Math 244 Periodic Forcing

Sections 3.8 Last time, we focused our attention on the spring-mass model that had no forcing. Today, we look at the special case of periodic forcing. No Damping, Periodic Forcing Rather than write the model as mu00 + ku = F (t), it is common practice to write the model as u00 + ω02 u = F (t). That way it is easy to read off the circular frequency of the homogenous part of the solution. Putting in the periodic forcing (we’ll use cosine) u00 + ω02 u = F0 cos(ωt) And, as long as the forcing function does not have the same frequency as the natural frequency, ω0 6= ω, then we can solve for the particular solution using complexification: yp = Aeiωt : F0 Aeiωt (−ω 2 + ω02 ) = F0 eiωt ⇒ A = 2 ω0 − ω 2 so the overall solution is: F0 y(t) = C1 cos(ω0 t) + C2 sin(ω0 t) + 2 cos(wt) w0 − w 2 It can be shown (Exercise 1) that, if y(0) = y 0 (0) = 0, then C1 = − ω02 F0 − ω2 C2 = 0 In this case, F0 (cos(wt) − cos(w0 t)) − w2 In the handout on the next page, we see what happens to this function. We see a series of graphs where w0 = 1 and w is changing, from w = 2 to w = 1.01. The values were w = 2, 1.5, 1.1, 1.01 y(t) = w02 This shows the phenomena known as beating. Beating occurs (formally) when there is no damping, and when the frequency for the driving force is very close to (but not equal to) the natural frequency. Using some trig identities, it is possible to rewrite y(t) as the following (p. 213 of the text): ! ! 2F0 (ω0 − ω)t (ω0 + ω)t sin sin ω02 − ω 2 2 2 As ω → ω0 , the period of the first sine gets larger and larger. In fact, we see the large, slower oscillating function in the images below- They represent 2F0 (ω0 − ω)t ± 2 sin 2 ω0 − ω 2 1 ! From the images, we see that ONE SINGLE BEAT has CIRCULAR FREQUENCY that is half of the sine wave, or |ω0 − ω|. From Beating to Resonance 10 Figure 1: Figure showing the function 1−ω 2 (cos(t) − cos(ωt)). The graphs show the result of taking ω = 2, ω = 1.5, ω = 1.1 and ω = 1.01. As w → 1, we see that the amplitude and period of the beats are getting larger and larger and larger! As a particular example, consider the upper right figure. This comes from taking ω0 = 1 and ω = 1.5. In that case, the circular frequency is 1/2, which makes the period of one beat: 2π/(1/2) = 4π ≈ 12.5 (examine the figure to double check). 2 Resonance What happens at ω = ω0 ? Something known as resonance. • Geometrically, we see that the period of the larger beat is infinitely long, with an infinitely large amplitude (draw graph). • Algebraically (from Method of Undetermined Coefficients), we see that the new guess for the particular solution: u00 + w02 u = F0 cos(w0 t) is yp (t) = tAeiω0 t . We could substitute this in and solve, or we can take the limit of our solution. • Using our previous solution, we take the limit using L’Hospital’s rule: lim w→w0 −F0 t sin(ωt) F0 F0 (cos(wt) − cos(w0 t)) = lim = t sin(ω0 t) 2 2 ω→ω0 w0 − w −2ω 2ω0 3 Bringing in Damping: The Full Model When we have the full damped system, then all solutions will tend to zero as t → ∞. How do we know that? Given mu00 + γu0 + ku = F0 cos(ωt) We will never have purely periodic solutions for yh , therefore, yp will never have the multiplication by t (by undetermined coefficients). HOWEVER, we will still analyze a slightly changed version of the system: u00 + pu0 + qu = cos(ωt) Now, we’ll feel free to make yp = Aeiωt , and substitute getting: Aeiωt (−ω 2 + iωp + q) = eiωt And we see that: 1 (q − ω 2 ) + iωp