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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
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Forced vibrations and resonance
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3 Coupled oscillators
3.1 Normal modes
4 Closing items
4.1 Module summary
4.2 Achievements
4.3 Exit test
Exit module
S570 V1.1
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.
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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.
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Forced vibrations and resonance
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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.
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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.
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Forced vibrations and resonance
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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, d2 x/dt2) 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.
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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 is
given by x(t) = A0 1sin1(ω1 t + 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?
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Question R4
(a) Given the trigonometric identity
cos (θ + φ ) = cos θ cos φ − sin θ sin φ
show that
cos 2 θ + sin 2 θ = 14and4 cos 2 θ − sin 2 θ = cos (2 θ ) .
(b) Use the trigonometric identity
sin ( θ + φ ) = sin θ cos φ + cos θ 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:
3 +1
5π
π
3 −1
cos   =
4and4 cos   =


 12 
2 2
2 2
12
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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.
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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❏
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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 Fd
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❏
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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.
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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 9 × 10 −141m 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.
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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 2 x due to friction and air resistance that opposes the
motion of the body.
A driving force F3x provided by an external agency.
P
m
rough surface
x
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).
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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.
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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
dx
F2 x = −b
(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
F3x = F0 1sin1(Ω1 t)
(2c)
☞
where F0 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.
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P
m
rough surface
x
Figure 14A simple driven
oscillator. The driving force has
been omitted from the figure
because its sign may change with
time.
Substituting Equations 2a, 2b and 2c into Equation 1,
F1x = −1kx
(Eqn 2a)
dx
= −b
dt
(Eqn 2b)
F3x = F0 1sin1(Ω1 t)
(Eqn 2c)
F2 x
m
d2x
= F1x + F2 x + F3x
dt 2
(Eqn 1)
we see that the behaviour of the sliding mass must now be described by the equation
d2x
dx
m 2 = −kx − b
+ F0 sin ( Ω t)
dt
dt
which can be rearranged to isolate the time-dependent driving term as follows
d2x
dx
m 2 +b
+ kx = F0 sin ( Ω t)
dt
dt
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(3)
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
d2x
dx
m 2 +b
+ kx = F0 sin ( Ω t)
(Eqn 3)
dt
dt
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 = F0/m
and the natural angular frequency ω 0 =
k m
☞
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:
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for the harmonically driven linearly damped oscillator:
d2x
dx
+γ
+ ω 02 x = a sin ( Ω t)
dt 2
dt
(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:
dx
d2x
+γ
+ ω 02 x = 0
2
dt
dt
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(5)
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:
d2x
+ ω 02 x = 0
dt 2
(6)
We will approach the solution of Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
2
dt
dt
(Eqn 4)
through the simpler Equation 6
and then Equation 5.
linearly damped oscillator:
dx
d2x
+γ
+ ω 02 x = 0
dt 2
dt
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(Eqn 5)
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 . ☞
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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) = A0 1e−γ1t/02 1sin1(ω1t + φ) 4where 4 ω = ω 02 − γ 2 4
(8)
P
m
x
rough surface
which is the general solution to Equation 5.
dx
d2x
linearly damped oscillator:
+γ
+ ω 02 x = 0
(Eqn 5)
Figure 14A simple driven
2
dt
dt
oscillator. The driving force has
If we want to determine how the sliding body of Figure 1 moves when been omitted from the figure
subjected to the three forces specified above we need to find a solution to because its sign may change with
time.
Equation 4,
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
dt 2
dt
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.
d2x
simple harmonic oscillator:
+ ω 02 x = 0
(Eqn 6)
dt 2
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This is because Equations 5 and 6
dx
d2x
linearly damped oscillator:
+γ
+ ω 02 x = 0
dt 2
dt
d2x
simple harmonic oscillator:
+ ω 02 x = 0
dt 2
(Eqn 5)
(Eqn 6)
are homogeneous differential equations in which each term is proportional to x or to one of its time
derivatives, whereas Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
dt 2
dt
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
d2x
dx
m 2 +b
+ kx = F0 sin ( Ω t)
(Eqn 3)
dt
dt
(see second-order differential equation in the Glossary for references), but we will not employ such methods
here.
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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
m
rough surface
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❏
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x
Figure 14A simple driven
oscillator. The driving force has
been omitted from the figure
because its sign may change with
time.
2.3 Steady state motion
d2x
dx
+γ
+ ω 02 x = a sin ( Ω t)
dt 2
dt
(Eqn 4)
Equation 4 represents a struggle. The oscillator, when disturbed, tends to oscillate with its natural angular
frequency ω0 , 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.
