Module P5.3 Forced vibrations and resonance

Transcription

1 F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module P5.3 Forced vibrations and resonance 1 Opening items 1.1 Module introduction 1.2 Fast track questions 1.3 Ready to study? 2 Driven oscillators 2.1 Qualitative discussion of driven oscillators and resonance 2.2 The equation of motion of a harmonically driven oscillator 2.3 Steady state motion 2.4 Steady state energy balance and power transfer 2.5 Transient motion 2.6 Resonance and frequency standards 3 Coupled oscillators 3.1 Normal modes 4 Closing items 4.1 Module summary 4.2 Achievements 4.3 Exit test Exit module

2 1 Opening items 1.1 Module introduction In order to sustain vibrations, energy must be supplied to a system to make up for the energy transferred out of the vibrating system due to dissipative forces or damping. The resulting vibrations are called forced vibrations. In a clock or watch, the pushes that maintain the vibrations are applied at the frequency at which the pendulum or balance wheel normally vibrates, i.e. at its natural frequency. Pushing at the natural frequency is the most efficient way to transfer energy into a vibrating system, as your childhood experiences on a swing should confirm. In Section 2 the equation of motion for the linearly damped oscillator is developed, both for an undriven oscillator and in the presence of a harmonic driving force. We examine the steady state motion of this latter system as the amplitude and frequency of the driving force is varied, considering the low and high frequency response and the resonance response. We then look at the energy balance in the steady state system, considering the kinetic, potential and total energies. The transient motion is distinguished from the steady state motion and examined briefly. Other mechanical and non-mechanical cases of driven oscillators are considered and the application of the principle of resonance absorption in clocks and frequency standards is discussed. The cases of quartz crystal oscillators, the caesium atomic clock and the potential use of frequency-stabilized lasers are considered briefly.

3 In Section 3 we look at the related situation where the exciting force is due to another oscillator and consider how two coupled oscillators interact. The important idea of normal modes is introduced, along with the idea of beats between these modes. Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the Fast track questions given in Subsection 1.2. If not, proceed directly to Ready to study? in Subsection 1.3.

4 1.2 Fast track questions Study comment Can you answer the following Fast track questions?. If you answer the questions successfully you need only glance through the module before looking at the Module summary (Subsection 5.1) and the Achievements listed in Subsection 5.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 5.3. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module. Question F1 Write down the equation of motion of a damped harmonic oscillator driven by an external force which varies sinusoidally (harmonically) with the time. Explain what is meant by the terms steady state motion and transient motion. Give a qualitative description of these motions and explain the relationship between them.

5 Question F2 What two parameters characterize the steady state motion of an oscillator which is driven harmonically at an angular frequency Ω? Make sketches indicating how these parameters vary with the angular frequency Ω of the driving force. With reference to your sketches, describe the main features of the phenomenon of resonance. Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items.

6 1.3 Ready to study? Study comment In order to study this module you will need to be familiar with the following terms: acceleration, Cartesian coordinate system, displacement, displacement time graph, energy, equilibrium, force, Hooke s law, mass, kinetic energy, Newton s laws of motion (including the use of the second law to formulate the equation of motion of a system), period, position, potential energy, power and velocity. You should also be familiar with simple harmonic motion (see Question R2), and aware of the general features of damped harmonic motion (though this is briefly reviewed in the module). Mathematically, you will need to be familiar with exponential and trigonometric functions and with the inverse trigonometric function arctan(x). You will also be required to use various trigonometric identities. The notation of differentiation (dx/dt, d 2 x/dt 2 ) is used freely throughout this module, and you must be familiar with it, however you are not actually required to evaluate derivatives for yourself. An averaging process is introduced which requires a knowledge of integration if it is to be fully understood, but again you are not required to evaluate any integrals for yourself. Simple differential equations are also introduced in the module but a familiarity with this topic is not assumed. Mathematical requirements are a familiarity with trigonometry (including angle, cosine, degree, periodic functions, radian, sine, tangent, trigonometric identities), the differentiation (including higher derivatives) and integration of polynomial and trigonometric functions. If you are unsure about any of these items you can review them now by reference to the Glossary, which will also indicate where in FLAP they are developed. The following Ready to study questions will allow you to establish whether you need to review some of the topics before embarking on this module.

7 Question R1 A body of mass m moves along the x-axis of a Cartesian coordinate system, subject to two forces F 1x and F 2x. If its instantaneous position is denoted by x(t), write down its equation of motion. Question R2 A simple harmonic oscillator moves in such a way that its displacement from its equilibrium position at time t is given by x(t) = A 0 1sin1(ω1t + φ). Explain the physical significance of the constants A 0, ω and φ. Question R3 A simple harmonic oscillator has period T = 5.001s. What is the frequency of the oscillator? What is the corresponding value of ω?

8 Question R4 (a) Given the trigonometric identity cos(θ + φ ) = cosθ cos φ sin θ sin φ show that cos 2 θ + sin 2 θ = 14and4 cos 2 θ sin 2 θ = cos(2θ). (b) Use the trigonometric identity sin (θ + φ ) = sin θ cos φ + cosθ sin φ to show that sin (θ π)= sin θ Question R5 Use the first identity in Question R4 and the simple values for the sine and cosine of π/4 and π/6 to show that: cos 5π 12 = and4 cos π 12 =

9 2 Driven oscillators 2.1 Qualitative discussion of driven oscillators and resonance Oscillating systems typically have one or more characteristic natural frequencies at which they vibrate when disturbed. These natural frequencies can be very useful. In the case of a simple pendulum for instance, the single natural frequency may be used for time keeping, as in a pendulum clock. However, when using oscillating systems in technical and scientific applications it is often necessary to suppress natural oscillations and replace them by externally imposed oscillations. An example of this kind arises in the case of loudspeakers. In order to reproduce a wide range of sounds, loudspeakers must be made to vibrate over a large range of frequencies, and if sounds are to be reproduced faithfully any natural frequencies associated with the loudspeaker must not be unduly accentuated. When we wish to produce oscillations in a system we do so by applying a driving force which has its own frequency associated with it. An oscillator responding to such a force is said to be a forced oscillator or a driven oscillator. Examples of forced oscillators are plentiful, even a device as simple as a child s playground swing can serve the purpose.

