FLEXIBLE LEARNING APPROACH TO PHYSICS ÊÊÊ … · FLAP P5.3 Forced vibrations and resonance COPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1 1 Opening items 1.1 Module introduction

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

FLAP P5.3 Forced vibrations and resonanceCOPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1

Module P5.3 Forced vibrations and resonance1 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


1 Opening items

1.1 Module introductionIn order to sustain vibrations, energy must be supplied to a system to make up for the energy transferred out ofthe 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 pendulumor balance wheel normally vibrates, i.e. at its natural frequency. Pushing at the natural frequency is the mostefficient way to transfer energy into a vibrating system, as your childhood experiences on a swing shouldconfirm.

In Section 2 the equation of motion for the linearly damped oscillator is developed, both for an undrivenoscillator and in the presence of a harmonic driving force. We examine the steady state motion of this lattersystem as the amplitude and frequency of the driving force is varied, considering the low and high frequencyresponse and the resonance response. We then look at the energy balance in the steady state system, consideringthe kinetic, potential and total energies. The transient motion is distinguished from the steady state motion andexamined briefly. Other mechanical and non-mechanical cases of driven oscillators are considered and theapplication of the principle of resonance absorption in clocks and frequency standards is discussed. The cases ofquartz crystal oscillators, the caesium atomic clock and the potential use of frequency-stabilized lasers areconsidered briefly.


In Section 3 we look at the related situation where the exciting force is due to another oscillator and considerhow two coupled oscillators interact. The important idea of normal modes is introduced, along with the idea ofbeats between these modes.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by thismodule and that you do not need to study it. If so, try the Fast track questions given in Subsection 1.2. If not, proceeddirectly to Ready to study? in Subsection 1.3.


1.2 Fast track questions

Study comment Can you answer the following Fast track questions?. If you answer the questions successfully you needonly glance through the module before looking at the Module summary (Subsection 5.1) and the Achievements listed inSubsection 5.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 5.3. If you havedifficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevantparts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised tostudy the whole module.

Question F1

Write down the equation of motion of a damped harmonic oscillator driven by an external force which variessinusoidally (harmonically) with the time. Explain what is meant by the terms steady state motion and transientmotion. Give a qualitative description of these motions and explain the relationship between them.


Question F2

What two parameters characterize the steady state motion of an oscillator which is driven harmonically at anangular frequency Ω? Make sketches indicating how these parameters vary with the angular frequency Ω of thedriving 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 routethrough 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 theClosing items.


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 asystem), 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 themodule). Mathematically, you will need to be familiar with exponential and trigonometric functions and with theinverse trigonometric function arctan(x). You will also be required to use various trigonometric identities. The notation ofdifferentiation (dx/dt, d2x/dt2) is used freely throughout this module, and you must be familiar with it, however you are notactually required to evaluate derivatives for yourself. An averaging process is introduced which requires a knowledge ofintegration if it is to be fully understood, but again you are not required to evaluate any integrals for yourself. Simpledifferential equations are also introduced in the module but a familiarity with this topic is not assumed. Mathematicalrequirements 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 andtrigonometric 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 toestablish whether you need to review some of the topics before embarking on this module.

Mike Tinker


Question R1

A body of mass m moves along the x-axis of a Cartesian coordinate system, subject to two forces F1x and F2x.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 isgiven by x(t) = A01sin1(ω1t + φ). Explain the physical significance of the constants A0, 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?


Question R4

(a) Given the trigonometric identity

cos(θ + φ ) = cosθ cos φ − sin θ sin φ

show that cos2 θ + sin2 θ = 14and4cos2 θ − sin2 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:

cos5π12

= 3 − 1

2 24and4cos

π12

= 3 + 1

2 2


2 Driven oscillators

2.1 Qualitative discussion of driven oscillators and resonanceOscillating systems typically have one or more characteristic natural frequencies at which they vibrate whendisturbed. These natural frequencies can be very useful. In the case of a simple pendulum for instance, the singlenatural frequency may be used for time keeping, as in a pendulum clock. However, when using oscillatingsystems in technical and scientific applications it is often necessary to suppress natural oscillations and replacethem by externally imposed oscillations. An example of this kind arises in the case of loudspeakers. In order toreproduce a wide range of sounds, loudspeakers must be made to vibrate over a large range of frequencies, and ifsounds are to be reproduced faithfully any natural frequencies associated with the loudspeaker must not beunduly accentuated.

When we wish to produce oscillations in a system we do so by applying a driving force which has its ownfrequency associated with it. An oscillator responding to such a force is said to be a forced oscillator or adriven oscillator. Examples of forced oscillators are plentiful, even a device as simple as a child’s playgroundswing can serve the purpose.


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 thetrue 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 accountthese various damping effects that dissipate the energy, and, provided they were relatively weak, would representthe 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 adisplacement–time graph for the first ten oscillations.4


Left to themselves the oscillations of the swing die away, but if you catch the swing at its highest point, at thecompletion of each oscillation, and push it back towards the equilibrium position, then the oscillations can besustained despite the damping. The swing may now be described as a driven damped oscillator, or simply aforced 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 youwould have to supply to sustain the oscillation. Assume that the driving force has a constant magnitude Fdduring the time you are in contact with the swing, and that you never apply the force when the swing’sdisplacement from its equilibrium position is negative. Your sketch should be guided by your answers to thefollowing questions: Is the motion periodic? What governs the pushing frequency? During what part of the cyclewould you push the swing? Is the motion symmetric about the equilibrium position?4

Mike Tinker


The driving force shown in the answer to Question T2 is similar to the force you really would supply if youwanted to keep the swing oscillating with the minimum effort. It is a periodic force with a frequency that closelymatches the natural frequency of the swing. Driving an oscillator in this way, at a frequency that is close to itsnatural frequency, can result in the efficient transfer of energy to the oscillator and may cause the amplitude ofthe oscillations to become very large. This is the condition known as resonance, an understanding of which isimportant 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 whenthe suspension bridge across which they were marching collapsed. The rhythmic marching of the soldiersexcited a natural swinging motion of the bridge, which brought about its destruction. Similarly, gusting wind inthe Tacoma Narrows in the north-western USA, in November 1940, caused a new suspension bridge to vibraterather in the same way as the metal slats of a venetian blind flutter when air rushes over them. Energy wassupplied to the bridge at its natural frequency and the amplitude of the vertical and then the twisting vibrationsbecame so large that the roadway ripped from its hangers and plunged into the water below. You may have seenthe classic film footage of this spectacular disaster. In the same way, tall buildings, radio aerials and factorychimneys can also sway in the wind and must be constructed so that resonance is avoided.


In contrast to these negative aspects of resonance, there are situations in which resonance is desirable and can beusefully 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 theopening. Organ pipes and some other wind instruments work in a similar way. The same principle is also widelyused in nature. Howler monkeys and some frogs and toads have large vocal sacs that allow forced oscillations ata near natural frequency to produce surprisingly loud sounds. In fact, such systems often resonate over quite awide 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 ofatoms and nuclei, where physical systems are described by quantum physics rather than classical physics. While quantumsystems are beyond the scope of this module, it is worth mentioning that useful insights into their behaviour can sometimesbe gained from classical ideas. For example, the strong absorption by atomic nuclei of γ-rays in the wavelength range from6 to 9 × 10−141m may be thought of as the resonant excitation of a vibration in which the protons and neutrons in a nucleusoscillate, in anti-phase, about the nuclear centre of mass.