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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 = F0 1sin1(Ω1 t)).
Having arrived at Equation 9 on the basis of physical arguments, rather than by straightforward mathematical
deduction from Equation 4,
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
dt 2
dt
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 (ω0 and γ1).
By doing this we should gain insight into our trial solution and either convince ourselves of its correctness or
expose its shortcomings.
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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 << ω0 ), then
when it is much larger than the natural angular frequency ( Ω >> ω 0 ), and finally when the two angular
frequencies 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
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
dt 2
dt
(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.
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Low frequency limit: W << w0
As we have just seen, our trial solution implies that the first two terms in Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
dt 2
dt
involve factors of Ω12 and Ω, respectively. If Ω << ω00 the first term (Ω12x) is small compared to the third term
(ω0 2 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) ≈ a sin ( Ω t)
in the low frequency limit:
 a 
x(t) =  2  sin ( Ω t)
 ω0 
(10)
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 02 ) , 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.
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High frequency limit: W >> w0
In this case the first term in Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
dt 2
dt
(Eqn 4)
is the important one, if the oscillator is lightly damped, since it involves a factor of Ω12 which will be large.
Thus a reasonable approximation to Equation 4 is:
d2x
≈ a sin ( Ω t)
dt 2
but in the steady state described by Equation 9
steady state motion
x(t) = A1sin1(Ω1t − δ1)
(Eqn 9)
we have
d2x
= − Ω 2 x(t)
dt 2
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Hence
a
x(t) ≈ −  2  sin ( Ω t)
Ω 
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)
in the high frequency limit:
(Eqn 9)
a
x(t) =  2  sin ( Ω t − π)
Ω 
(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.
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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 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,
steady state motion
x(t) = A1sin1(Ω1t − δ1)
(Eqn 9)
and setting Ω = ω0 , Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
2
dt
dt
(Eqn 4)
now becomes
− ω 02 x + γω 0 A cos (ω 0 t − δ ) + ω 02 x = a sin ( Ω t)
On the left-hand side the first and last terms cancel, so that:
γ1ω0 A1cos1(ω0 t − δ1) = a1sin1(Ω1 t)
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We can use the trigonometric identity cos1θ = sin1(π/2 + θ) to rewrite this in the form
γ1ω0 A1sin1(π/2 + ω0 t − δ1) = a1sin1(Ω1 t)
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, it
must also be the case that γω0 A = a. It then follows from Equation 9
steady state motion
x(t) = A1sin1(Ω1t − δ1)
(Eqn 9)
that
in the resonance limit:
 a  
π
x(t) = 
 sin  Ω t − 
2
 γω 0 
(12)
where the amplitude at resonance is given by A = a γ ω 0 which exceeds the amplitude in the low frequency
a ω 02 ω 0
limit by a factor
=
. We will call this factor the quality factor or Q-factor of the lightly damped
γω 0 a
γ
oscillator.
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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 = ω0 /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❏
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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.
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
2
dt
dt
(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.
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When this is done we find:
A(ω 02 − Ω 2 )sin ( Ω t − δ ) + Aγ Ω cos ( Ω t − δ )
= a sin δ cos ( Ω t − δ ) + a cos δ sin ( Ω t − δ )
(13)
☞
Examine Equation 13 carefully. Notice that the time t occurs in two different sinusoidal functions.
steady state motion
d2x
dt 2
+γ
x(t) = A1sin1(Ω1t − δ1)
(Eqn 9)
dx
+ ω 02 x = a sin ( Ω t)
dt
(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(Ω1 t − δ1) on each side. Equating the two sets of coefficients leads
immediately to the pair of equations:
Aγ Ω
sin δ =
(14a)
a
A(ω 02 − Ω 2 )
cos δ =
(14b)
a
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Squaring and adding these equations to eliminate δ, through the identity sin2 δ + cos2 δ = 1, gives the amplitude
as:
general case amplitude A =
a
(ω 02
−
Ω 2 )2
+ ( γ Ω )2
Dividing Equation 14a by 14b
Aγ Ω
sin δ =
a
A(ω 02 − Ω 2 )
cos δ =
a
(Eqn 14a)
(Eqn 14b)
leads to:
tan δ =
γΩ
sin δ
=
cos δ ω 02 − Ω 2
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(15)
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Therefore we can write
 γΩ 
general case phase lag: δ = arctan  2

 ω0 − Ω 2 
general case amplitude
A=
(16)
a
(ω 02
−
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 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.
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180
phase lag δ/degrees
A/(a/ω 02 )
10.0
7.5
5.0
2.5
135
90
45
0
0
0
1
driving frequency Ω /ω 0
2
Q = 7.5
Q=5
Q = 2.5
Q = 10
0
1
driving frequency Ω/ω0
2
Figure 24Graphs of amplitude A (expressed as a multiple of Figure 34Graphs of phase lag δ (expressed in degrees)
a ω 02 ) against driving frequency Ω (expressed as a multiple against driving frequency Ω (expressed as a multiple of ω0 )
for oscillators of different Q.
of ω ) for oscillators of different Q. The Q-value for each
0
oscillator is given by the peak value of the response curve.
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10.0
A/(a/ω 02 )
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
ωres and the corresponding resonance frequency
fres = ωres/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.
7.5
5.0
2.5
0
0
1
driving frequency Ω /ω 0
2
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 each
oscillator is given by the peak value of the response curve.
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180
phase lag δ/degrees
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 = ω0 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.
135
90
45
0
Q = 7.5
Q=5
Q = 2.5
Q = 10
0
1
driving frequency Ω/ω0
2
Figure 34Graphs of phase lag δ (expressed in degrees)
against driving frequency Ω (expressed as a multiple of ω0 )
for oscillators of different Q.
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Question T7
Use Equations 15 and 16
a
general case amplitude
A=
general case phase lag:
 γΩ 
δ = arctan  2