10 Imagine a playground swing with a small child seated on it. If you pull the swing back a little from equilibrium, into a raised position, and then release it from rest, it will oscillate back and forth about its equilibrium position. To a first approximation, the swing will behave like an ideal pendulum, oscillating with constant total energy, and may be represented by a simple harmonic oscillator. Of course, this is only a crude approximation to the true behaviour; the swing will actually lose energy due to dissipative forces, such as friction and air resistance, and the oscillations will eventually die away. A more accurate description of the swing would take into account these various damping effects that dissipate the energy, and, provided they were relatively weak, would represent the swing as a lightly damped harmonic oscillator with an amplitude that decreases gradually with time. Question T1 If the amplitude of the oscillations of the swing decays to half its initial value after five oscillations, sketch a displacement time graph for the first ten oscillations.4

11 Left to themselves the oscillations of the swing die away, but if you catch the swing at its highest point, at the completion of each oscillation, and push it back towards the equilibrium position, then the oscillations can be sustained despite the damping. The swing may now be described as a driven damped oscillator, or simply a forced oscillator, driven by the force you provide during part of each cycle of oscillation. Question T2 Sketch a possible displacement time graph for this case, superimposing on your sketch a graph of the force you would have to supply to sustain the oscillation. Assume that the driving force has a constant magnitude F d during the time you are in contact with the swing, and that you never apply the force when the swing s displacement from its equilibrium position is negative. Your sketch should be guided by your answers to the following questions: Is the motion periodic? What governs the pushing frequency? During what part of the cycle would you push the swing? Is the motion symmetric about the equilibrium position?4

12 The driving force shown in the answer to Question T2 is similar to the force you really would supply if you wanted to keep the swing oscillating with the minimum effort. It is a periodic force with a frequency that closely matches the natural frequency of the swing. Driving an oscillator in this way, at a frequency that is close to its natural frequency, can result in the efficient transfer of energy to the oscillator and may cause the amplitude of the oscillations to become very large. This is the condition known as resonance, an understanding of which is important in many fields of science and technology. The large amplitude forced oscillations that occur near to the natural frequency of a system, due to resonance, can be disastrous. At Angers in France, in 1850, almost half of a column of about 500 soldiers were killed when the suspension bridge across which they were marching collapsed. The rhythmic marching of the soldiers excited a natural swinging motion of the bridge, which brought about its destruction. Similarly, gusting wind in the Tacoma Narrows in the north-western USA, in November 1940, caused a new suspension bridge to vibrate rather in the same way as the metal slats of a venetian blind flutter when air rushes over them. Energy was supplied to the bridge at its natural frequency and the amplitude of the vertical and then the twisting vibrations became so large that the roadway ripped from its hangers and plunged into the water below. You may have seen the classic film footage of this spectacular disaster. In the same way, tall buildings, radio aerials and factory chimneys can also sway in the wind and must be constructed so that resonance is avoided.

13 In contrast to these negative aspects of resonance, there are situations in which resonance is desirable and can be usefully exploited. Many examples of this kind occur in the production and detection of sound. For instance, it is quite easy to produce sound by applying a driving force to a small body of air and causing it to resonate. A tone of about 2001Hz can be obtained from an empty one-pint milk bottle, simply by blowing across the opening. Organ pipes and some other wind instruments work in a similar way. The same principle is also widely used in nature. Howler monkeys and some frogs and toads have large vocal sacs that allow forced oscillations at a near natural frequency to produce surprisingly loud sounds. In fact, such systems often resonate over quite a wide range of frequencies, a point to which we will return later in the module. Aside The phenomenon of resonance is not confined to macroscopic systems; examples abound in the microscopic world of atoms and nuclei, where physical systems are described by quantum physics rather than classical physics. While quantum systems are beyond the scope of this module, it is worth mentioning that useful insights into their behaviour can sometimes be gained from classical ideas. For example, the strong absorption by atomic nuclei of γ-rays in the wavelength range from 6 to m may be thought of as the resonant excitation of a vibration in which the protons and neutrons in a nucleus oscillate, in anti-phase, about the nuclear centre of mass.

14 2.2 The equation of motion of a harmonically driven oscillator Another simple example of a forced oscillator is shown in Figure 1. A body of mass m is attached to one end of a horizontal spring, the other end of which is attached to a fixed point P. The body can slide back and forth along a straight line, which we will take to be the x-axis of a system of Cartesian coordinates, and is subject to three forces all acting in the x-direction (they may be positive or negative): A restoring force F 1x due to the spring that tends to return the body to its equilibrium position. A damping force F 2x due to friction and air resistance that opposes the motion of the body. A driving force F 3x provided by an external agency. P rough surface x m Figure 14A simple driven oscillator. The driving force has been omitted from the figure because its sign may change with time. In the absence of any driving force, the system would have a natural equilibrium configuration in which the spring was neither extended nor compressed. If we take the equilibrium position of the mass to be the origin of the coordinate system, we can say that the mass s displacement from equilibrium at time t is given by its instantaneous position coordinate x1(t).