2.2 The equation of motion of a harmonically driven oscillator

m

xrough surface

P

Figure 14A simple drivenoscillator. The driving force hasbeen omitted from the figurebecause its sign may change withtime.

Another simple example of a forced oscillator is shown in Figure 1. A body ofmass m is attached to one end of a horizontal spring, the other end of which isattached to a fixed point P. The body can slide back and forth along a straightline, 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 positiveor negative):

A restoring force F1x due to the spring that tends to return the body to itsequilibrium position.A damping force F2x due to friction and air resistance that opposes themotion of the body.A driving force F3x provided by an external agency.

In the absence of any driving force, the system would have a natural equilibrium configuration in which thespring was neither extended nor compressed. If we take the equilibrium position of the mass to be the origin ofthe coordinate system, we can say that the mass’s displacement from equilibrium at time t is given by itsinstantaneous position coordinate x1(t).


Using calculus notation, it follows that at any time t the velocity and acceleration of the mass are, respectively

vx = dx

dt4and4

ax = dvx

dt= d2 x

dt2

Using Newton’s second law of motion we can then say that the sliding body must obey an equation of motion ofthe form

md2 x

dt2= F1x + F2 x + F3x (1)

To make further progress we now need to relate the forces that act on the sliding body to the time t or to thebody’s instantaneous position x. This is easily done provided we idealize the system to some extent. First, let ussuppose 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.


Second, let us assume that damping forces which oppose the motion have magnitudes that are proportional to theinstantaneous speed of the sliding body, so that

F2 x = −bdx

dt(2b)

where b is a constant that characterises the dissipative forces.

m

xrough surface

P


Third, let us suppose that the driving force is periodic, and has the relativelysimple form

F3x = F01sin1(Ω1t) (2c)

where F0 is the maximum magnitude that the driving force attains, and Ω is theangular frequency of the driving force. Note that the angular frequency Ω isexternally imposed and is not necessarily related in any way to the naturalfrequency of the system in Figure 1.


Substituting Equations 2a, 2b and 2c into Equation 1,

F1x = −1kx (Eqn 2a)

F2 x = −bdx

dt(Eqn 2b)

F3x = F01sin1(Ω1t) (Eqn 2c)

md2 x

dt2= F1x + F2 x + F3x (Eqn 1)

we see that the behaviour of the sliding mass must now be described by the equation

md2 x

dt2= −kx − b

dx

dt+ F0 sin (Ω t)

which can be rearranged to isolate the time-dependent driving term as follows

md2 x

dt2+ b

dx

dt+ kx = F0 sin (Ω t) (3)


A driving force that depends sinusoidally on time is sometimes described as a harmonic driving force, and adamping force with magnitude proportional to velocity is sometimes called a linear damping force. Equation 3

md2 x

dt2+ b

dx

dt+ kx = F0 sin (Ω t) (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 differssomewhat from Equation 3. One of the most common variants is obtained by dividing both sides of Equation 3by the mass m, and then introducing

the damping constant γ = b/m

the maximum magnitude of the driving acceleration a = F0/m

and the natural angular frequency ω0 = k m

The last of these is the angular frequency the oscillator would have if it were just a simple harmonic oscillatorwithout any damping or driving.

Substituting these expressions into Equation 3 we can say that:


for the harmonically driven linearly damped oscillator:

d2 x

dt2+ γ dx

dt+ ω0

2 x = a sin (Ω t) (4)

Equations of this kind, involving an independent variable (t), a dependent variable (x(t)) and derivatives of thedependent 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 alinear equation) and the derivative of highest order that it contains is a second derivative. Equation 4 is thereforeclassified 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 sideof Equation 4. So, if we set a = 0 we obtain the equation of motion for an undriven linearly damped oscillator:

linearly damped oscillator:d2 x

dt2+ γ dx

dt+ ω0

2 x = 0 (5)


If, in addition, we suppose that there is no damping, so γ = 0, the equation of motion becomes that of a simpleharmonic oscillator:

simple harmonic oscillator:d2 x

dt2+ ω0

2 x = 0 (6)

We will approach the solution of Equation 4

d2 x

dt2+ γ dx

dt+ ω0

2 x = a sin (Ω t) (Eqn 4)

through the simpler Equation 6

and then Equation 5.


dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)


First we look at Equation 6.


dt2+ ω0

2 x = 0 (Eqn 6)

Now, you should be familiar with simple harmonic motion in one-dimension, and you should know that it cangenerally be described by an equation of the form

x(t) = A01sin1(ω0t + 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

vx = dx

dt= A0ω0 cos(ω0t + φ )

and ax = d2 x

dt2= − A0ω0

2 sin (ω0t + φ )

and hence confirm that Equation 7 is a solution to Equation 6. To a mathematician, the presence of two arbitraryconstants (A0 and φ) in Equation 7 also shows that it is the general solution to Equation 6, i.e. every solution toEquation 6 can be written in the form of Equation 7 by assigning suitable values to A0 and φ.


m

xrough surface

P


In a similar way, but with more mathematical manipulation, it can be shownthat damped harmonic motion, with its exponentially decaying amplitude canbe described by an expression of the form

x(t) = A01e−γ1t/021sin1(ω1t + φ) 4where 4ω = ω02 − γ 2 4 (8)

which is the general solution to Equation 5.


dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)

If we want to determine how the sliding body of Figure 1 moves whensubjected to the three forces specified above we need to find a solution toEquation 4,

d2 x

dt2+ γ dx

dt+ ω0


i.e. we need to find an expression relating the instantaneous position x of the body to the time t, that satisfiesEquation 4 and any other condition that the oscillator is known to satisfy. Solving Equation 4 is actuallysomewhat more complicated than solving Equations 5 and 6.


dt2+ ω0

2 x = 0 (Eqn 6)


This is because Equations 5 and 6


dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)


dt2+ ω0

2 x = 0 (Eqn 6)

are homogeneous differential equations in which each term is proportional to x or to one of its timederivatives, whereas Equation 4

d2 x

dt2+ γ dx

dt+ ω0


is an inhomogeneous differential equation because it contains the driving term, a1sin1(Ωt), which does notinvolve the displacement at all, but does depend explicitly on the time. The mathematical procedure for solvingsuch equations is described in detail in the maths strand of FLAP, where it is applied to Equation 3

md2 x

dt2+ b

dx


(see second-order differential equation in the Glossary for references), but we will not employ such methodshere.


In this module we will adopt a more physical approach in which we will use physical arguments to determine theoverall form of a solution, and mathematical techniques to work out the details. Methods of this kind are muchused by physicists.

m

xrough surface

P


Question T3

Devise and describe two simple systems that, like the system in Figure 1, arecapable of oscillatory motion and which may be acted upon by an externallyproduced periodic driving force.4

Question T4

A student claims that the action of an external driving force must eventuallylead 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 otherdissipative effect, the sliding body will gain energy continuously and theoscillation will grow without limit, at least until the spring fractures. Do youagree with this argument?4


2.3 Steady state motiond2 x

dt2+ γ dx

dt+ ω0


Equation 4 represents a struggle. The oscillator, when disturbed, tends to oscillate with its natural angularfrequency ω0, but it is being driven at a different angular frequency, Ω. It is clear that the driving force musteventually win this struggle, because the oscillator loses energy due to damping and is only sustained inperpetuity by the external agency that provides the driving force. The motion during the struggle is usuallycalled the transient motion and its finite duration is determined by the time taken for free oscillations to decayto a negligible amplitude. This time is of the order of 2π/γ, as you will see in Subsection 2.5, where we considerthe transient motion in more detail. In this subsection we will be concerned with the steady state motion thatpersists 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 islikely to be harmonic motion with the same angular frequency as the driving force. However, there is no reasonwhy it should be exactly in-phase with the driving force, indeed on physical grounds you might well expect thedisplacement to lag somewhat behind the force that causes it.