 ω0 − Ω 2 
(ω 02
−
Ω 2 )2
+ ( γ Ω )2
(Eqn 15)
(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 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 when
the angular frequency of the driving force is: (a) 501s−1, (b) 1001s−1 and (c) 2001s−1.4❏
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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 = k m so we have:
Etot = E = Ekin + Epot =
2
2
1  dx 
1
1
dx
1
m
+ kx 2 = m   + mω 02 x 2
2  dt 
2
2  dt 
2
(17)
In the steady state we can use Equation 9 to express this total energy in terms of t rather than x, giving:
1
1
E = mA 2 Ω 2 cos 2 ( Ω t − δ ) + mω 02 A 2 sin 2 ( Ω t − δ )
2
2
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For the harmonically driven oscillator:
1
mA 2 [ Ω 2 cos 2 ( Ω t − δ ) + ω 02 sin 2 ( Ω t − δ )]
2
total energy
E=
kinetic energy
Ekin =
1
mA 2 Ω 2 cos 2 ( Ω t − δ )
2
(19)
potential energy
Epot =
1
mA 2 ω 02 sin 2 ( Ω t − δ )
2
(20)
(18)
Equations 18 to 20 have several important consequences for the driven oscillator:
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total energy
o
o
E=
1
mA 2 [ Ω 2 cos 2 ( Ω t − δ ) + ω 02 sin 2 ( Ω t − δ )]
2
The instantaneous total energy E of the driven oscillator depends on time unless Ω = ω 0 , in which case
1
E = mA 2 ω 02
2
The average total energy over a complete cycle, denoted by 〈1E1〉, does not depend on time. In fact it can be
shown that
〈E〉 = 〈Ekin 〉 + 〈Epot 〉 =
o
o
(Eqn 18)
1
1
1
mA 2 Ω 2 + mA 2 ω 02 = mA 2 ( Ω 2 + ω 02 )
4
4
4
(21)
☞
We also see from Equation 21 that (unless Ω = ω0) 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 << ω 00 ) potential energy
dominates, while at high frequency ( Ω >> ω0) 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.
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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
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(Eqn 3)
It follows that the instantaneous power delivered by this particular force is
P(t) = F3x
dx
dx
dx
= m(ω 02 − Ω 2 )x
+ mγ  
 dt 
dt
dt
2
and the corresponding average power over a full period will be
〈P〉 = m(ω 02 − Ω 2 ) x
dx
dx
+ mγ  
 dt 
dt
2
Now, as you will shortly be asked to demonstrate, x
dx
= 0.
dt
Anticipating this result,
dx
P = mγ  
 dt 
2
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(22)
S570 V1.1
However if we identify the damping force F2 x = −mγ
dx
P = mγ  
 dt 
dx
, it is clear that we can rewrite this last expression
dt
2
(Eqn 22)
as
P = − F2 x
dx
dt
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
above.
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dx
= 0 , as claimed
dt
Using Equation 22
dx
P = mγ  
 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
m  dx 
m〈vx2 〉 =
2
2  dt 
2
so that Equation 22 can be written in the form
dx
〈P〉 = mγ  
 dt 
2
= 2 γ 〈Ekin 〉
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and then use Equation 21,
〈E〉 = 〈Ekin 〉 + 〈Epot 〉 =
which gives 〈Ekin 〉 =
dx
〈P〉 = mγ  
 dt 
1
1
1
mA 2 Ω 2 + mA 2 ω 02 = mA 2 ( Ω 2 + ω 02 )
4
4
4
(Eqn 21)
1
mA 2 Ω 2 , to rewrite this
4
2
= 2 γ 〈Ekin 〉
as
〈P〉 =
1
mγ ( AΩ )2
2
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(23)
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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
1
〈P〉 = mγ ( AΩ )2
(Eqn 23)
2
that the average rate of energy transfer at resonance will be
ma 2
〈Pres 〉 =
(24)
2γ
and substituting this into Equation 23 we see that at any driving frequency Ω
〈P〉 = 