15 Using calculus notation, it follows that at any time t the velocity and acceleration of the mass are, respectively dx dv d2x vx = 4and4 a x = x = 2 dt dt dt Using Newton s second law of motion we can then say that the sliding body must obey an equation of motion of the form d2x m 2 = F1x + F2 x + F3x (1) dt To make further progress we now need to relate the forces that act on the sliding body to the time t or to the body s instantaneous position x. This is easily done provided we idealize the system to some extent. First, let us suppose that the spring is ideal, which means that it obeys Hooke s law perfectly, so that F1x = 1kx (2a) where k is the spring constant that characterizes the spring. FLAP P5.3 Forced vibrations and resonance COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

16 Second, let us assume that damping forces which oppose the motion have magnitudes that are proportional to the instantaneous speed of the sliding body, so that F 2 x = b dx (2b) dt where b is a constant that characterises the dissipative forces. Third, let us suppose that the driving force is periodic, and has the relatively simple form F P 3x = F 0 1sin1(Ω1t) (2c) m where F 0 is the maximum magnitude that the driving force attains, and Ω is the angular frequency of the driving force. Note that the angular frequency Ω is externally imposed and is not necessarily related in any way to the natural frequency of the system in Figure 1. rough surface Figure 14A simple driven oscillator. The driving force has been omitted from the figure because its sign may change with time. x

17 Substituting Equations 2a, 2b and 2c into Equation 1, F 1x = 1kx (Eqn 2a) F 2 x = b dx dt (Eqn 2b) F 3x = F 0 1sin1(Ω1t) (Eqn 2c) m d 2 x dt = F 2 1x + F 2 x + F 3x (Eqn 1) we see that the behaviour of the sliding mass must now be described by the equation m d 2 x dt 2 dx = kx b dt + F 0 sin (Ωt) which can be rearranged to isolate the time-dependent driving term as follows m d 2 x dt + b dx 2 dt + kx = F 0 sin (Ωt) (3)

18 A driving force that depends sinusoidally on time is sometimes described as a harmonic driving force, and a damping force with magnitude proportional to velocity is sometimes called a linear damping force. Equation 3 m d 2 x dt + b dx 2 dt + kx = F 0 sin (Ωt) (Eqn 3) may therefore be described as the equation of motion of a harmonically driven linearly damped oscillator. This equation arises in a number of physical contexts, though it is often presented in a form that differs somewhat from Equation 3. One of the most common variants is obtained by dividing both sides of Equation 3 by the mass m, and then introducing the damping constant γ = b/m the maximum magnitude of the driving acceleration a = F 0 /m and the natural angular frequency ω 0 = km The last of these is the angular frequency the oscillator would have if it were just a simple harmonic oscillator without any damping or driving. Substituting these expressions into Equation 3 we can say that:

19 for the harmonically driven linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (4) Equations of this kind, involving an independent variable (t), a dependent variable (x(t)) and derivatives of the dependent variable with respect to the independent variable, are known as differential equations. From a mathematical point of view, Equation 4 involves only the first power of x and its derivatives (i.e. it is a linear equation) and the derivative of highest order that it contains is a second derivative. Equation 4 is therefore classified as a linear second-order differential equation. From a physical point of view, it is clear that the driving force is represented by the term on the right-hand side of Equation 4. So, if we set a = 0 we obtain the equation of motion for an undriven linearly damped oscillator: linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = 0 (5)

20 If, in addition, we suppose that there is no damping, so γ = 0, the equation of motion becomes that of a simple harmonic oscillator: simple harmonic oscillator: d 2 x dt 2 + ω 0 2 x = 0 (6) We will approach the solution of Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) through the simpler Equation 6 and then Equation 5. linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = 0 (Eqn 5)

21 First we look at Equation 6. d2x + ω 02 x = 0 (Eqn 6) dt 2 Now, you should be familiar with simple harmonic motion in one-dimension, and you should know that it can generally be described by an equation of the form simple harmonic oscillator: x(t) = A0 1sin1(ω0 t + φ) (7) where the amplitude A0 is a constant that determines the maximum displacement from the equilibrium position, and the phase constant φ determines the initial position of the oscillator at t = 0, since x(0) = A01sin1φ. If you know how to differentiate the function that appears in Equation 7 you should be able to show that dx vx = = A0 ω 0 cos (ω 0 t + φ ) dt d2x and a x = 2 = A0 ω 02 sin (ω 0 t + φ ) dt and hence confirm that Equation 7 is a solution to Equation 6. To a mathematician, the presence of two arbitrary constants (A0 and φ) in Equation 7 also shows that it is the general solution to Equation 6, i.e. every solution to Equation 6 can be written in the form of Equation 7 by assigning suitable values to A0 and φ. FLAP P5.3 Forced vibrations and resonance COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

22 In a similar way, but with more mathematical manipulation, it can be shown that damped harmonic motion, with its exponentially decaying amplitude can be described by an expression of the form x(t) = A 0 1e γ1t/ 0 2 1sin1(ωt 1 + φ) 4where 4 ω = ω 02 γ 2 4 (8) which is the general solution to Equation 5. linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = 0 (Eqn 5) If we want to determine how the sliding body of Figure 1 moves when subjected to the three forces specified above we need to find a solution to Equation 4, d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) P rough surface x m Figure 14A simple driven oscillator. The driving force has been omitted from the figure because its sign may change with time. i.e. we need to find an expression relating the instantaneous position x of the body to the time t, that satisfies Equation 4 and any other condition that the oscillator is known to satisfy. Solving Equation 4 is actually somewhat more complicated than solving Equations 5 and 6. d simple harmonic oscillator: 2 x dt + ω x = 0 (Eqn 6)