For these reasons we will assume that the steady state motion of the harmonically driven linearly dampedoscillator 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 motionand the driving force (F3x = F01sin1(Ω1t)).Having arrived at Equation 9 on the basis of physical arguments, rather than by straightforward mathematicaldeduction from Equation 4,

d2 x

dt2+ γ dx

dt+ ω0


we must treat it with caution. As is usual in such circumstances we will regard Equation 9 as a trial solution ofEquation 4 and investigate its physical implications. In particular, we will use it to determine the way in whichthe values of A and δ depend on the characteristics of the driving force (a and Ω0) and the oscillator (ω0 and γ1).By doing this we should gain insight into our trial solution and either convince ourselves of its correctness orexpose its shortcomings.


We will start these investigations by considering separately the steady state behaviour of the oscillator when theangular frequency of the driving force is first much smaller than the natural angular frequency (Ω << ω0), thenwhen it is much larger than the natural angular frequency (Ω >> ω0), and finally when the two angularfrequencies are equal (Ω = ω0) 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

d2 x

dt2+ γ dx

dt+ ω0


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 beinvestigated.


Low frequency limit: W << w0

As we have just seen, our trial solution implies that the first two terms in Equation 4

d2 x

dt2+ γ dx

dt+ ω0


involve factors of Ω12 and Ω, respectively. If Ω << ω00 the first term (Ω12x) is small compared to the third term(ω0

2x). For light damping (small γ1) the second term will also be small compared with the third term (except verynear x = 0). As a first approximation we can therefore neglect these first two terms, so that Equation 4 becomes:

ω02 x(t) ≈ a sin (Ω t)

in the low frequency limit: x(t) = a

ω02

sin (Ω t) (10)

This confirms that the oscillator is moving at the driving angular frequency Ω, as expected. In terms of the trialsolution it also implies that, in the low frequency limit, the amplitude is A = a ω0

2( ) , which is independent ofthe driving frequency, and the displacement is in phase with the driving force; i.e. δ = 0. In practice, this limitusually results in small amplitude oscillations in phase with the driving force.


High frequency limit: W >> w0

In this case the first term in Equation 4

d2 x

dt2+ γ dx

dt+ ω0


is the important one, if the oscillator is lightly damped, since it involves a factor of Ω 12 which will be large.Thus a reasonable approximation to Equation 4 is:

d2 x

dt2≈ a sin (Ω t)

but in the steady state described by Equation 9


we have

d2 x

dt2= −Ω 2 x(t)


Hence x(t) ≈ − a

Ω 2

sin (Ω t)

Using one of the trigonometric identities discussed in Question R4, we can rewrite this in a form that is directlycomparable to Equation 9:


in the high frequency limit: x(t) = a

Ω 2

sin (Ω t − π) (11)

Thus the oscillator still moves with the angular frequency of the driving force, but it is now in anti-phase withthat force, i.e. δ = π, and its amplitude is A = a Ω 2( ) , which does depend on Ω and decreases rapidly as Ωincreases.


Resonance limit: W ª w0

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 ω00. For our present purposes we will assume that the

damping is light and that resonance implies Ω = ω0. Using the steady state expression for x(t) from Equation 9,


and setting Ω = ω0, Equation 4

d2 x

dt2+ γ dx

dt+ ω0


now becomes

−ω02 x + γω0 A cos(ω0t − δ ) + ω0

2 x = a sin (Ω t)

On the left-hand side the first and last terms cancel, so that:

γ1ω0A1cos1(ω0t − δ1) = a1sin1(Ω1t)


We can use the trigonometric identity cos1θ = sin1(π/2 + θ) to rewrite this in the form

γ1ω0A1sin1(π/2 + ω0t − δ1) = a1sin1(Ω1t)

If this relation is to hold true for all values of t, then it must be the case that Ω = ω0 and δ = π/2. In addition, itmust also be the case that γω0A = a. It then follows from Equation 9


that

in the resonance limit: x(t) = a

γω0

sin Ω t − π2

(12)

where the amplitude at resonance is given by A = a γ ω0 which exceeds the amplitude in the low frequency

limit by a factor a

γω0

ω02

a= ω0

γ. We will call this factor the quality factor or Q-factor of the lightly damped

oscillator.


We will have more to say about the Q-factor in the next subsection, but it worth pointing out now that itprovides 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 itwould 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 thedriving force, and the amplitude is given by Q × (a ω0

2 ) where Q = ω0/γ.

Question T5

Three lightly damped oscillators A, B and C are driven by the same harmonic force. The natural frequency ofoscillator A is much higher than the frequency of the driving force, and that of B much lower. Oscillator C has anatural frequency very close to the driving frequency. Make a sketch showing the time dependence of thedriving force and the displacements of the three oscillators in their steady state motion.4


Question T6

Using the three limiting cases as guidance, summarize how the amplitude and phase lag of the steady statemotion of the driven oscillator change as the driving frequency is increased from a low value, through thenatural frequency, to a high value.4

Our considerations of the frequency limits of the solution have led to sensible predictions, in accord withexperimental results. Accordingly, we are encouraged to believe that Equation 9


is indeed a solution to Equation 4.

d2 x

dt2+ γ dx

dt+ ω0


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 intoEquation 4.


When this is done we find:

A(ω02 − Ω 2 )sin (Ω t − δ ) + Aγ Ω cos(Ω t − δ )

= a sin δ cos(Ω t − δ ) + a cosδ sin (Ω t − δ )(13)

Examine Equation 13 carefully. Notice that the time t occurs in two different sinusoidal functions.


d2 x

dt2+ γ dx

dt+ ω0


For Equation 9 to be a solution of Equation 4, it is necessary that Equation 13 should be an identity, validfor all t. This is true if, and only if, the coefficients of cos1(Ω1t − δ1) on each side of the equation are identical; thesame must be true of the coefficients of sin1(Ω1t − δ1) on each side. Equating the two sets of coefficients leadsimmediately to the pair of equations:

sin δ = Aγ Ωa

(14a)

cosδ = A(ω02 − Ω 2 )

a(14b)


Squaring and adding these equations to eliminate δ, through the identity sin2δ + cos2δ = 1, gives the amplitudeas:

general case amplitude A = a

(ω02 − Ω 2 )2 + (γ Ω )2

(15)

Dividing Equation 14a by 14b

sin δ = Aγ Ωa

(Eqn 14a)

cosδ = A(ω02 − Ω 2 )

a(Eqn 14b)

leads to:

tan δ = sin δcosδ

= γ Ωω0

2 − Ω 2


Therefore we can write

general case phase lag: δ = arctanγ Ω

ω02 − Ω 2

(16)


(ω02 − Ω 2 )2 + (γ Ω )2

(Eqn 15)

Graphs of A and δ as functions of angular frequency, as calculated from Equations 15 and 16, are shown inFigures 2 and 3, respectively, for oscillators with Q = 2.5, 5, 7.5 and 10. It may be seen that they exhibit theexpected qualitative behaviour.