Since the forcing function is a cosine function, from our handout, we had that the amplitude R and phase angle δ for yp are given by: A= 1 R= 2 |(q − ω ) + iwp| −1 δ = tan ωp q − ω2 ! Computing these, we can say that the forcing function then has the form yp = R cos(ωt − δ) Now, suppose that we can “tune” the value of ω, so that we’ll fix the other model parameters p, q. In that case, R becomes a function of ω: R= q 1 (q − ω 2 )2 + p2 ω 2 And we can ask: Is there a value of ω that maximizes the amplitude of the forced response, R? To find the max in this case, we’ll differentiate R, then set the derivative to zero. Before we do that, notice the following: 1 = (f (ω))−1/2 R= q f (ω) If we set dR dω dR 1 1 f 0 (ω) = − (f (ω))−3/2 f 0 (ω) = − dω 2 2 (f (ω))3/2 ⇒ = 0, we just have f 0 (ω) = 0- So that simplifies our computation by quite a bit. f (ω) = (q − ω 2 )2 + p2 ω 2 ⇒ df = 2(q − ω 2 )(−2ω) + p2 · 2ω = 0 dω q 2 Solving for only the positive ω, we get ω = 2q−p . It is at this maximizer that we’ll extend 2 the idea of resonance to cover... Here, the amplitude of the response can be VERY LARGE, even though damping is present. 4 Homework Replaces Section 3.8 1. Solve the IVP u00 + ω02 u = F0 cos(ωt), u(0) = 0 and u0 (0) = 0, if ω 6= ω0 . 2. Show that the period q of motion of an undamped vibration of a mass hanging from a vertical spring is 2π L/g. NOTE: We defined L to be the length of the spring stretched from its natural length to equilibrium. 3. Convert the following to R cos(ω0 t − δ). (a) cos(9t) − sin(9t) (c) −2π cos(πt) − π sin(πt) (b) 2 cos(3t) + sin(3t) (d) 5 sin(t/2) − cos(t/2) 4. Suppose u00 + 4u = cos(2.8t). This function exhibits beating. (i) Give the frequence and amplitude of a beat, and (ii) Give (only) the particular part of the solution. HINT: Did you notice the relationship between the constant in front of the particular solution and the constant in front of the product of sines on page 1? 5. Same question as before, but with u00 + 9u = cos(3.1t). 6. Same question as before, but with u00 + u = cos(1.3 t) 7. Find the solution to u00 + 9u = cos(3t) with zero initial conditions. 8. Find the general solution using complex exponentials: y 00 + 3y 0 + 2y = cos(t). 9. Consider u00 + pu0 + qu = cos(ωt). In the notes at the bottom of p. 4, we got that s ω= 2q − p2 2 Thinking of p as damping, if the damping is very very small, then approximately what value of ω will result in a very large amplitude response? 10. Consider u00 + u0 + 2u = cos(t). Find the amplitude and the phase angle of the particular part of the solution (or, for the forced response). You do NOT need to get the homogeneous part of the solution. 11. Consider u00 + u0 + 2u = cos(ωt). Find the value of ω that will maximize the amplitude of the response. NOTE: I don’t want you to memorize the value of ω. Rather, find the amplitude R, then differentiate to find where the derivative is zero. Remember our shortcut (dealing with f (ω)). 5 12. Pictured below are the graphs of several solutions to the differential equation: y 00 + py 0 + qy = cos(ωt) Match the figure to the choice of parameters. Choice (A) (B) (C) (D) b 5 0 0 2 c 3 2 1 1 ω 1 1 1 3 13. Write the forced response to the ODE below as R cos(ωt − δ): u00 + u0 + 2u = cos(3t) 14. Suppose we can tune the value of q rather than the value of ω in the differential equation (where ω = 3): u00 + u0 + qu = cos(3t) Find the value of q that will maximize the amplitude of the forced response. 6