23 This is because Equations 5 and 6 linearly damped oscillator: d 2 x dt + γ dx 2 dt + ω 0 2 x = 0 (Eqn 5) simple harmonic oscillator: d 2 x dt + ω x = 0 (Eqn 6) are homogeneous differential equations in which each term is proportional to x or to one of its time derivatives, whereas Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) is an inhomogeneous differential equation because it contains the driving term, a1sin1(ωt), which does not involve the displacement at all, but does depend explicitly on the time. The mathematical procedure for solving such equations is described in detail in the maths strand of FLAP, where it is applied to Equation 3 m d 2 x dt 2 + b dx dt + kx = F 0 sin (Ωt) (Eqn 3) (see second-order differential equation in the Glossary for references), but we will not employ such methods here.

24 In this module we will adopt a more physical approach in which we will use physical arguments to determine the overall form of a solution, and mathematical techniques to work out the details. Methods of this kind are much used by physicists. Question T3 Devise and describe two simple systems that, like the system in Figure 1, are capable of oscillatory motion and which may be acted upon by an externally produced periodic driving force.4 P rough surface x m Question T4 A student claims that the action of an external driving force must eventually lead to an increase in the mechanical energy of the sliding body of Figure 1. The student therefore concludes that in the absence of friction, or any other dissipative effect, the sliding body will gain energy continuously and the oscillation will grow without limit, at least until the spring fractures. Do you agree with this argument?4 Figure 14A simple driven oscillator. The driving force has been omitted from the figure because its sign may change with time.

25 2.3 Steady state motion d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) Equation 4 represents a struggle. The oscillator, when disturbed, tends to oscillate with its natural angular frequency ω 0, but it is being driven at a different angular frequency, Ω. It is clear that the driving force must eventually win this struggle, because the oscillator loses energy due to damping and is only sustained in perpetuity by the external agency that provides the driving force. The motion during the struggle is usually called the transient motion and its finite duration is determined by the time taken for free oscillations to decay to a negligible amplitude. This time is of the order of 2π/γ, as you will see in Subsection 2.5, where we consider the transient motion in more detail. In this subsection we will be concerned with the steady state motion that persists after the transient motion has decayed. The steady state motion of the driven oscillator can be expected to be in sympathy with the driving force, so it is likely to be harmonic motion with the same angular frequency as the driving force. However, there is no reason why it should be exactly in-phase with the driving force, indeed on physical grounds you might well expect the displacement to lag somewhat behind the force that causes it.

26 For these reasons we will assume that the steady state motion of the harmonically driven linearly damped oscillator is described by an expression of the form: steady state motion x(t) = A1sin1(Ω1t δ1) (9) where A represents the amplitude of the steady state motion, and δ the phase lag between the steady state motion and the driving force (F 3x = F 0 1sin1(Ω1t)). Having arrived at Equation 9 on the basis of physical arguments, rather than by straightforward mathematical deduction from Equation 4, d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) we must treat it with caution. As is usual in such circumstances we will regard Equation 9 as a trial solution of Equation 4 and investigate its physical implications. In particular, we will use it to determine the way in which the values of A and δ depend on the characteristics of the driving force (a and Ω0) and the oscillator (ω 0 and γ1). By doing this we should gain insight into our trial solution and either convince ourselves of its correctness or expose its shortcomings.

27 We will start these investigations by considering separately the steady state behaviour of the oscillator when the angular frequency of the driving force is first much smaller than the natural angular frequency (Ω << ω 0 ), then when it is much larger than the natural angular frequency (Ω >> ω 0 ), and finally when the two angular frequencies are equal (Ω = ω 0 ) and the oscillator is near to resonance. Using Equation 9, steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) express the first two terms of Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) in terms of Ω, A, t and δ. Now let us look at the first of the frequency regimes in which our trial solution to Equation 4 is to be investigated.

28 Low frequency limit: W << w 0 As we have just seen, our trial solution implies that the first two terms in Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) involve factors of Ω1 2 and Ω, respectively. If Ω << ω0 0 the first term (Ω1 2 x) is small compared to the third term (ω 02 x). For light damping (small γ1) the second term will also be small compared with the third term (except very near x = 0). As a first approximation we can therefore neglect these first two terms, so that Equation 4 becomes: ω 02 x(t) asin (Ωt) a in the low frequency limit: x(t) = ω 2 sin (Ωt) (10) 0 This confirms that the oscillator is moving at the driving angular frequency Ω, as expected. In terms of the trial solution it also implies that, in the low frequency limit, the amplitude is A = a ω 2 ( 0 ), which is independent of the driving frequency, and the displacement is in phase with the driving force; i.e. δ = 0. In practice, this limit usually results in small amplitude oscillations in phase with the driving force.

29 High frequency limit: W >> w 0 In this case the first term in Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) is the important one, if the oscillator is lightly damped, since it involves a factor of Ω1 2 which will be large. Thus a reasonable approximation to Equation 4 is: d 2 x dt 2 asin (Ωt) but in the steady state described by Equation 9 steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) we have d 2 x dt 2 = Ω2 x(t)

30 Hence x(t) a sin (Ωt) Ω 2 Using one of the trigonometric identities discussed in Question R4, we can rewrite this in a form that is directly comparable to Equation 9: steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) in the high frequency limit: x(t) = a Ω 2 sin (Ωt π) (11) Thus the oscillator still moves with the angular frequency of the driving force, but it is now in anti-phase with that force, i.e. δ = π, and its amplitude is A = ( a Ω 2 ), which does depend on Ω and decreases rapidly as Ω increases.

