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Erich Poppitz

Lecture 10: Driven Oscillations ▯ relevant reading: Morin 4.4 ▯ oscillation dynamics when you have driving forces Forced Oscillations ▯ Consider a system that has restoring forces and damping forces but also has a driving force that is periodic. So we have a driving force F(t) such that F(t) = F(t + T) = F(t + 2T)for some T: Fourier’s theorem tells us that such a forcing can be decomposed into sinusoids, and for linear systems, the solutions can be linearly super- posed. ▯ So we are motivated to think of the problem where the driving force is a single sinusoid:  + 2 _ + ! x = F0 cos! t (1) 0 m d with the idea that we can solve for any periodic forcing by expressing F(t) as a Fourier series and summing over the solutions. ▯ Notice that the above equation is INHOMOGENEOUS. That means we are going to have to do something a little di▯erent to ▯nd the solution. It turns out there are 2 pieces to the solution for inhomogeneous equa- tions. We call 1 part the particular solution \x " and the other part p the homogeneous solution \x h. The general solution is then written: x = x h x p We’ll start with the particular solution: ▯ For the homogeneous ODEs we were guessing solutions that were ex- ponentials (including sinusoids) because those are the only functions whose derivatives are proportional to the functions themselves and 1 hence could add to 0. But now we don’t want them to add to 0, we want them to add to the right hand side. For sinusoidal driving forces, that means we want the solutions to add up to the sinusoid. The best starting guess is therefore a sinusoid of the same frequency as the inhomogeneous term. For equation (1) that would be: x = Acos(! t) p d ▯ However, before we do anything we might notice that this won’t work. This is because the x _pwill be proportional to sin(! td and there will be no other sin terms in the equation. Lets try it. Take derivatives and plug into equation (1) gives: 2 2 F0 ▯! dcos(! t)d▯ 2 ! Asid(! t) +d! Acos(0 t) = d m cos(! d) Equating the cos terms gives: 2 2 F 0 ▯! dA + ! 0 = m Equating the sin terms gives: ▯2 ! d = 0 From the second equation, A = 0 so we didn’t get a solution. ▯ The way to ▯x this is to realize we are going to get the sin term in the derivatives and therefore to include a sin term in our initial guess: So instead, we try: x p Acos(! t) d B sin(! t) d Taking derivatives and plugging into equation (1) gives: ▯! (Acos(! t) d B sin(! t))d+ 2 (▯! Asind t + ! Bdcos! d)+ d d 2 2 F0 !0(Acos(! td + B sin(! td) = cos(!dt) m Equating the cos terms gives: 2 2 F0 ▯! dA + 2 ! d + ! A0= m Equating the sin terms gives: 2 2 ▯! Bd▯ 2 ! A d ! B =00 ▯ We can solve these 2 equations for A and B. We get: 2 !d B = A ) ! 0 ! d " # F !2 ▯ !2 A = 0 0 d ) m (! ▯ ! ) + (2 ! ) 2 0 d d " # F 0 2 !d B = 2 2 2 2 m (! 0 ! )d+ (2 ! ) d We have therefore found our particular solution: " # " # F0 ! ▯ ! 2 F0 2 ! d x p 0 d cos! d+ sin! d m (! 0 ! ) + (2 ! ) d2 m (! 0 ! ) + (2 ! ) d2 d d ▯ Note that we could have chosen to use complex exponentials rather than sin and cos in our particular solution. For example you could: { write cos! tdas [e i!dt+ e ▯i!d]=2 on the right hand side of the
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