γ AΩ  2
〈Pres 〉
a 
(25a)
☞
According to Equation 14a γ1AΩ1/a = sin1δ, so we can also write the above result in the form:
〈P〉 = 〈Pres 〉 sin 2 δ
(25b)
☞
For a given oscillator, driven at a given frequency Ω , the quantity sin2 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.
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Question T9
180
1.00
δ
〈P〉
〈Pres〉
0.75
135
〈P〉
0.50
〈Pres〉
90
0.25
45
0
0
ω1
ω0
1
ω2
ω0
phase lag δ /degrees
The way in which 〈1P1〉 varies with Ω for an
oscillator with Q = ω0 /γ = 1 can be seen from
Figure 4. Note that by plotting values of
〈1P1〉/〈1Pres1〉 against Ω /ω00 we are effectively
measuring 〈1P1〉 in units of 〈1Pres1〉 and Ω in units
of ω00 . 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〉/〈1Pres1〉 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 = ω0 and δ = 90°.
2 Ω/ω0
Figure 44Graphs of 〈1P1〉/〈1Pres1〉 and δ against Ω1/ω0 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〉 = maAω0 /2, but in both the low frequency limit and the high frequency limit the
power transfer averages to zero.4❏
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may be written in the form
ma 2 ma 2 Q
〈Pres 〉 =
=
2γ
2ω 0
180
1.00
δ
〈P〉
〈Pres〉
0.75
135
〈P〉
0.50
〈Pres〉
90
0.25
45
0
0
ω1
ω0
1
ω2
ω0
phase lag δ /degrees
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 = ω0 /γ, Equation 24
ma 2
〈Pres 〉 =
(24)
2γ
2 Ω/ω0
Figure 44Graphs of 〈1P1〉/〈1Pres1〉 and δ against Ω1/ω0 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.
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In addition to its height, given by Equation 25,
γ AΩ 
〈Pres 〉
a 
(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
ω
ω
half-power points 1 and 2 at which
ω0
ω0
〈P〉 = 12 〈Pres 〉 . In order that they should
correspond to half-power points, ω 1 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 .
135
〈P〉
0.50
〈Pres〉
90
0.25
45
0
0
ω1
ω0
1
ω2
ω0
2 Ω/ω0
Figure 44Graphs of 〈1P1〉/〈1Pres1〉 and δ against Ω1/ω0 for an oscillator
with Q = 1.
This is illustrated in Figure 4.
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δ
〈P〉
〈Pres〉
0.75
phase lag δ /degrees
〈P〉 = 

180
1.00
2
S570 V1.1
The corresponding frequency range
resonance absorption bandwidth
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δ
〈P〉
〈Pres〉
0.75
135
〈P〉
0.50
〈Pres〉
90
0.25
45
∆f = (ω2 − ω1 )/2π
is called the resonance absorption bandwidth
of the oscillator. This too depends on the Qfactor of the oscillator, since it can be shown
that ω2 − ω 1 = γ, (see Question T11) implying
that
180
1.00
0
0
ω1
ω0
ω2
ω0
1
phase lag δ /degrees
The average power absorption 〈1P1〉 attains its
maximum at the resonance frequency f0 , but it
remains significant throughout the range of
angular driving frequencies between the halfpower points.
2 Ω/ω0
Figure 44Graphs of 〈1P1〉/〈1Pres1〉 and δ against Ω1/ω0 for an oscillator
with Q = 1.
∆f =
ω 2 − ω1
γ
ω
f
=
= 0 = 0
2π
2π 2πQ
Q
S570 V1.1
(26)
The resonance absorption bandwidth therefore
becomes narrower as the Q-factor increases,
and it follows from Equation 26
10.0
Q = 10
Q = 7.5
7.5
resonance absorption bandwidth
ω − ω1
γ
ω
f
∆f = 2
=
= 0 = 0
2π
2π 2πQ
Q
(Eqn 26)
that a simple expression for the Q-factor is
Q = f0 /∆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.
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2ω0〈P〉
ma2
Q=5
5.0
Q = 2.5
2.5
0
0
1
2 Ω/ω0
Figure 54Graphs of 2 ω0 〈1P1〉/(m0a 2 ) against Ω1/ω00 for oscillators of
different Q.
S570 V1.1
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:
2

 Ω
ω  
〈P〉
= 1 + Q 2 
− 0 
Ω  
〈Pres 〉 
 ω0

−1
and use this to obtain expressions for ω1 and ω02 . Hence show that ω2 − ω1 = γ, as claimed above.4❏
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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 xss(t). In contrast, the general solution of the homogeneous differential equation,
linearly damped oscillator:
dx
d2x
+γ
+ ω 02 x = 0
2
dt
dt
(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 (ω t + φ )
(27)
☞
where B and φ are arbitrary constants determined by the initial conditions of the motion, and
γ2
ω = ω 02 −
.
4
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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,
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
dt 2
dt
so
dx
d 2 xss
+ γ ss + ω 02 xss = a sin ( Ω t)
2
dt
dt
but the transient motion of Equation 27
transient motion:
x tr (t) = B e − γ t 2 cos (ω t + φ )
(Eqn 27)
is a solution to Equation 5,
linearly damped oscillator:
so
dx
d2x
+γ
+ ω 02 x = 0
2
dt
dt
d 2 x tr
dx tr
+ ω 02 x tr = 0
2 +γ
dt
dt
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(Eqn 5)
adding corresponding terms in these two equations,
 d 2 xss d 2 x tr 
 dxss + dx tr  + ω (x + x ) = a sin ( Ω t)
0 ss
tr