10.0

7.5

5.0

2.5

00 1 2

driving frequency Ω ω/ 0

A/(a

/ω02 )

Figure 24Graphs of amplitude A (expressed as a multiple of

a ω 02 ) against driving frequency Ω (expressed as a multiple

of ω0) for oscillators of different Q. The Q-value for eachoscillator is given by the peak value of the response curve.

180

135

90

45

00 1 2

phas

e la

g δ/d

egre

es

Q = 10Q = 2.5

Q = 5Q = 7.5

driving frequency Ω/ω0

Figure 34Graphs of phase lag δ (expressed in degrees)

against driving frequency Ω (expressed as a multiple of ω0)for oscillators of different Q.


10.0

7.5

5.0

2.5

00 1 2

driving frequency Ω ω/ 0

A/(a

/ω02 )

Figure 24Graphs of amplitude A (expressed as a multiple of

a ω 02 ) against driving frequency Ω (expressed as a multiple

of ω0) for oscillators of different Q. The Q-value for eachoscillator is given by the peak value of the response curve.

Notice that the maximum amplitude response inFigure 2 occurs at a value of ω that is slightly lessthan ω0, the effect being more marked for the moreheavily damped oscillators with lower values of Q. Wenormally define the resonance angular frequencyωres and the corresponding resonance frequencyfres = ωres/2π as the frequency at which the system hasa maximum response. This means that theresonance frequency of a damped oscillator is slightlyless than the natural frequency of the undampedoscillator.


180

135

90

45

00 1 2

phas

e la

g δ/d

egre

es

Q = 10Q = 2.5

Q = 5Q = 7.5

driving frequency Ω/ω0

Figure 34Graphs of phase lag δ (expressed in degrees)

against driving frequency Ω (expressed as a multiple of ω0)for oscillators of different Q.

The natural frequency of the undamped system canbe found from measurements on the damped systemby noting the frequency at which the phase lagbecomes 90°. The curves in Figures 2 and 3 are notsymmetric, since the low and high frequency limitsare different, but the phase lag is antisymmetric about90°, for which Ω = ω0 exactly. Notice also that thegradient in the phase lag curves as the angularfrequency goes through resonance is steeper forhigher values of Q , i.e. for oscillators with lighterdamping.

Mike Tinker


Question T7

Use Equations 15 and 16


(ω02 − Ω 2 )2 + (γ Ω )2

(Eqn 15)

general case phase lag: δ = arctanγ Ω

ω02 − Ω 2

(Eqn16)

to confirm the results obtained at the beginning of this subsection for the low frequency, high frequency andresonance limits.4

Question T8

An oscillator with ω0 = 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 whenthe angular frequency of the driving force is: (a) 501s−1, (b) 1001s−1 and (c) 2001s−1.4


2.4 Steady state energy balance and power transferEquation 9


indicates that the oscillator, in its steady state, undergoes pure harmonic motion with the same angularfrequency, Ω, as the driving force. However, when the behaviour is compared with that of a simple, undampedharmonic 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 ω0

2 = k m so we have:

Etot = E = Ekin + Epot = 12

mdx

dt

2

+ 12

kx2 = 12

mdx

dt

2

+ 12

mω02 x2 (17)

In the steady state we can use Equation 9 to express this total energy in terms of t rather than x, giving:

E = 12

mA2Ω 2 cos2 (Ω t − δ ) + 12

mω02 A2 sin2 (Ω t − δ )


For the harmonically driven oscillator:

total energy E = 12

mA2 Ω 2 cos2 (Ω t − δ ) + ω02 sin2 (Ω t − δ )[ ] (18)

kinetic energy Ekin = 12

mA2Ω 2 cos2 (Ω t − δ ) (19)

potential energy Epot = 12

mA2ω02 sin2 (Ω t − δ ) (20)

Equations 18 to 20 have several important consequences for the driven oscillator:


total energy E = 12

mA2 Ω 2 cos2 (Ω t − δ ) + ω02 sin2 (Ω t − δ )[ ] (Eqn 18)

o The instantaneous total energy E of the driven oscillator depends on time unless Ω = ω0, in which case

E = 12

mA2ω02

o The average total energy over a complete cycle, denoted by ⟨ 1E1⟩ , does not depend on time. In fact it can beshown that

⟨E⟩ = ⟨Ekin ⟩ + ⟨Epot ⟩ = 14

mA2Ω 2 + 14

mA2ω02 = 1

4mA2 (Ω 2 + ω0

2 ) (21)

o We also see from Equation 21 that (unless Ω = ω0) the averages of kinetic and potential energies over a fullperiod are not equal, as they are in free SHM. In the low frequency limit (Ω << ω00) potential energydominates, while at high frequency (Ω >> ω0) the energy is almost entirely kinetic.

o Since the average total energy over a cycle is constant, the driving force must provide precisely sufficientenergy over a full period to compensate for the energy lost due to damping over a full period.


It is interesting to look at this last result from another point of view, by investigating the rate of energy transferin more detail. When a general one-dimensional force Fx, acts on a body along the line of the force withinstantaneous velocity dvx/d0t, the instantaneous rate of energy transfer to the body is just the instantaneouspower

P(t) = Fx (t)dx

dt

but, from Equation 3,

md2 x

dt2+ b

dx


the driving force of the oscillator, F3x = F01sin1(Ω1t) is given by

F3x = md2 x

dt2+ mω0

2 x + mγ dx

dt

so in the steady state, when x(t) = A 1sin1(Ω1t − δ0),

F3x = m(ω02 − Ω 2 )x + mγ dx

dt


It follows that the instantaneous power delivered by this particular force is

P(t) = F3xdx

dt= m(ω0

2 − Ω 2 )xdx

dt+ mγ dx

dt

2

and the corresponding average power over a full period will be

⟨P⟩ = m(ω02 − Ω 2 ) x

dx

dt+ mγ dx

dt

2

Now, as you will shortly be asked to demonstrate, xdx

dt= 0 .

Anticipating this result,

P = mγ dx

dt

2

(22)


However if we identify the damping force F2 x = −mγ dx

dt, it is clear that we can rewrite this last expression

P = mγ dx

dt

2

(Eqn 22)

as

P = − F2 xdx

dt

Hence, as claimed, the rate at which energy is transferred to the oscillator by the driving force, averaged over afull period, is equal to the rate at which energy is transferred from the oscillator by damping, averaged over thesame period.

Formulate a convincing physical argument to show that over a full period of oscillation xdx

dt= 0 , as claimed

above.


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

⟨Ekin ⟩ = 1

2m⟨vx

2 ⟩ = m

2dx

dt

2

so that Equation 22 can be written in the form

⟨P⟩ = mγ dx

dt

2

= 2γ ⟨Ekin ⟩


and then use Equation 21,

⟨E⟩ = ⟨Ekin ⟩ + ⟨Epot ⟩ = 14

mA2Ω 2 + 14

mA2ω02 = 1

4mA2 (Ω 2 + ω0

2 ) (Eqn 21)

which gives ⟨Ekin ⟩ = 14

mA2Ω 2 , to rewrite this

⟨P⟩ = mγ dx

dt

2

= 2γ ⟨Ekin ⟩

as

⟨P⟩ = 12

mγ (AΩ )2 (23)


As we saw in the last subsection, in the case of a lightly damped oscillator, resonance occurs when Ω = ω0,which implies that A = a/(γ1ω00) and δ = 90°. It follows from Equation 23

⟨P⟩ = 12

mγ (AΩ )2 (Eqn 23)

that the average rate of energy transfer at resonance will be

⟨Pres ⟩ = ma2

2γ(24)

and substituting this into Equation 23 we see that at any driving frequency Ω

⟨P⟩ = γ AΩa

2

⟨Pres ⟩ (25a)

According to Equation 14a γ1AΩ1/a = sin1δ, so we can also write the above result in the form:

⟨P⟩ = ⟨Pres ⟩ sin2 δ (25b)

For a given oscillator, driven at a given frequency Ω , the quantity sin2δ in Equation 25b is called thepower factor and measures the ratio of the average power absorbed at the driving frequency to the averagepower absorbed at resonance.