31 Resonance limit: W ª w 0 The driving angular frequency at which resonance occurs depends on the damping constant, but for light damping it is very close to the natural angular frequency ω0 0. For our present purposes we will assume that the damping is light and that resonance implies Ω = ω 0. Using the steady state expression for x(t) from Equation 9, steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) and setting Ω = ω 0, Equation 4 d 2 x dt + γ dx 2 dt + ω 0 2 x = asin (Ωt) (Eqn 4) now becomes ω 02 x + γω 0 Acos(ω 0 t δ ) + ω 02 x = asin (Ωt) On the left-hand side the first and last terms cancel, so that: γ1ω 0 A1cos1(ω 0 t δ1) = a1sin1(ω1t)

32 We can use the trigonometric identity cos1θ = sin1(π/2 + θ) to rewrite this in the form γ1ω 0 A1sin1(π/2 + ω 0 t δ1) = a1sin1(ω1t) If this relation is to hold true for all values of t, then it must be the case that Ω = ω 0 and δ = π/2. In addition, it must also be the case that γω 0 A = a. It then follows from Equation 9 steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) that a in the resonance limit: x(t) = γω 0 sin Ωt π 2 (12) where the amplitude at resonance is given by A = a γω 0 which exceeds the amplitude in the low frequency limit by a factor oscillator. a ω 2 0 γω 0 a = ω 0. We will call this factor the quality factor or Q-factor of the lightly damped γ

33 We will have more to say about the Q-factor in the next subsection, but it worth pointing out now that it provides a useful way of characterizing oscillators that will be used throughout the rest of this module. For a very lightly damped oscillator, the Q-factor is usually very large. Indeed, in the absence of damping it would be infinite and the response of the system would be unbounded1 1to destruction. For a lightly damped oscillator at resonance the phase of the displacement lags 90 behind that of the driving force, and the amplitude is given by Q (a ω 02 ) where Q = ω 0 /γ. Question T5 Three lightly damped oscillators A, B and C are driven by the same harmonic force. The natural frequency of oscillator A is much higher than the frequency of the driving force, and that of B much lower. Oscillator C has a natural frequency very close to the driving frequency. Make a sketch showing the time dependence of the driving force and the displacements of the three oscillators in their steady state motion.4

34 Question T6 Using the three limiting cases as guidance, summarize how the amplitude and phase lag of the steady state motion of the driven oscillator change as the driving frequency is increased from a low value, through the natural frequency, to a high value.4 Our considerations of the frequency limits of the solution have led to sensible predictions, in accord with experimental results. Accordingly, we are encouraged to believe that Equation 9 steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) is indeed a solution to Equation 4. d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) Let us press on with this line of investigation. The most direct way to obtain the precise dependencies of A and δ on Ω is to substitute Equation 9 into Equation 4.

35 When this is done we find: A(ω 02 Ω 2 )sin (Ωt δ ) + AγΩcos(Ωt δ ) = asin δ cos(ωt δ ) + a cosδ sin (Ωt δ ) (13) Examine Equation 13 carefully. Notice that the time t occurs in two different sinusoidal functions. steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) For Equation 9 to be a solution of Equation 4, it is necessary that Equation 13 should be an identity, valid for all t. This is true if, and only if, the coefficients of cos1(ω1t δ1) on each side of the equation are identical; the same must be true of the coefficients of sin1(ω1t δ1) on each side. Equating the two sets of coefficients leads immediately to the pair of equations: sin δ = AγΩ (14a) a cosδ = A(ω 0 2 Ω 2 ) (14b) a

36 Squaring and adding these equations to eliminate δ, through the identity sin 2 δ + cos 2 δ = 1, gives the amplitude as: general case amplitude A = a (ω 02 Ω 2 ) 2 + (γω) 2 (15) Dividing Equation 14a by 14b sin δ = AγΩ a cosδ = A(ω 0 2 Ω 2 ) a leads to: tan δ = sin δ cosδ = γω ω 02 Ω 2 (Eqn 14a) (Eqn 14b)

37 Therefore we can write γω general case phase lag: δ = arctan ω 02 Ω 2 (16) general case amplitude A = a (ω 02 Ω 2 ) 2 + (γω) 2 (Eqn 15) Graphs of A and δ as functions of angular frequency, as calculated from Equations 15 and 16, are shown in Figures 2 and 3, respectively, for oscillators with Q = 2.5, 5, 7.5 and 10. It may be seen that they exhibit the expected qualitative behaviour.

38 A/(a/ω 0 2 ) phase lag δ/degrees Q = 2.5 Q = 7.5 Q = 5 Q = driving frequency Ω/ ω driving frequency Ω/ω 0 Figure 24Graphs of amplitude A (expressed as a multiple of a ω 02 ) against driving frequency Ω (expressed as a multiple of ω 0 ) for oscillators of different Q. The Q-value for each oscillator is given by the peak value of the response curve. Figure 34Graphs of phase lag δ (expressed in degrees) against driving frequency Ω (expressed as a multiple of ω 0 ) for oscillators of different Q.

39 Notice that the maximum amplitude response in Figure 2 occurs at a value of ω that is slightly less than ω 0, the effect being more marked for the more heavily damped oscillators with lower values of Q. We normally define the resonance angular frequency ω res and the corresponding resonance frequency f res = ω res /2π as the frequency at which the system has a maximum response. This means that the resonance frequency of a damped oscillator is slightly less than the natural frequency of the undamped oscillator. A/(a/ω 0 2 ) driving frequency Ω/ ω0 Figure 24Graphs of amplitude A (expressed as a multiple of a ω 02 ) against driving frequency Ω (expressed as a multiple of ω 0 ) for oscillators of different Q. The Q-value for each oscillator is given by the peak value of the response curve.