2 +
2  +γ
dt
dt 
dt 
 dt
and using the mathematical property that the derivative of a sum is a sum of derivatives:
d(xss + x tr )
d 2 (xss + x tr )
+γ
+ ω 02 (xss + x tr ) = a sin ( Ω t)
2
dt
dt
This is just Equation 4
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
dt 2
dt
(Eqn 4)
with (x ss + xtr) in place of x. Hence (xss + xtr) 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) = xss (t) + x tr (t) = A sin ( Ω t − δ ) + Be − γ t 2 cos (ω t + φ )
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(28)
☞
Equation 28,
x(t) = xss (t) + x tr (t)
= A sin ( Ω t − δ ) + Be − γ t 2 cos (ω t + 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, xtr, in Equation 28
becomes negligible for t >> 2π/γ, so that when
t >> 2π/γ
displacement x(t)
4
0
−2
−4
x(t) ≈ xss (t) = A cos (ω t − δ )
0
10
20
30
40
50
time t
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.
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2
Figure 64Displacement–time graphs for the same
oscillator with three different sets of initial conditions
(different choices of B and φ).
S570 V1.1
Question T12
The constants B and φ that appear in Equation 28
x(t) = xss (t) + x tr (t) = A sin ( 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❏
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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.
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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.
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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,
ω − ω1
γ
ω
f
resonance absorption bandwidth
∆f = 2
=
= 0 = 0
(Eqn 26)
2π
2π 2πQ
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 24 × 36001s, or about 10−4. 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 f0/∆f = 104.
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Our next example is a quartz crystal oscillator, such as is used in a watch. This is another mechanical
oscillation, 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 vibrate
at the driver frequency, rather like a loudspeaker, but piezoelectric crystals can respond at much higher driver
frequencies than is possible for a conventional loudspeaker. In the quartz crystal oscillator the mechanical
oscillations 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 circuit
much sharper. The oscillations are then counted and this is converted into a time display. Such watches are
capable of an order of magnitude better time-keeping than the pendulum clock and so the required Q-value is
about 105 . High quality quartz-controlled clocks can do rather better than this but our next example shows a
spectacular improvement on this.
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The caesium atomic clock is so precise that its oscillations are used to establish and maintain our fundamental
unit 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 the
place to go into details of this process but suffice to say that caesium atoms can absorb electromagnetic radiation
at a frequency of around 91921MHz, in a resonant process. ☞ For our purpose here the important point is that
the sharpness of the atomic resonance can be made so great that the full resonance absorption bandwidth is as
low 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 the
frequency of such clocks can then be made reproducible to about 1 part in 1013. Two such clocks will keep the
same time to within about 3 seconds in a million years! Unfortunately, these clocks are of a size which would fill
a medium-sized room and you would not want one strapped around your wrist. Their use is confined to
establishing the international time standard and in calibrating other more portable clocks. Such high precision is
needed in certain scientific experiments. For example, caesium clocks flown in aircraft have enabled the direct
observation of the tiny time changes for moving clocks, predicted by Einstein’s special theory of relativity.
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Even the caesium clock could be made to look rather crude when compared to future clocks that may be based
on frequency-stabilized lasers. In a laser the oscillators are effectively an ensemble of atoms in a cavity, rather
than individual atoms, and the driver is an electromagnetic wave (light), trapped within the same cavity, between
mirrors. The atoms absorb and re-emit the wave within the cavity but because they are all in communication
with the same wave they all absorb and emit with well-defined phases, rather than with random phases as would
be the case if they were acting individually. Such phase-organized light is called coherent light. The process by
which an incoming wave drives an atom to emit is known as stimulated emission and it is the counterpart to
absorption. If the stimulated emission exceeds the absorption then this will give rise to laser action, or Light
Amplification by Stimulated Emission of Radiation. Within the cavity of a laser oscillator there is a single giant
amplitude oscillation with a well-defined phase and an extremely well-defined frequency. The frequency output
bandwidth for the laser is very much narrower than would be the case if the same atoms were placed outside the
cavity, as in an ordinary lamp.
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Lasers have many advantages over conventional light sources but the one that concerns us here is this very
narrow bandwidth. It gives rise to an enormous value for the Q-factor of the oscillator. The full resonance
bandwidth of a laser oscillator can be made to be only a few Hz, even when the laser is operating in the visible
spectrum, at frequencies of around 5 × 1014 Hz. If such a laser could also have its frequency stabilized
sufficiently to capitalize on this low bandwidth then the resulting frequency-stabilized laser could have a