ω0

1.00

0.75

0.50

0.25

00 1ω1

ω0

ω2 Ω/ω02

45

90

135

180phase lag δ/degrees

⟨P⟩⟨Pres⟩

⟨P⟩⟨Pres⟩

δ

Figure 44Graphs of ⟨ 1P1⟩/⟨1Pres1⟩ and δ against Ω1/ω0 for an oscillatorwith Q = 1.

The way in which ⟨ 1P1⟩ varies with Ω for anoscillator with Q = ω0/γ = 1 can be seen fromFigure 4. Note that by plotting values of⟨ 1P1⟩/⟨ 1Pres1⟩ against Ω /ω00 we are effectivelymeasuring ⟨ 1P1⟩ in units of ⟨ 1Pres1⟩ and Ω in unitsof ω00. The graph also shows how the phase lagδ between the driving force and thedisplacement varies with Ω for this Q = 1oscillator, making it possible to see how⟨ 1P1⟩/⟨ 1Pres1⟩ varies with δ. As you can see theaverage power transfer per cycle is greatest atresonance, and in the lightly damped case thisoccurs when Ω = ω0 and δ = 90°.

Question T9

Show that in the resonance limit, for a lightly damped oscillator, the average power transferred by the drivingforce in each full cycle is ⟨ 1P1⟩ = maAω0/2, but in both the low frequency limit and the high frequency limit thepower transfer averages to zero.4


ω0

1.00

0.75

0.50

0.25

00 1ω1

ω0

ω2 Ω/ω02

45

90

135


⟨P⟩⟨Pres⟩

⟨P⟩⟨Pres⟩

δ


Figure 4 was drawn for an oscillator with therather small Q-factor of 1. It is interesting toenquire how the situation would be changed ifan oscillator with a larger value of Q had beenconsidered. (For a fixed natural frequency,higher values of Q correspond to lighterdamping.) Using the light damping result,Q = ω0/γ, Equation 24

⟨Pres ⟩ = ma2

2γ(24)

may be written in the form

⟨Pres ⟩ = ma2

2γ= ma2Q

2ω0

showing that for oscillators of the same natural frequency and mass, driven by the same force, the average powertransfer over a full cycle at resonance is proportional to the Q-value. Thus Q acts as an amplification factor forthe power absorbed. This explains why very lightly damped oscillators resonate more powerfully than heavilydamped ones.


ω0

1.00

0.75

0.50

0.25

00 1ω1

ω0

ω2 Ω/ω02

45

90

135


⟨P⟩⟨Pres⟩

⟨P⟩⟨Pres⟩

δ


In addition to its height, given by Equation 25,

⟨P⟩ = γ AΩa

2

⟨Pres ⟩ (Eqn 25a)

the power resonance curve of Figure 4 is alsocharacterized by its width. This may be definedas the (positive) difference between the

half-power points ω1

ω0and

ω2

ω0 at which

⟨P⟩ = 12 ⟨Pres ⟩ . In order that they should

correspond to half-power points, ω1 must be

the driving angular frequency at which δ1= 45°and ω02 that at which δ = 135°. At both of these

values the power factor is 0.5, so that ⟨ 1P1⟩exceeds ⟨ 1Pres1⟩/2 in the range ω1 < Ω < ω2.

This is illustrated in Figure 4.


ω0

1.00

0.75

0.50

0.25

00 1ω1

ω0

ω2 Ω/ω02

45

90

135


⟨P⟩⟨Pres⟩

⟨P⟩⟨Pres⟩

δ


The average power absorption ⟨ 1P1⟩ attains itsmaximum at the resonance frequency f0, but itremains significant throughout the range ofangular driving frequencies between the half-power points.

The corresponding frequency range

∆f = (ω2 − ω1)/2π

is called the resonance absorption bandwidthof the oscillator. This too depends on the Q-factor of the oscillator, since it can be shownthat ω2 − ω1 = γ, (see Question T11) implyingthat

resonance absorption bandwidth ∆f = ω2 − ω1

2π= γ

2π= ω0

2πQ= f 0

Q(26)


10.0

7.5

5.0

2.5

00 1 2

2ω0⟨P⟩ma2

Q = 10Q = 7.5

Q = 5

Q = 2.5

Ω/ω0

Figure 54Graphs of 2ω0⟨ 1P1⟩/(m0a2) against Ω1/ω00 for oscillators ofdifferent Q.

The resonance absorption bandwidth thereforebecomes narrower as the Q-factor increases,and it follows from Equation 26

resonance absorption bandwidth

∆f = ω2 − ω1

2π= γ

2π= ω0

2πQ= f 0

Q

(Eqn 26)

that a simple expression for the Q-factor isQ = f0/∆f. It follows that for high Q oscillatorsthe power resonance is proportionally very talland narrow. Power resonance curves for a fewQ-values are shown in Figure 5.


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⟩⟨Pres ⟩

= 1 + Q2 Ωω0

− ω0

Ω

2

−1

and use this to obtain expressions for ω1 and ω02. Hence show that ω2 − ω1 = γ, as claimed above.4


2.5 Transient motionThe steady state motion, expressed by Equation 9,


is just one possible motion for a harmonically driven oscillator. In the present subsection we will refer to thesteady state motion as xss(t). In contrast, the general solution of the homogeneous differential equation,


dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)

that describes an undriven linearly damped oscillator is a transient motion of the form:

transient motion: xtr (t) = B e−γ t 2 cos(ω t + φ ) (27)

where B and φ are arbitrary constants determined by the initial conditions of the motion, and

ω = ω02 − γ 2

4.


Interestingly this kind of transient motion is also relevant to a driven oscillator. To see why this is so note thatthe steady state motion is a solution to Equation 4,

d2 x

dt2+ γ dx

dt+ ω0


so

d2 xss

dt2+ γ dxss

dt+ ω0

2 xss = a sin (Ω t)

but the transient motion of Equation 27

transient motion: xtr (t) = B e−γ t 2 cos(ω t + φ ) (Eqn 27)

is a solution to Equation 5,


dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)

sod2 xtr

dt2 + γ dxtr

dt+ ω0

2 xtr = 0


adding corresponding terms in these two equations,

d2 xss

dt2 + d2 xtr

dt2

+ γ dxss

dt+ dxtr

dt

+ ω0 (xss + xtr ) = a sin (Ω t)

and using the mathematical property that the derivative of a sum is a sum of derivatives:

d2 (xss + xtr )dt2

+ γ d(xss + xtr )dt

+ ω02 (xss + xtr ) = a sin (Ω t)

This is just Equation 4

d2 x

dt2+ γ dx

dt+ ω0


with (xss + xtr) in place of x. Hence (xss + xtr) is also a solution of the inhomogeneous differential equation thatdescribes a damped driven harmonic oscillator. The full solution, including both the steady state and thetransient motion is therefore:

x(t) = xss (t) + xtr (t) = Asin (Ω t − δ ) + Be−γ t 2 cos(ω t + φ ) (28)


4

2

0

−2

0−4

10 20 30 40 50time t

disp

lace

men

t x(t)

Figure 64Displacement–time graphs for the sameoscillator with three different sets of initial conditions(different choices of B and φ).

