP is As a  , the amplitude of the beats  ∞. When a and  are not close, the amplitude is smaller and the beats are not noticeable. When a = , resonance occurs. envelope of beats freq = (a -  )/2  ampl = 2F 0 /(a 2 -  2 ) freq = (a +  )/2  ampl = 1

Exercise Use the formula on the previous slide to do p. 419, #17. Use the guess-and-test method to try to solve p. 419, #13. Why doesn’t it work? Look up the solution to p. 419, #13 in the back of the book. Verify that it is a solution.

Resonance What happens when a =  ? We can’t just plug a =  into the formula. However, we can use L’Hospital’s Rule to find the limit of the envelope function as a  . so a particular solution when a =  is and the general solution is

Exercise Finish p. 419, #13. p. 420, #21