Q-factor of around 1014, with a potential improvement factor of 104 over the caesium clock. The technology for
this is under development and stabilized lasers could well be the reference clocks of the future.
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3 Coupled oscillators
In 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 now
take a look at what happens when the driving force is replaced by another oscillator.
3.1 Normal modes
As a specific example, consider two identical oscillators, consisting of
equal masses A and B, resting on a smooth horizontal surface and
connected by identical springs to two rigid walls, as shown in Figure 7.
The two masses are also connected together by another spring, which we
will suppose to be much weaker than the two main ones. Initially the two
masses are at rest at their equilibrium positions.
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A
B
x1
x2
Figure 74A model of two coupled
oscillators.
Now let us move A from its equilibrium position, while holding B fixed,
A
B
and then let go of both. A will immediately start to oscillate, but because
of the coupling between the two masses, it will excite an oscillation of
x1
x2
increasing amplitude in B. There is only a limited amount of energy to
supply these oscillations, so as the amplitude of B increases that of A
must decrease. Surprisingly, this exchange of energy does not stop when Figure 74A model of two coupled
the two amplitudes are equal, but continues until A is stationary and B is oscillators.
moving with the full amplitude that A had at the start. The process is then reversed, with the B amplitude
decreasing and that of A increasing until the starting configuration is reached and the whole process is repeated
ad infinitum, or until damping reduces the vibrational energy to zero.
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Figure 8 shows how the two displacements vary with time.
x1
✦ Are these the sort of time variations we should expect for two
harmonic oscillators?
A
t
x2
B
Figure 84The transfer of energy between
two coupled oscillators.
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t
The overall motion of the two coupled oscillators looks quite
complicated and neither oscillator moves with simple harmonic
motion. However, the motion is periodic and this encourages us to
ask whether there is any concealed simple harmonic motion in this
system. As you will soon see, the answer is yes, and these motions
are remarkably simple in this case1—1once we expose them.
x1
A
x2
B
Figure 84The transfer of energy between
two coupled oscillators.
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t
t
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, with
one mass first oscillating vigorously then becoming quiescent, while the
other does the same but with opposite periods of activity, as the energy is
switched back and forth. However, there are two particular patterns of
relative motion in which the motion continues unchanged, with neither
mass changing its energy with time.
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A
B
x1
x2
Figure 74A model of two coupled
oscillators.
In one of these motions
the two identical masses
move together, in phase,
with equal amplitude, as
shown in Figure 9a.
The distance between the
two
will
remain
unchanged, so the
coupling spring will have
no effect on the motion,
and the frequency of
oscillation will be just
the same as if the two
were uncoupled. In the
other motion the two
masses move in opposite
directions with equal
amplitude, as shown in
Figure 9b.
x1
A
A
B
x1
x2
t
x2
B
(a)
t
x1
A
A
B
x1
−x 2
x2
B
(b)
Figure 94(a) Masses moving in phase; (b) masses moving out of phase.
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t
t
✦ Would you expect this second oscillation to have the same frequency as the first? If not, will the frequency
be higher or lower?
These two forms of steady state oscillation are simple harmonic motions, with constant amplitudes, and are
known as the two normal modes of this system. Such normal modes always appear whenever two or more
oscillators are coupled together. Any motion of the system can be analysed in terms of an appropriate
superposition of these normal modes of the system, as we will see shortly, so the normal modes can be
considered as the basic SHM building blocks of any oscillation of the system.
These two normal modes may then be identified through the mass displacements x 1 (t) and x2(t) as:
for
ω = ω1 = ω00
for
ω = ω02
x1(t) = x 0 1cos1(ω00 t) and x2(t) = x 0 1cos1(ω00 t)
x1(t) = x 0 1cos1(ω02 t) and x2(t) = −x01cos1(ω02 t)
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(29a)
(29b)
We will now show that the complicated motions depicted in Figure 8
x1
A
t
x2
B
Figure 84The transfer of energy between
two coupled oscillators.
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t
and the simple normal
modes in Figure 9 are
related. To do this we
need to invoke the idea
of the superposition of
two simple harmonic
motions.
x1
A
A
B
x1
x2
t
x2
B
(a)
t
x1
A
A
B
x1
−x 2
x2
B
(b)
Figure 94(a) Masses moving in phase; (b) masses moving out of phase.
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t
t
The principle of superposition states that:
When several oscillations are added, the resultant displacement at any time is the sum of the displacements
due 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 two
normal modes of Equation 29.
ω = ω1 = ω00
ω = ω02
x1(t) = x 0 1cos1(ω00 t) and x2(t) = x 0 1cos1(ω00 t)
x1(t) = x 0 1cos1(ω02 t) and x2(t) = −x01cos1(ω02 t)
We obtain:
x1 (t) = 12 [x 0 cos (ω 0 t) + x 0 cos (ω 2 t)]
x 2 (t) = 12 [x 0 cos (ω 0 t) − x 0 cos (ω 2 t)]
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(Eqn 29a)
(Eqn 29b)
Using an identity from Question R4 we can write these two
x1 (t) = 12 [x 0 cos (ω 0 t) + x 0 cos (ω 2 t)]
x 2 (t) = 12 [x 0 cos (ω 0 t) − x 0 cos (ω 2 t)]
as
x1 (t) = x 0 cos