Equation 28,

x(t) = xss (t) + xtr (t)

= Asin (Ω t − δ ) + Be−γ t 2 cos(ω t + φ )

gives a complete representation of the motion of thisdriven oscillator, including the transient phase.Whatever the starting conditions, as embodied in thevalues of B and φ, the transient term, xtr, in Equation 28becomes negligible for t >> 2π/γ, so that when

t >> 2π/γ x(t) ≈ xss (t) = A cos(ω t − δ )

This is illustrated in Figure 6, which shows thedisplacement–time graphs of a driven oscillator withvarious starting conditions. The transient behaviour andthe onset of steady state motion can both be clearlyseen.


Question T12

The constants B and φ that appear in Equation 28

x(t) = xss (t) + xtr (t) = Asin (Ω t − δ ) + Be−γ t 2 cos(ω t + 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


2.6 Resonance and frequency standardsSo far we have developed the mathematical model of the driven oscillator by thinking of mechanical oscillationsand mechanical resonance. As we mentioned in the introduction there are many other examples of resonancefrom non-mechanical oscillator systems. The electrical response of a tuned circuit or the resonant response of atuned cavity or an atom or nucleus to an electromagnetic wave are cases in point. The damping is caused by anyprocess 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 flowsthrough any circuit resistance. While these other systems are not always amenable to the simple linearly dampedmathematical model introduced here, the model enables us to appreciate the processes involved.


We can describe all these systems in terms of resonance, with an appropriate Q-factor implied by the frequencyresponse shown on the power absorption curve. In some cases these systems have Q-factors which are verymuch larger than those of the mechanical systems considered thus far and so the frequency response is a greatdeal narrower and more selective. If the frequency of an oscillator is narrowly defined then the period of theoscillation is also narrowly defined and the oscillator can be used as a clock. For this reason, resonant systemsfind widespread application in the maintenance of frequency or time standards. The higher the Q-factor of theoscillator 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 andquoting the Q-factors that typify them.


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 asfriction. The damping of the system is minimized and an estimate for the Q-factor achieved can be found fromEquation 26,

resonance absorption bandwidth ∆f = ω2 − ω1

2π= γ

2π= ω0

2πQ= f 0

Q(Eqn 26)

knowing the typical performance of a good mechanical clock. If the clock is to gain or lose by no more than 101sa day then this is a fractional time error of 101s in 24 × 36001s, or about 10−4. The fractional frequency error is thesame as the fractional time error and if we take a typical frequency error to be the full width of the power curveat half height (i.e. ∆0f1) then Equation 26 gives the required Q-value as f0/∆f = 104.


Our next example is a quartz crystal oscillator, such as is used in a watch. This is another mechanicaloscillation, but on a smaller size scale and at a much higher frequency (a few tens of kHz). Some materials(quartz, for example) have the property that an applied electric field causes mechanical stresses in the material.They are said to be piezoelectric materials and an oscillating voltage applied across them causes them to vibrateat the driver frequency, rather like a loudspeaker, but piezoelectric crystals can respond at much higher driverfrequencies than is possible for a conventional loudspeaker. In the quartz crystal oscillator the mechanicaloscillations of the crystal are maintained by electrical oscillations in a circuit which drives the crystal.The crystal resonance stabilizes the frequency of these oscillations and makes the resonance of the tuned circuitmuch sharper. The oscillations are then counted and this is converted into a time display. Such watches arecapable of an order of magnitude better time-keeping than the pendulum clock and so the required Q-value isabout 105. High quality quartz-controlled clocks can do rather better than this but our next example shows aspectacular improvement on this.


The caesium atomic clock is so precise that its oscillations are used to establish and maintain our fundamentalunit of time, the second. The oscillations here are rather far removed from mechanical oscillations;they correspond to a characteristic internal transition frequency within single caesium atoms. This is not theplace to go into details of this process but suffice to say that caesium atoms can absorb electromagnetic radiationat a frequency of around 91921MHz, in a resonant process. For our purpose here the important point is thatthe sharpness of the atomic resonance can be made so great that the full resonance absorption bandwidth is aslow as 11Hz. With a basic resonance frequency of 9192 MHz this gives a Q-factor of about 1010!

Sophisticated technology is used to stabilize the frequency to within a small fraction of this bandwidth and thefrequency of such clocks can then be made reproducible to about 1 part in 1013. Two such clocks will keep thesame time to within about 3 seconds in a million years! Unfortunately, these clocks are of a size which would filla medium-sized room and you would not want one strapped around your wrist. Their use is confined toestablishing the international time standard and in calibrating other more portable clocks. Such high precision isneeded in certain scientific experiments. For example, caesium clocks flown in aircraft have enabled the directobservation of the tiny time changes for moving clocks, predicted by Einstein’s special theory of relativity.


Even the caesium clock could be made to look rather crude when compared to future clocks that may be basedon frequency-stabilized lasers. In a laser the oscillators are effectively an ensemble of atoms in a cavity, ratherthan individual atoms, and the driver is an electromagnetic wave (light), trapped within the same cavity, betweenmirrors. The atoms absorb and re-emit the wave within the cavity but because they are all in communicationwith the same wave they all absorb and emit with well-defined phases, rather than with random phases as wouldbe the case if they were acting individually. Such phase-organized light is called coherent light. The process bywhich an incoming wave drives an atom to emit is known as stimulated emission and it is the counterpart toabsorption. If the stimulated emission exceeds the absorption then this will give rise to laser action, or LightAmplification by Stimulated Emission of Radiation. Within the cavity of a laser oscillator there is a single giantamplitude oscillation with a well-defined phase and an extremely well-defined frequency. The frequency outputbandwidth for the laser is very much narrower than would be the case if the same atoms were placed outside thecavity, as in an ordinary lamp.


Lasers have many advantages over conventional light sources but the one that concerns us here is this verynarrow bandwidth. It gives rise to an enormous value for the Q-factor of the oscillator. The full resonancebandwidth of a laser oscillator can be made to be only a few Hz, even when the laser is operating in the visiblespectrum, at frequencies of around 5 × 1014 Hz. If such a laser could also have its frequency stabilizedsufficiently to capitalize on this low bandwidth then the resulting frequency-stabilized laser could have aQ-factor of around 1014, with a potential improvement factor of 104 over the caesium clock. The technology forthis is under development and stabilized lasers could well be the reference clocks of the future.


3 Coupled oscillatorsIn the previous section we considered how an oscillator is affected by an oscillatory (harmonic) driving force.This force has been taken to be so robust that its driving mechanism is not affected by the oscillator. Let us nowtake a look at what happens when the driving force is replaced by another oscillator.

3.1 Normal modes

A B

x2x1

Figure 74A model of two coupledoscillators.

As a specific example, consider two identical oscillators, consisting ofequal masses A and B, resting on a smooth horizontal surface andconnected by identical springs to two rigid walls, as shown in Figure 7.The two masses are also connected together by another spring, which wewill suppose to be much weaker than the two main ones. Initially the twomasses are at rest at their equilibrium positions.