40 The natural frequency of the undamped system can be found from measurements on the damped system by noting the frequency at which the phase lag becomes 90. The curves in Figures 2 and 3 are not symmetric, since the low and high frequency limits are different, but the phase lag is antisymmetric about 90, for which Ω = ω 0 exactly. Notice also that the gradient in the phase lag curves as the angular frequency goes through resonance is steeper for higher values of Q, i.e. for oscillators with lighter damping. phase lag δ/degrees Q = 7.5 Q = 5 Q = Q = driving frequency Ω/ω 0 Figure 34Graphs of phase lag δ (expressed in degrees) against driving frequency Ω (expressed as a multiple of ω 0 ) for oscillators of different Q.

41 Question T7 Use Equations 15 and 16 general case amplitude A = a (ω 02 Ω 2 ) 2 + (γω) 2 (Eqn 15) γω general case phase lag: δ = arctan ω 02 Ω 2 (Eqn16) to confirm the results obtained at the beginning of this subsection for the low frequency, high frequency and resonance limits.4 Question T8 An oscillator with ω 0 = 1001s 1 and Q = 5 is driven by a harmonically varying force of maximum magnitude 21N. If the mass of the oscillator is 0.21kg, calculate the amplitude and phase lag of the steady state oscillations when the angular frequency of the driving force is: (a) 501s 1, (b) 1001s 1 and (c) 2001s 1.4

42 2.4 Steady state energy balance and power transfer Equation 9 steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) indicates that the oscillator, in its steady state, undergoes pure harmonic motion with the same angular frequency, Ω, as the driving force. However, when the behaviour is compared with that of a simple, undamped harmonic oscillator, important differences emerge, particularly regarding the energy of the oscillator. As usual, the total energy is the sum of the kinetic and potential energies, and ω 02 = km so we have: E tot = E = E kin + E pot = 1 2 m dx dt 2 kx 2 = 1 2 m dx dt 2 mω 0 2 x 2 (17) In the steady state we can use Equation 9 to express this total energy in terms of t rather than x, giving: E = 1 2 ma2 Ω 2 cos 2 (Ωt δ ) mω 0 2 A 2 sin 2 (Ωt δ )

43 For the harmonically driven oscillator: total energy E = 1 [ 2 ma2 Ω 2 cos 2 (Ωt δ ) + ω 02 sin 2 (Ωt δ )] (18) kinetic energy E kin = 1 2 ma2 Ω 2 cos 2 (Ωt δ ) (19) potential energy E pot = 1 2 ma2 ω 02 sin 2 (Ωt δ ) (20) Equations 18 to 20 have several important consequences for the driven oscillator:

44 total energy E = 1 [ 2 ma2 Ω 2 cos 2 (Ωt δ ) + ω 02 sin 2 (Ωt δ )] (Eqn 18) o The instantaneous total energy E of the driven oscillator depends on time unless Ω = ω 0, in which case E = 1 2 ma2 ω 0 2 o The average total energy over a complete cycle, denoted by 1E1, does not depend on time. In fact it can be shown that E = E kin + E pot = 1 4 ma2 Ω ma2 ω 02 = 1 4 ma2 (Ω 2 + ω 02 ) (21) o o We also see from Equation 21 that (unless Ω = ω 0 ) the averages of kinetic and potential energies over a full period are not equal, as they are in free SHM. In the low frequency limit (Ω << ω0 0 ) potential energy dominates, while at high frequency (Ω >> ω 0 ) the energy is almost entirely kinetic. Since the average total energy over a cycle is constant, the driving force must provide precisely sufficient energy over a full period to compensate for the energy lost due to damping over a full period.

45 It is interesting to look at this last result from another point of view, by investigating the rate of energy transfer in more detail. When a general one-dimensional force F x, acts on a body along the line of the force with instantaneous velocity dvx/d0t, the instantaneous rate of energy transfer to the body is just the instantaneous power dx P(t) = Fx (t) dt but, from Equation 3, d2x dx +b + kx = F0 sin ( Ω t) 2 dt dt the driving force of the oscillator, F3x = F0 1sin1(Ω1 t) is given by m dx d2x + mω 02 x + mγ dt 2 dt so in the steady state, when x(t) = A1sin1(Ω1t δ0), F3x = m F3x = m(ω 02 Ω 2 )x + mγ dx dt FLAP P5.3 Forced vibrations and resonance COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1 (Eqn 3)

46 It follows that the instantaneous power delivered by this particular force is dx P(t) = F 3x dt = m(ω 0 2 Ω 2 )x dx dt + mγ dx 2 dt and the corresponding average power over a full period will be P =m(ω 02 Ω 2 ) x dx 2 dx + mγ dt dt Now, as you will shortly be asked to demonstrate, x dx dt Anticipating this result, = 0. P = mγ dx dt 2 (22)

47 However if we identify the damping force F 2 x = mγ dx, it is clear that we can rewrite this last expression dt as P = mγ dx dt P = F 2x dx dt 2 (Eqn 22) Hence, as claimed, the rate at which energy is transferred to the oscillator by the driving force, averaged over a full period, is equal to the rate at which energy is transferred from the oscillator by damping, averaged over the same period. Formulate a convincing physical argument to show that over a full period of oscillation x dx dt above. = 0, as claimed

48 Using Equation 22 P = mγ dx dt 2 (Eqn 22) we can now obtain a useful explicit expression for the average rate of energy transfer over a full period. There are many ways of doing this, one is to note that E kin = 1 2 m v x 2 = m 2 dx 2 dt so that Equation 22 can be written in the form P =mγ dx dt 2 = 2γ E kin

