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math2930_2011fa_week07_IDE_Sol_Strogatz

# math2930_2011fa_week07_IDE_Sol_Strogatz - vital-«r 2.1 2.2...

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Unformatted text preview: vital-«r. 2.1. 2.2 2.3 ._-_p..—mn TC-‘L ’lDe-Jl‘ Fen We l r CUR-Fl L'l’u ‘19-‘8- Interactive Differential Equations Assume a) at 600. Note that we can find xp, the particular solution for Equation (2), by using the method of undetermined coefficients. As your textbook will show, xh is a sinusoidal function of frequency (00, and xp is a sinusoidal function of frequency 0). Use the software to observe that the position function x is the sum of these two sinusoidal functions of different frequencies. Using three colors or dotted and dashed lines, sketch and clearly label the graphs of these functions be10w. xi (0. lltSU-SCI‘ safekis cm 00 bull/t I: my». e'l'el‘s M I-_Q_ 7.“ clause.) t Now set m = 1, k = 1, and a) = 0.5 rad/ sec. Let the values of a) approach coo. For what values of to do beats begin to appear? What happens as 0) gets closer and closer to (00? Describe what happens to the beats. Do they have a greater amplitude? Is their frequency increasing or decreasing? Remember that the scale on the vertical axis is changing as you change 00. A5 ()0 OFF/MM l, x lad/nun l‘llCC Hal's: wags lb er cw L99]. . - bed- lhc fem—9‘ _ Now we get to the exciting part! Set a) : m0 = 1. The effects you observe are called resenance. ra‘bcp. Rather than expanding the scale to accommodate vibrations of huge amplitude, the screen shows the consequences of a ”real” resonance effect. To see what happened mathematically, we must solve Equation (2) using the method of undetermined coefficients. Now 3:? = Atcos(w0t) + Btsin(w0t) for some appropriate A and B. As your textbook will show, this process yields the solution x — P0 p _ meo tude of the oscillations as time increases? tsin(aJ0t]. Sketch a graph of the particular solution below. What happens to the ampli— ‘Hie “Mplitoole <3?an (mend) in ‘h'me. The amplitude of the oscillations increases with time. 90 Interactive Differential Equations W completely general, in the following sense: if m, k, and F0 are all nonzero, we can always convert Equation (1) to an equation of the form 2 i2£+2bﬁ+x=coswt (2) dt dt by rescaling time and x appropriately. (For now, take that on faith. If you don’t believe that anything so wonderful could possibly be true, see the last question at the end of this lab.) The upshot is that we can get away with studying Equation (2), which has only two parameters instead of five, and yet we aren’t sacrificing any generality! 1.1 Show that Equation (2) has a particular solution (known as the forced response) given by xF,(t) =Acos(cot- qb) . (3) where the amplification factor A is given by f " 1 A = —— \/(1— w2)2 +4bzco2 (4) and the phase lag q) is ll’ = tan—1[12_b::2 J . ' (5) The solution is obtained by the method of undetermined coefficients, or with complex algebra, and is found in almost all textbooks. 2. Amplitude Response Our goal in this part is to understand the meaning of Equation (4). Open the Vibrations: Amplitude Response tool. The schematic shows a mass on a spring, along with a dashpot that damps the oscillations of the mass. The whole system is being driven by a piston that moves the top of the spring up and down. (More intuitively, imagine holding the top of the spring in your hand, and jiggling it up and down periodically.) The time series graph shows x(t), the solution of Equation (2), for the initial conditions 96(0) 2 0, 55(0) = 0, and for the values of b and a) chosen on the sliders. In the upper-right part of the screen, the curve (4) is plotted as a function of a) , for a given value of b. This graph of A vs. (0 is known as the amplitude response curve. 2.1 Set b = 0.25. Then move the slider for to back and forth, and notice that a horizontal line is being drawn on the time series graph at a height given by A. Now choose a) a 1. Click on the [Start] button to see the resulting solution x(t). How is the eventual amplitude of the oscillations in x(t) related to A ? Explain your observations, using the concepts of homogeneous and particular solu- tions. Thefamplitude of x{t) approaches A. This is expected, since the solution of Equation (2) is x(t) = xh(t) + xp(t), where the homogeneous solution xh(t) -—§ 0 as tincreases. Hence eventually x(t) x xi, (t) = Acos(a)t — gi) , and therefore the amplitude of x(t) approaches A. Lab 12 Forced Vibrations: Advanced Topics 91 2.2 For E: = 0.25 and a) = 1, what is the value of A predicted by (4)? Does this agree with what yOu observe in x(t)? According to Equation (4), A = 2 when b = 0.25 and a) = 1. This agrees with the eventual amplitude of ‘ x(t). 2.3 Increase the driving frequency to a) = 2. How does x(t) change, compared to the previous results? The frequency of x(t) increases to co = 2, and its amplitude decreases. So far you have held the damping strength 19 fixed at b = 0.25. Now explore what happens when you vary b. 2.4 As you decrease i9 below 0.25, describe what happens to the shape of the amplitude response curve. In particular, what happens to the height of the peak in the curve? Also, roughly estimate the value of a) at which the peak occurs. As b is reduced below 0.25, the curve becomes more sharply peaked, and the height of the peak increases. The peak occurs near a) = 1. 2.5 Now suppose you increase 17. What happens to the height and location of the peak in the curve? As b increases, the peak height decreases and its location shifts to smaller values of (a. For b larger than about 0.7, the curve is monotone decreasing, and the peak occurs at a) = 0. 2.6 Perhaps you have noticed that the peak in the amplitude response curve occurs near a) = 1 when E) is small, but shifts to the left for larger b. By maximizing Equation (4) with respect to co, find a formula for the value of a) at which the peak occurs. The straightforward way is to differentiate Equation (4) and then set the result equal to zero. Here’s an easier way that avoids differentiating the square root. The maximum occurs when the denominator of Equation (4) is a minimum, and hence when (11((1— ((9)2 + 4192602) = O. This occurs at to repeal, =V1—2b2. Here we are assuming that 2212 S 1, that is, b S % m 0.707. Otherwise the peak l . occurs at a) = 0. 2.7 Find the height of the peak, as a function of b. This quantity is known as Q, the quality factor of the system. Give an approximate formula for Q if b is small. Substituting wpeak = \ll- 2172 into Equation (4), we find Q = 4 . Hence Q a i for small b. 2W1 - b2 21’ Conduﬂon We have seen that if the damping b is small, the amplification factor A becomes large when a) z 1. This is the phenomenon of resonance: if a vibrating system is weakly damped, it exhibits a large ampliﬁcation factor when driven at a frequency near its natural frequency (which is a) = 1 because of the way we have scaled the parameters). The strength of the resonance is commonly expressed in terms of the peak amplifi- cation factor, that is, the quality factor Q of the system. For instance, electrical engineers try to design radio ampliﬁers with a high Q, to allow for precise tuning and strong amplification of faint signals. ...
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