A B

x2x1


Now let us move A from its equilibrium position, while holding B fixed,and then let go of both. A will immediately start to oscillate, but becauseof the coupling between the two masses, it will excite an oscillation ofincreasing amplitude in B. There is only a limited amount of energy tosupply these oscillations, so as the amplitude of B increases that of Amust decrease. Surprisingly, this exchange of energy does not stop whenthe two amplitudes are equal, but continues until A is stationary and B ismoving with the full amplitude that A had at the start. The process is then reversed, with the B amplitudedecreasing and that of A increasing until the starting configuration is reached and the whole process is repeatedad infinitum, or until damping reduces the vibrational energy to zero.


x1

t

x2

t

A

B

Figure 84The transfer of energy betweentwo coupled oscillators.

Figure 8 shows how the two displacements vary with time.

Are these the sort of time variations we should expect for twoharmonic oscillators?


x1

t

x2

t

A

B


The overall motion of the two coupled oscillators looks quitecomplicated and neither oscillator moves with simple harmonicmotion. However, the motion is periodic and this encourages us toask whether there is any concealed simple harmonic motion in thissystem. As you will soon see, the answer is yes, and these motionsare remarkably simple in this case1—1once we expose them.


A B

x2x1


When we disturb the system of Figure 7 in different ways we notice that,as we have described, usually the state of motion changes with time, withone mass first oscillating vigorously then becoming quiescent, while theother does the same but with opposite periods of activity, as the energy isswitched back and forth. However, there are two particular patterns ofrelative motion in which the motion continues unchanged, with neithermass changing its energy with time.


A Bt

t

x1

x2

A

B

A Bt

t

x1

x2

A

B

(a)

(b)

x2x1

−x2x1

Figure 94(a) Masses moving in phase; (b) masses moving out of phase.

In one of these motionsthe two identical massesmove together, in phase,with equal amplitude, asshown in Figure 9a.The distance between thetwo wi l l remainunchanged, so thecoupling spring will haveno effect on the motion,and the frequency ofoscillation will be justthe same as if the twowere uncoupled. In theother motion the twomasses move in oppositedirections with equalamplitude, as shown inFigure 9b.


Would you expect this second oscillation to have the same frequency as the first? If not, will the frequencybe higher or lower?

These two forms of steady state oscillation are simple harmonic motions, with constant amplitudes, and areknown as the two normal modes of this system. Such normal modes always appear whenever two or moreoscillators are coupled together. Any motion of the system can be analysed in terms of an appropriatesuperposition of these normal modes of the system, as we will see shortly, so the normal modes can beconsidered as the basic SHM building blocks of any oscillation of the system.

These two normal modes may then be identified through the mass displacements x1(t) and x2(t) as:

for ω = ω1 = ω00 x1(t) = x01cos1(ω00t) and x2(t) = x01cos1(ω00t) (29a)

for ω = ω02 x1(t) = x01cos1(ω02t) and x2(t) = −x01cos1(ω02t) (29b)


x1

t

x2

t

A

B


We will now show that the complicated motions depicted in Figure 8


A Bt

t

x1

x2

A

B

A Bt

t

x1

x2

A

B

(a)

(b)

x2x1

−x2x1

Figure 94(a) Masses moving in phase; (b) masses moving out of phase.

and the simple normalmodes in Figure 9 arerelated. To do this weneed to invoke the ideaof the superposition oftwo simple harmonicmotions.


The principle of superposition states that:

When several oscillations are added, the resultant displacement at any time is the sum of the displacementsdue to each oscillation at that time.

Now let us construct an oscillation representing 50% of mode 1 and 50% of mode 2 by superposing the twonormal modes of Equation 29.

ω = ω1 = ω00 x1(t) = x01cos1(ω00t) and x2(t) = x01cos1(ω00t) (Eqn 29a)

ω = ω02 x1(t) = x01cos1(ω02t) and x2(t) = −x01cos1(ω02t) (Eqn 29b)

We obtain:

x1(t) = 12 [x0 cos(ω0t) + x0 cos(ω2t)]

x2 (t) = 12 [x0 cos(ω0t) − x0 cos(ω2t)]


Using an identity from Question R4 we can write these two

x1(t) = 12 [x0 cos(ω0t) + x0 cos(ω2t)]

x2 (t) = 12 [x0 cos(ω0t) − x0 cos(ω2t)]

as

x1(t) = x0 cosω2 − ω0

2t

cos

ω2 + ω0

2t

(30a)

x2 (t) = x0 sinω2 − ω0

2t

sin

ω2 + ω0

2t

(30b)

Question T13

Verify Equations 30a and 30b.4


x1

t

x2

t

A

B


Since ω0 and ω2 are different, but usually not greatly different, eachof the right-hand sides in these equations

x1(t) = x0 cosω2 − ω0

2t

cos

ω2 + ω0

2t

(Eqn 30a)

x2 (t) = x0 sinω2 − ω0

2t

sin

ω2 + ω0

2t

(Eqn 30b)

is a product of a slowly oscillating term, involving (ω02 − ω0), and arapidly oscillating one involving (ω02 + ω0).

If we refer back to Figure 8 we can see that this slowly oscillatingterm can be interpreted as describing the overall slowly changingamplitude of the motion of each mass, often referred to as beating, at the difference frequency of the two modes.The rapid oscillation of each mass is represented by the mean frequency of the two modes.

Question T14

After what time does the mass A come to rest, with all the motion passed to mass B? How long is it before A ismoving again with maximum amplitude and with B stationary?4


Normal modes appear in any situation where we have oscillations coupled together, and when there is a largenumber of oscillators1—1an extreme case is the atoms in a crystal, the forms of oscillation and their frequenciescan become extremely complex. However, the main points to bear in mind are:

Any set of oscillators coupled together will result in normal modes which carry out simple harmonicmotion.

The frequencies of the normal modes are almost always different from the frequencies of the uncoupledoscillators.


4 Closing items

4.1 Module summary1 Freely oscillating systems are characterised by natural frequencies of oscillation. The action of damping

forces will generally modify the frequency of an oscillator, and cause its amplitude to diminish with time.Oscillations in damped systems may be sustained by means of an externally applied driving force; suchoscillators are called forced or driven oscillators. Under appropriate conditions, a periodic driving forcewith a period close to the natural period of an oscillator can cause a driven oscillator to develop oscillationsof relatively large amplitude; this condition is known as resonance.

2 The equation of motion of a harmonically driven, linearly damped oscillator is a (inhomogeneous)differential equation of the form:

d2 x

dt2+ γ dx

dt+ ω0


If the driving force is absent (a = 0) the remaining (homogeneous) differential equation describes alinearly damped oscillator:

d2 x

dt2+ γ dx

dt+ ω0

2 x = 0 (Eqn 5)


and if γ = 0, we obtain the equation of motion of a simple harmonic oscillator

d2 x

dt2+ ω0

2 x = 0 (Eqn 6)

3 The full solution of the driven oscillator equation can be regarded as the sum of two parts; one describingthe transient motion which effectively decays to zero after a time of order 2π/γ, and one describing thepersisting steady state motion that is sustained by the driving force.