49 and then use Equation 21, E = E kin + E pot = 1 4 ma2 Ω ma2 ω 02 = 1 4 ma2 (Ω 2 + ω 02 ) (Eqn 21) which gives E kin = 1 4 ma2 Ω 2, to rewrite this P =mγ dx dt 2 = 2γ E kin as P = 1 2 mγ(aω)2 (23)

50 As we saw in the last subsection, in the case of a lightly damped oscillator, resonance occurs when Ω = ω 0, which implies that A = a/(γ1ω0 0 ) and δ = 90. It follows from Equation 23 P = 1 2 mγ(aω)2 (Eqn 23) that the average rate of energy transfer at resonance will be P res = ma2 2γ (24) and substituting this into Equation 23 we see that at any driving frequency Ω P = γaω 2 P a res (25a) According to Equation 14a γ1aω1/a = sin1δ, so we can also write the above result in the form: P = P res sin 2 δ (25b) For a given oscillator, driven at a given frequency Ω, the quantity sin 2 δ in Equation 25b is called the power factor and measures the ratio of the average power absorbed at the driving frequency to the average power absorbed at resonance.

51 The way in which 1P1 varies with Ω for an oscillator with Q = ω 0 /γ = 1 can be seen from Figure 4. Note that by plotting values of 1P1 / 1P res 1 against Ω/ω0 0 we are effectively measuring 1P1 in units of 1P res 1 and Ω in units of ω0 0. The graph also shows how the phase lag δ between the driving force and the displacement varies with Ω for this Q = 1 oscillator, making it possible to see how 1P1 / 1P res 1 varies with δ. As you can see the average power transfer per cycle is greatest at resonance, and in the lightly damped case this occurs when Ω = ω 0 and δ = 90. Question T9 P P res P P res ω 1 1 ω 2 2 Ω/ω 0 ω 0 ω 0 Figure 44Graphs of 1P1 / 1P res 1 and δ against Ω1/ω 0 for an oscillator with Q = 1. Show that in the resonance limit, for a lightly damped oscillator, the average power transferred by the driving force in each full cycle is 1P1 = maaω 0 /2, but in both the low frequency limit and the high frequency limit the power transfer averages to zero.4 δ phase lag δ/degrees

52 Figure 4 was drawn for an oscillator with the rather small Q-factor of 1. It is interesting to enquire how the situation would be changed if an oscillator with a larger value of Q had been considered. (For a fixed natural frequency, higher values of Q correspond to lighter damping.) Using the light damping result, Q = ω 0 /γ, Equation 24 P res = ma2 (24) 2γ may be written in the form P res = ma2 2γ = ma2 Q 2ω 0 P P res P P res ω 1 1 ω 2 2 Ω/ω 0 ω 0 ω 0 Figure 44Graphs of 1P1 / 1P res 1 and δ against Ω1/ω 0 for an oscillator with Q = 1. showing that for oscillators of the same natural frequency and mass, driven by the same force, the average power transfer over a full cycle at resonance is proportional to the Q-value. Thus Q acts as an amplification factor for the power absorbed. This explains why very lightly damped oscillators resonate more powerfully than heavily damped ones. δ phase lag δ/degrees

53 In addition to its height, given by Equation 25, P = γaω 2 P a res (Eqn 25a) the power resonance curve of Figure 4 is also characterized by its width. This may be defined as the (positive) difference between the half-power points ω 1 and ω 2 at which ω 0 ω 0 P = 1 2 P res. In order that they should correspond to half-power points, ω 1 must be the driving angular frequency at which δ1= 45 and ω0 2 that at which δ = 135. At both of these values the power factor is 0.5, so that 1P1 exceeds 1P res 1 /2 in the range ω 1 < Ω < ω 2. This is illustrated in Figure 4. P P res P P res ω 1 1 ω 2 2 Ω/ω 0 ω 0 ω 0 Figure 44Graphs of 1P1 / 1P res 1 and δ against Ω1/ω 0 for an oscillator with Q = 1. δ phase lag δ/degrees

54 The average power absorption 1P1 attains its maximum at the resonance frequency f 0, but it remains significant throughout the range of angular driving frequencies between the halfpower points. The corresponding frequency range f = (ω 2 ω 1 )/2π is called the resonance absorption bandwidth of the oscillator. This too depends on the Q- factor of the oscillator, since it can be shown that ω 2 ω 1 = γ, (see Question T11) implying that P P res P P res ω 1 1 ω 2 2 Ω/ω 0 ω 0 ω 0 Figure 44Graphs of 1P1 / 1P res 1 and δ against Ω1/ω 0 for an oscillator with Q = 1. δ phase lag δ/degrees resonance absorption bandwidth f = ω 2 ω 1 2π = γ 2π = ω 0 2πQ = f 0 Q (26)

55 The resonance absorption bandwidth therefore becomes narrower as the Q-factor increases, and it follows from Equation 26 resonance absorption bandwidth f = ω 2 ω 1 2π = γ 2π = ω 0 2πQ = f 0 Q (Eqn 26) that a simple expression for the Q-factor is Q = f 0 / f. It follows that for high Q oscillators the power resonance is proportionally very tall and narrow. Power resonance curves for a few Q-values are shown in Figure 5. 2ω 0 P ma Q = 10 Q = 7.5 Q = 5 Q = Ω/ω 0 Figure 54Graphs of 2ω 0 1P1 /(m0a 2 ) against Ω1/ω0 0 for oscillators of different Q.