ω2 − ω0 
ω + ω0 
t cos 2
t



2
2
(30a)
x 2 (t) = x 0 sin 

ω2 − ω0   ω2 + ω0 
t sin
t
 

2
2
(30b)
Question T13
Verify Equations 30a and 30b.4❏
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☞
Since ω 0 and ω2 are different, but usually not greatly different, each
of the right-hand sides in these equations
ω − ω0 
ω + ω0 
x1 (t) = x 0 cos 2
t cos 2
t
(Eqn 30a)




2
2
x 2 (t) = x 0 sin 

ω2 − ω0   ω2 + ω0 
t sin
t
 

2
2
(Eqn 30b)
x1
A
x2
B
is a product of a slowly oscillating term, involving (ω02 − ω 0 ), and a
rapidly oscillating one involving (ω02 + ω0).
t
t
Figure 84The transfer of energy between
If we refer back to Figure 8 we can see that this slowly oscillating two coupled oscillators.
term can be interpreted as describing the overall slowly changing
amplitude 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 is
moving again with maximum amplitude and with B stationary?4❏
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Normal modes appear in any situation where we have oscillations coupled together, and when there is a large
number of oscillators1—1an extreme case is the atoms in a crystal, the forms of oscillation and their frequencies
can 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 harmonic
motion.
The frequencies of the normal modes are almost always different from the frequencies of the uncoupled
oscillators.
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4 Closing items
4.1 Module summary
1
2
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; such
oscillators are called forced or driven oscillators. Under appropriate conditions, a periodic driving force
with a period close to the natural period of an oscillator can cause a driven oscillator to develop oscillations
of relatively large amplitude; this condition is known as resonance.
The equation of motion of a harmonically driven, linearly damped oscillator is a (inhomogeneous)
differential equation of the form:
dx
d2x
+γ
+ ω 02 x = a sin ( Ω t)
(Eqn 4)
2
dt
dt
If the driving force is absent (a = 0) the remaining (homogeneous) differential equation describes a
linearly damped oscillator:
dx
d2x
+γ
+ ω 02 x = 0
(Eqn 5)
dt 2
dt
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3
4
and if γ = 0, we obtain the equation of motion of a simple harmonic oscillator
d2x
+ ω 02 x = 0
(Eqn 6)
dt 2
The full solution of the driven oscillator equation can be regarded as the sum of two parts; one describing
the transient motion which effectively decays to zero after a time of order 2π/γ, and one describing the
persisting steady state motion that is sustained by the driving force.
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 = ma1sin1(Ω1t)) and the
oscillator’s displacement. The steady state motion is therefore independent of the starting conditions and
can be written as
x(t) = A1sin1(Ω1t − δ)
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(Eqn 9)
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5
In the steady state, the amplitude A and the phase lag δ both depend on the driving frequency Ω and are
given by:
a
A=
(Eqn 15)
2
2
(ω 0 − Ω )2 + ( γ Ω )2
and
 γΩ 
δ = arctan  2

