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Section 4.3 UNDAMPED FORCING AND RESONANCE
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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
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The solution The general solution to is Explain where each term in the solution comes from. What methods were used to obtain the solution?
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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!
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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
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Homogeneous solution Solve which is the DE for the _________________________. The general solution is undamped harmonic oscillator
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One solution to nonhomogeneous DE Remember that we just need one solution to Guess: Plug in to DE: So = 4/3 and
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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
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Exercise p. 419, #11
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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
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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.
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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
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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.
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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
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Exercise Finish p. 419, #13. p. 420, #21
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