56 Question T10 Describe two features of the performance of an oscillator that are determined by its quality factor Q.4 Question T11 Show that the power factor of an oscillator may be written in the form: P P res = 1 + Ω Q2 ω 0 ω 0 Ω 2 1 and use this to obtain expressions for ω 1 and ω0 2. Hence show that ω 2 ω 1 = γ, as claimed above.4

57 2.5 Transient motion The steady state motion, expressed by Equation 9, steady state motion x(t) = A1sin1(Ω1t δ1) (Eqn 9) is just one possible motion for a harmonically driven oscillator. In the present subsection we will refer to the steady state motion as x ss (t). In contrast, the general solution of the homogeneous differential equation, linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = 0 (Eqn 5) that describes an undriven linearly damped oscillator is a transient motion of the form: transient motion: x tr (t) = B e γ t 2 cos(ωt + φ ) (27) where B and φ are arbitrary constants determined by the initial conditions of the motion, and ω = ω 02 γ 2 4.

58 Interestingly this kind of transient motion is also relevant to a driven oscillator. To see why this is so note that the steady state motion is a solution to Equation 4, so d 2 x dt 2 d 2 x ss dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) + γ dx ss dt + ω 02 x ss = asin (Ωt) but the transient motion of Equation 27 transient motion: x tr (t) = B e γ t 2 cos(ωt + φ ) (Eqn 27) is a solution to Equation 5, linearly damped oscillator: d 2 x dt 2 + γ dx dt + ω 0 2 x = 0 (Eqn 5) so d 2 x tr dt 2 + γ dx tr dt + ω 0 2 x tr = 0

59 adding corresponding terms in these two equations, d 2 x ss dt 2 + d 2 x tr dt 2 + γ dx ss + dx tr dt dt + ω 0 (x ss + x tr ) = asin (Ωt) and using the mathematical property that the derivative of a sum is a sum of derivatives: d 2 (x ss + x tr ) + γ d(x ss + x tr ) + ω dt 2 02 (x ss + x tr ) = asin (Ωt) dt This is just Equation 4 d 2 x dt 2 + γ dx dt + ω 0 2 x = asin (Ωt) (Eqn 4) with (x ss + x tr ) in place of x. Hence (x ss + x tr ) is also a solution of the inhomogeneous differential equation that describes a damped driven harmonic oscillator. The full solution, including both the steady state and the transient motion is therefore: x(t) = x ss (t) + x tr (t) = Asin (Ωt δ ) + Be γ t 2 cos(ωt + φ ) (28)

60 Equation 28, x(t) = x ss (t) + x tr (t) = Asin (Ωt δ ) + Be γ t 2 cos(ωt + φ ) gives a complete representation of the motion of this driven oscillator, including the transient phase. Whatever the starting conditions, as embodied in the values of B and φ, the transient term, x tr, in Equation 28 becomes negligible for t >> 2π/γ, so that when t >> 2π/γ x(t) x ss (t) = A cos(ωt δ ) This is illustrated in Figure 6, which shows the displacement time graphs of a driven oscillator with various starting conditions. The transient behaviour and the onset of steady state motion can both be clearly seen. displacement x(t) time t Figure 64Displacement time graphs for the same oscillator with three different sets of initial conditions (different choices of B and φ).

61 Question T12 The constants B and φ that appear in Equation 28 x(t) = x ss (t) + x tr (t) = Asin (Ωt δ ) + Be γ t 2 cos(ωt + φ ) (Eqn 28) are the arbitrary constants that are determined by the starting conditions of the oscillator. Why can t the constants A and δ be regarded in the same way?4

62 2.6 Resonance and frequency standards So far we have developed the mathematical model of the driven oscillator by thinking of mechanical oscillations and mechanical resonance. As we mentioned in the introduction there are many other examples of resonance from non-mechanical oscillator systems. The electrical response of a tuned circuit or the resonant response of a tuned cavity or an atom or nucleus to an electromagnetic wave are cases in point. The damping is caused by any process through which energy can be transferred out of the vibrational system, often through heating. For example, in an electrical circuit the dissipation arises from the heating effect of the current as it flows through any circuit resistance. While these other systems are not always amenable to the simple linearly damped mathematical model introduced here, the model enables us to appreciate the processes involved.

63 We can describe all these systems in terms of resonance, with an appropriate Q-factor implied by the frequency response shown on the power absorption curve. In some cases these systems have Q-factors which are very much larger than those of the mechanical systems considered thus far and so the frequency response is a great deal narrower and more selective. If the frequency of an oscillator is narrowly defined then the period of the oscillation is also narrowly defined and the oscillator can be used as a clock. For this reason, resonant systems find widespread application in the maintenance of frequency or time standards. The higher the Q-factor of the oscillator the greater the potential for using the resonant oscillator as a well-defined frequency standard. Here we cannot describe these systems in any detail but it is worth mentioning a few important examples and quoting the Q-factors that typify them.

64 We begin with a simple familiar mechanical oscillator, the pendulum clock. The pendulum oscillations are sustained by energy input to compensate for dissipative processes, such as friction. The damping of the system is minimized and an estimate for the Q-factor achieved can be found from Equation 26, resonance absorption bandwidth f = ω 2 ω 1 = γ 2π 2π = ω 0 2πQ = f 0 (Eqn 26) Q knowing the typical performance of a good mechanical clock. If the clock is to gain or lose by no more than 101s a day then this is a fractional time error of 101s in s, or about The fractional frequency error is the same as the fractional time error and if we take a typical frequency error to be the full width of the power curve at half height (i.e. 0f1) then Equation 26 gives the required Q-value as f 0 / f = 10 4.