4 Once it has entered the steady state, the driven oscillator acquires the same frequency as the driving force,though there is generally a constant phase lag between the driving force (F3x = ma 1sin1(Ω1t)) and theoscillator’s displacement. The steady state motion is therefore independent of the starting conditions andcan be written as

x(t) = A1sin1(Ω1t − δ) (Eqn 9)


5 In the steady state, the amplitude A and the phase lag δ both depend on the driving frequency Ω and aregiven by:

A = a

(ω02 − Ω 2 )2 + (γ Ω )2

(Eqn 15)

and δ = arctanγ Ω

ω02 − Ω 2

(Eqn 16)

The amplitude is almost independent of Ω at low frequencies, attains a maximum value at resonance, whichdefines the resonance frequency, and then decreases at high frequency. The phase lag is small at lowfrequency, 90° at the undamped natural frequency, which is near resonance and approaches 180° at highfrequency. In particular;

in the low frequency limit: x(t) = a

ω02

sin (Ω t) (Eqn 10)

in the high frequency limit: x(t) = a

Ω 2

sin (Ω t − π) (Eqn 11)


and in the resonance limit (Ω = ω0 for the case of light damping):

x(t) = a

γω0

sin Ω t − π2

(Eqn 12)

In the latter case the resonance amplitude exceeds the low frequency amplitude by the oscillator’squality factor Q = ω0/γ.

6 In the steady state, the total energy, kinetic energy and potential energy are all functions of time, and aregiven by

total energy E = 12

mA2 Ω 2 cos2 (Ω t − δ ) + ω02 sin2 (Ω t − δ )[ ] (Eqn 18)

kinetic energy Ekin = 12

mA2Ω 2 cos2 (Ω t − δ ) (Eqn 19)

potential energy Epot = 12

mA2ω02 sin2 (Ω t − δ ) (Eqn 20)


However, the average value of each of these quantities, taken over a full period of oscillation, has a constantvalue, as given by

⟨E⟩ = ⟨Ekin ⟩ + ⟨Epot ⟩ = 14

mA2Ω 2 + 14

mA2ω02 = 1

4mA2 (Ω 2 + ω0

2 ) (Eqn 21)

The average total energy is mostly potential energy below resonance, and mostly kinetic energy aboveresonance. At resonance, in the case of light damping (when Ω = ω0), the total energy is independent oftime, potential and kinetic energy contribute equally to the total, and

E = ⟨E⟩ = 12

mA2ω02

7 In the steady state, the average power ⟨ 1P1⟩ transferred to the oscillator from the driving force, is exactlyequal to the average power dissipated by the damping force.


8 The average power absorption, ⟨ 1P1⟩ , exhibits resonance behaviour, i.e. it rises to a maximum when thedriving frequency is close to the undamped natural frequency. The height of the maximum is proportional toQ, and its width is proportional to Q−1. Thus, more lightly damped oscillators (characterized by largervalues of Q), have taller and narrower resonance peaks. The average power transfer at resonance is

⟨Pres ⟩ = ma2

2γ= ma2Q

2ω0(Eqn 24)

and the resonance absorption bandwidth is:

∆f = ∆ω2π

= ω2 − ω1

2π= γ

2π= ω0

2πQ= f 0

Q(Eqn 26)

9 One important application of resonance absorption is to establish oscillators whose frequencies are verywell-defined; these can constitute frequency and time standards.

10 The transient motion for the driven oscillator can be represented by:

xtr (t) = B e−γ t 2 cos(ω t + φ ) (Eqn 27)

where B and φ are arbitrary constants determined by the initial conditions of the motion, and

ω = ω02 − γ 2

4.


11 The general solution to the equation of motion for the harmonically driven, linearly damped harmonicoscillator may be written

x(t) = xss (t) + xtr (t) = Asin (Ω t − δ ) + Be−γ t 2 cos(ω t + φ ) (Eqn 28)

12 When two oscillators are coupled together, the energy of oscillation may be exchanged continually betweenthem.

13 The simple harmonic motions of a pair of identical coupled oscillators are the normal modes, consisting ofin-phase and out-of-phase motions of the two. These can be represented as:

for ω = ω1 = ω00 x1(t) = x01cos1(ω00t) and x2(t) = x01cos1(ω00t) (Eqn 29a)

for ω = ω02 x1(t) = x01cos1(ω02t) and x2(t) = −x01cos1(ω02t) (Eqn 29b)

When these two normal modes are superposed the resulting oscillation has the average frequency of the twomodes and the superposed amplitude shows beats at the difference frequency:

x1(t) = x0 cosω2 − ω0

2t

cos

ω2 + ω0

2t

(Eqn 30a)

x2 (t) = x0 sinω2 − ω0

2t

sin

ω2 + ω0

2t

(Eqn 30b)


14 Any set of oscillators coupled together will result in normal modes which carry out simple harmonicmotion. The frequencies of the normal modes are almost always different from the frequencies of theuncoupled oscillators.


4.2 AchievementsHaving completed this module, you should be able to:

A1 Define the terms that are emboldened and flagged in the margins of the module.

A2 Describe what is meant by a driven oscillator, explaining its main response features in steady state motionand transient motion and how these may be varied, giving specific examples.

A3 Write down the equation of motion of a harmonically driven linearly damped oscillator.

A4 Derive the properties of the steady state motion from this equation.

A5 Account for the main features of resonant behaviour in terms of the frequency dependence of the amplitudeand phase lag in the steady state motion.

A6 Show that the average value of the total energy of the oscillator is constant, and that the proportion ofkinetic energy increases from zero at low frequency to 50% at resonance and approaches 100% at highfrequency.

A7 Show that the average power absorbed from the driving force is equal to the average power dissipated bydamping.


A8 Discuss the resonance phenomenon in terms of the graph of average power absorbed against drivingfrequency, showing (in the case of light damping) that it possesses a peak of height proportional to Q andwidth proportional to Q−1, centred on the natural frequency.

A9 Describe the importance of resonance and the Q-factor of an oscillator in relation to frequency standards,illustrating this with some examples.

A10 Describe the possible motions that follow when two identical oscillators are coupled together, and explainhow they can be described in terms of the normal modes of the system.

Study comment You may now wish to take the Exit test for this module which tests these Achievements.If you prefer to study the module further before taking this test then return to the Module contents to review some of thetopics.


4.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions each of which testsone or more of the Achievements.

Question E1

(A3 and A4)4An oscillator consists of a mass of 0.021kg suspended from a spring with spring constant 501N1m−1.It is subject to a damping force of the form −bvx, where vx is the velocity and b = 0.51kg1s−1, and is driven by aharmonically varying force of amplitude 41N and frequency 5501rpm. Calculate the amplitude and phase lag ofthe resulting steady state vibration.

Question E2

(A6)4For the oscillator in Question E1 calculate the average potential energy, the average kinetic energy andthe average power absorbed.


Question E3

(A2, A5, A8 and A7)4Under what practical circumstances does one try to minimize the amplitude of a forcedvibration? What methods can one use to do this?

Question E4

(A2, A5, A8 and A7)4In designing a loudspeaker one wants to have as ‘flat’ a response as possible to differentfrequencies so the amplitude is independent of the driver frequency. What principles should one bear in mindwhen trying to achieve this? What unavoidable disadvantage accompanies the process?

Question E5

(A2, A5, A8 and A9)4Under what practical circumstances does one try to maximize the amplitude of a forcedvibration? How can one achieve this?


Question E6

(A10)4Two identical oscillators are connected together by weak coupling. Initially they are at rest.

(a) Describe what happens when one is made to oscillate.

(b) How does this behaviour change if the coupling is made stronger?

(c) What simple harmonic motions are possible in the system?

Study comment This is the final Exit test question. When you have completed the Exit test go back to Subsection 1.2 andtry the Fast track questions if you have not already done so.

If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave ithere.

Documents

FLEXIBLE LEARNING APPROACH TO PHYSICS ÊÊÊ … · FLAP P5.3 Forced vibrations and resonance COPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1 1 Opening items 1.1 Module introduction