 ω0 − Ω 2 
(Eqn 16)
The amplitude is almost independent of Ω at low frequencies, attains a maximum value at resonance, which
defines the resonance frequency, and then decreases at high frequency. The phase lag is small at low
frequency, 90° at the undamped natural frequency, which is near resonance and approaches 180° at high
frequency. In particular;
in the low frequency limit:
 a 
x(t) =  2  sin ( Ω t)
 ω0 
(Eqn 10)
in the high frequency limit:
a
x(t) =  2  sin ( Ω t − π)
Ω 
(Eqn 11)
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and in the resonance limit (Ω = ω0 for the case of light damping):
 a  
π
x(t) = 
 sin  Ω t − 
2
 γω 0 
6
(Eqn 12)
In the latter case the resonance amplitude exceeds the low frequency amplitude by the oscillator’s
quality factor Q = ω0/γ.
In the steady state, the total energy, kinetic energy and potential energy are all functions of time, and are
given by
1
total energy
E = mA 2 [ Ω 2 cos 2 ( Ω t − δ ) + ω 02 sin 2 ( Ω t − δ )] (Eqn 18)
2
1
kinetic energy
Ekin = mA 2 Ω 2 cos 2 ( Ω t − δ )
(Eqn 19)
2
1
potential energy Epot = mA 2 ω 02 sin 2 ( Ω t − δ )
(Eqn 20)
2
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7
However, the average value of each of these quantities, taken over a full period of oscillation, has a constant
value, as given by
1
1
1
〈E〉 = 〈Ekin 〉 + 〈Epot 〉 = mA 2 Ω 2 + mA 2 ω 02 = mA 2 ( Ω 2 + ω 02 )
(Eqn 21)
4
4
4
The average total energy is mostly potential energy below resonance, and mostly kinetic energy above
resonance. At resonance, in the case of light damping (when Ω = ω0 ), the total energy is independent of
time, potential and kinetic energy contribute equally to the total, and
1
E = 〈E〉 = mA 2 ω 02
2
In the steady state, the average power 〈1P1〉 transferred to the oscillator from the driving force, is exactly
equal to the average power dissipated by the damping force.
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8
The average power absorption, 〈1P1〉, exhibits resonance behaviour, i.e. it rises to a maximum when the
driving frequency is close to the undamped natural frequency. The height of the maximum is proportional to
Q, and its width is proportional to Q −1. Thus, more lightly damped oscillators (characterized by larger
values of Q), have taller and narrower resonance peaks. The average power transfer at resonance is
ma 2 ma 2 Q
〈Pres 〉 =
=
(Eqn 24)
2γ
2ω 0
and the resonance absorption bandwidth is:
γ
ω
∆ω ω 2 − ω 1
f
∆f =
=
=
= 0 = 0
2π
2π
2π 2πQ
Q
(Eqn 26)
9
One important application of resonance absorption is to establish oscillators whose frequencies are very
well-defined; these can constitute frequency and time standards.
10 The transient motion for the driven oscillator can be represented by:
x tr (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
γ2
ω = ω 02 −
.
4
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Forced vibrations and resonance
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S570 V1.1
11 The general solution to the equation of motion for the harmonically driven, linearly damped harmonic
oscillator may be written
x(t) = xss (t) + x tr (t) = A sin ( 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 between
them.
13 The simple harmonic motions of a pair of identical coupled oscillators are the normal modes, consisting of
in-phase and out-of-phase motions of the two. These can be represented as:
for ω = ω1 = ω00
x1(t) = x 0 1cos1(ω00 t) and x2(t) = x 0 1cos1(ω00 t)
(Eqn 29a)
for ω = ω02
x1(t) = x 0 1cos1(ω02 t) and x2(t) = −x01cos1(ω02 t)
(Eqn 29b)
When these two normal modes are superposed the resulting oscillation has the average frequency of the two
modes and the superposed amplitude shows beats at the difference frequency:
ω − ω0 
ω + ω0 
x1 (t) = x 0 cos 2
t cos 2
t
(Eqn 30a)




2
2
x 2 (t) = x 0 sin 

ω2 − ω0   ω2 + ω0 
t sin
t
 

2
2
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Forced vibrations and resonance
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(Eqn 30b)
S570 V1.1
14 Any set of oscillators coupled together will result in normal modes which carry out simple harmonic
motion. The frequencies of the normal modes are almost always different from the frequencies of the
uncoupled oscillators.
FLAP P5.3
Forced vibrations and resonance
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THE OPEN UNIVERSITY
S570 V1.1
4.2 Achievements
Having 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 motion
and 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 amplitude
and 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 of
kinetic energy increases from zero at low frequency to 50% at resonance and approaches 100% at high
frequency.
A7 Show that the average power absorbed from the driving force is equal to the average power dissipated by
damping.
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Forced vibrations and resonance
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A8 Discuss the resonance phenomenon in terms of the graph of average power absorbed against driving
frequency, showing (in the case of light damping) that it possesses a peak of height proportional to Q and
width 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 explain
how 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 the
topics.
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Forced vibrations and resonance
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S570 V1.1
4.3 Exit test
Study comment
Having completed this module, you should be able to answer the following questions each of which tests
one 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 a
harmonically varying force of amplitude 41N and frequency 5501rpm. Calculate the amplitude and phase lag of
the resulting steady state vibration.
Question E2
(A6)4For the oscillator in Question E1 calculate the average potential energy, the average kinetic energy and
the average power absorbed.
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Forced vibrations and resonance
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Question E3
(A2, A5, A8 and A7)4Under what practical circumstances does one try to minimize the amplitude of a forced
vibration? 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 different
frequencies so the amplitude is independent of the driver frequency. What principles should one bear in mind
when 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 forced
vibration? How can one achieve this?
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Forced vibrations and resonance
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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 and
try 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 it
here.
FLAP P5.3
Forced vibrations and resonance
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THE OPEN UNIVERSITY
S570 V1.1