Resonance in a Simple Dynamic System BRIEF ABSTRACT A study of a n oscillator in which we attempt to ascertain the maximum frequency response, where frequency response is the amplitude of the output signal as a function of the frequency of the input signal to a second order differential equation modeling a spring-dashpot system. GENERAL INFORMATION FileName: RESONATE Full title: Resonance in a Simple Dynamic System Developer: Jerry Fine Department of Mechanical Engineering Rose-Hulman Institute of Technology Terre Haute, IN 47803 Phone: 812-877-8353. Email: fine@nextwork.rose-hulman.edu FAX: 812-877-3198 Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Background Information on Simple Harmonic Oscillator The simple harmonic oscillator is probably the most popular model in mathematical physics. Here is a sketch of the concept as it would be thought of by a mechanical engineer. The system has mass, elasticity, and damping. It's being driven by a sinusoidal input force, F = sin(wt), and its output is x(t). Your Problem The system shown above is modeled by the following differential equation. It is obtained by applying Newton's law of motion, F = m a, to the mass. In the following expression prime (') denotes differentiation with respect to time. Therefore, velocity, v=x'(t), and acceleration a = x''(t). x''(t) + 0.2 x'(t) + 4.4 x(t) = sin(wt) The steady state solution to the differential equation is the one that remains in effect long after the system has started and the transient terms have died out. (Transient refers to initial start-up behavior which dies out after a time) It can be shown that a steady state solution to this differential equation is x(t) = A(w) sin(w t) - B(w) cos(w t) 44000 - 10000 w2 where A(w) = ----------------------------------- 193600 - 87600 w2 + 10000 w4 2000 w and B(w) = ----------------------------------- 193600 - 87600 w2 + 10000 w4 Your problem consists in three parts: (1) Verify that the solution given above satisfies the differential equation which models the system. (2) Find the amplitude of the response as a function of w. This is called the frequency response. Plot this response. (3) Find the frequency, w, of the driving force which produces the largest amplitude in the response, x(t). This is called the resonant frequency. KEYWORDS Simple harmonic oscillator, 2nd order differential equation, frequency response, resonance, trigonometric identities TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem shows students a famous identity often used in working with steady state solutions of 2nd order differential equations. The identity takes a function consisting of a sine and a cosine and re-writes it as either a sine or cosine with an amplitude and phase shift. (The problem emphasizes the amplitude feature). This identity leads directly to frequency response, and particularly resonance, which is a prominent feature of the mathematical model. Note that the solution to the differential equation is given. There is no solving of differential equations in the problem. THE IDENTITY IS CLEARLY MARKED AS A STEP IN THE SUGGESTED SOLUTION. Prerequisites Basic knowledge of physics: spring, mass and dashpot systems would help. A feel for what a steady state solution is would be very useful. As already mentioned, the student need not solve the differential equation, only verify that the given solution will work. The main prerequisite is comfort in working with trigonometric functions and identities. Time allotment - time management The first part of the problem is a very tedious exercise without a computer algebra system. With a CAS, it should be the work of a few minutes. The second part of the problem is the hardest. To save time, it will probably be necessary to hint at the trigonometric identity. the third part of the problem involves finding the maximum value of a function. With a CAS it should be very quick. Estimated time for the whole exercise (if hint is given) about 50 minutes. Expectations The student may have some trouble realizing that the amplitude of the response is a function of the driving frequency. In other words, the physical point of the problem may not be obvious and some demonstration or example might be useful. Sometimes just bringing a mass on a spring to class and having the students shake it with various frequencies can be helpful. Future payoffs The problem will make the students more comfortable in working with differential equations. The student will also become more familiar with an important physical phenomenon. Extensions One extension would to create a parameter identification problem. Given the frequency response, find the coefficients in the differential equation. Another, simpler, extension would involve the study of the phase shift as a function of frequency. Reference "Modeling, Analysis and Control of Dynamic Systems" by William J. Palm. John Wiley and Sons, New York, 1983. ISBN 0-471-05800-9. Chapter 5. POSSIBLE SOLUTION Differential Equation ee = x''[t] + .2 x'[t] + 4.4 x[t] == Sin[w t] 4.4 x[t] + 0.2 x'[t] + x''[t] == Sin[t w] Solution to Differential Equation x(t) = A(w) sin(w t) - B(w) cos(w t) 44000 - 10000 w2 where A(w) = ----------------------------------- 193600 - 87600 w2 + 10000 w4 2000 w where B(w) = ----------------------------------- 193600 - 87600 w2 + 10000 w4 X = ( (44000 - 10000 w^2) Sin[ w #]/ ( 193600 - 87600 w^2 + 10000 w^4 ) - 2000 w Cos[ w #] / ( 193600 - 87600 w^2 + 10000 w^4 ) )&; Verification of the Solution X''[t] + 2/10 X'[t] + 440/100 X[t] //Simplify Sin[t w] Identification of the Coefficients A(w) and B(w) B = Coefficient[ X[t], Cos[w t] ] -2000 w ---------------------------- 2 4 193600 - 87600 w + 10000 w A = Coefficient[ X[t], Sin[ w t] ] 2 44000 - 10000 w ---------------------------- 2 4 193600 - 87600 w + 10000 w Step May be Given as a Hint!!!!! If we start with x(t) = A(w) sin(w t) - B(w) cos(w t) we can re-write this as A(w) B(w) x(t) = Q(w) { ------- sin(w t) - ----------- cos(w t) } Q(w) Q(w) where we define Q(w) = [ A(w)2 + B(w)2]1/2 Next , let A(w) / Q(w) = cos(f) and B(w) / Q(w) = sin(f). So f(w) = tan-1( B(w)/ A(w)) and we have x(t) = Q(w) sin( w t - f(w)) so the function Q(w) is our amplitude. Calculation of Amplitude as function of Driving Frequency Amp[w_] = Sqrt[ A^2 + B^2] 2 4000000 w Sqrt[------------------------------- + 2 4 2 (193600 - 87600 w + 10000 w ) 2 2 (44000 - 10000 w ) -------------------------------] 2 4 2 (193600 - 87600 w + 10000 w ) Plots of Amplitude Response Plot[Amp[w], {w,0,8}] -Graphics- Note the peak amplitude in the response which occurs at a frequency of approximately 2.09 radians per second. At this frequency the dynamic system is resonating. This corresponds to tuning in an electrical circuit.. The input to the system is "matched" to the system's own internal properties in a way that produces maximum output. The following two plots show different features of the plot above. Plot[ Amp[w], {w,1.8,2.2}] -Graphics- Plot[ Amp[w], {w,0,100}, PlotRange -> All] -Graphics- Find Resonant Frequency FindRoot[ Amp'[w] == 0, {w,2.1}] {w -> 2.09284} ISSUES IN SOLUTION A major issue is how to answer the question, "Where did the steady state solution x(t), and in particular A(w) and B(w) come from?" At this point it becomes necessary to explain that this kind of solution may be found in a number of ways which are covered in a course on ordinary differential equations. Suffice it to (1): Check that the solution works, and (2) Find out something useful about the behavior given the solution. This problem accomplishes these two goals, but does not go any further.