enrgy, damping and resonance

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 P5.2 Energy, damping and resonance in harmonc motionCOPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1

Module P5.2 Energy, damping and resonance in harmonic motion1 Opening items

1.1 Module introduction

1.2 Fast track questions

1.3 Ready to study?

2 Energy in oscillating systems

2.1 Gravitational potential energy for the simple pendulum

2.2 Strain potential energy in a stretched or compressed spring

2.3 Total potential energy for a mass suspended on a spring

2.4 Energy oscillations in SHM

3 Damped and driven harmonic oscillators

3.1 The mechanisms of damping: friction

3.2 Frictional forces as dissipative forces in mechanical SHM

3.3 Lightly damped harmonic motion

3.4 Qualitative discussion of general damping

3.5 Qualitative discussion of forced vibrationsresonance

4 Closing items

4.1 Module summary

4.2 Achievements

4.3 Exit test

Exit module


1 Opening items

1.1 Module introductionWhen an object vibrates there are always some points in its oscillatory motion where it is momentarily at restand other points at which it is moving with maximum speed. Consequently, the kinetic energy of the system issometimes zero and sometimes a maximum; if the motion is periodic then so too is the kinetic energy. However,if the oscillator is isolated from its surroundings, so that the principle of energy conservation applies, its totalenergy will be constant. Such an oscillator must therefore have potential energy and this too must be periodic inorder that the sum of kinetic and potential energy can be constant; the potential energy is maximum when thekinetic energy is minimum and vice versa. Section 2 deals with the energy in vibrating mechanical systems,particularly systems in one-dimensional simple harmonic motion (SHM); among other things it provides amathematical expression for the kinetic and potential energy in an isolated simple harmonic oscillator.

It is a matter of common experience that the amplitude of any mechanical vibration tends to decrease with time,until the motion eventually stops. The total energy of the vibrating system is clearly not constant in thissituation. However, the principle of energy conservation still applies on a larger scale, so the vibrating systemmust be transferring energy elsewhere, and cannot be treated as an isolated system. When a vibrating systemloses mechanical energy in this way it is said to be damped. Any vibrating system will have some damping,although it may be very small.


The damping of vibrations is of great technological importance and many engineers spend time designingsystems to have particular levels of damping1—1either low or high or optimal. For example, in a mechanicalclock the oscillations of the balance wheel or pendulum should be as lightly damped as possible, to minimize theenergy input needed to sustain the oscillations. In contrast, a car suspension system uses heavily damped shock-absorbers to prevent the car and its passengers being driven into excessive vertical oscillations or being jolted todestruction by bumps in the road surface; the main function of the suspension system is to absorb and dissipate(as heat) the energy from unwanted vibrations. Electrical meters and weighing balances are other exampleswhere the correct level of damping is important1—1too much damping and the instrument will be too slow torespond to a changing reading, too little damping and the instrument will oscillate about the true reading.Section 3 describes damped harmonic motion. It provides a mathematical treatment of lightly damped harmonicmotion, in which the amplitude of the oscillation decays gradually, and a qualitative discussion of more heavilydamped oscillations. The module closes with a brief qualitative introduction to driven oscillations andresonance.


Although the mathematical expressions introduced in this module are developed in the context of mechanicaloscillations, the results are equally applicable to many other situations, including the oscillations of electriccharge in circuits1—1some of these applications are covered elsewhere in FLAP.

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 4.1) and the Achievements listed inSubsection 4.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 4.3. 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

An object of mass 0.201kg is suspended from the end of a spring with a spring constant of 1251N1m−1 and is setinto simple harmonic motion (SHM). Calculate the angular frequency ω of the resulting oscillation, in theabsence of damping. The system is then damped and the magnitude of the damping force is linearly proportionalto the speed, with a damping constant of 1.41s−1. Explain what this statement means. Calculate the Q-factor forthis oscillator.


Question F2

The object described in Question F1 is displaced by 0.11m from its equilibrium position and then released.Sketch graphs showing how (i) the amplitude, (ii) the kinetic energy, (iii) the potential energy, and (iv) the totalenergy vary with time for (a) the undamped situation, and (b) the damped situation as given in Question F1.

Study comment

Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module andto 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, component (of a vector),displacement, energy transfer (or work done) by a force, equilibrium, kinetic energy, mass, momentum, Newton’s laws ofmotion, potential energy, principle of energy conservation, speed, tension, velocity and weight. In addition you should befamiliar with the general description of one-dimensional simple harmonic motion (SHM) and the concepts of amplitude,angular frequency, cycle, force constant, frequency, pendulum, period, phase, phase constant, restoring force and springconstant. The module also assumes you are familiar with the expressions for displacement, velocity and restoring force inone-dimensional SHM. These expressions are developed elsewhere in FLAP and, if necessary, you can review them throughthe Glossary entry for simple harmonic motion. The module also requires familiarity with the following mathematicalconcepts: exponential function, gradient, linear function, modulus, natural logarithm, parabola, quadratic function and

trigonometric function. The trigonometric identities sin2θ = 12 [1 − cos1(2θ)] and sin2θ + cos2θ = 1 are used, together with the

approximation cos1θ = 1 − θ12/2 for small θ. You do not need to be fully conversant with differentiation in order to study thismodule, but you should be familiar with the calculus notation dx/dt used to represent the rate of change of x with respect to t.If you are unsure about any of these items you should refer to the Glossary, which will also indicate where in FLAP they aredeveloped. The following Ready to study questions will allow you to establish whether you need to review some of the topicsbefore embarking on this module.


Question R1

Write down the values of the phase constant φ that make the function

x(t) = A1cos1(ω00t + φ)

equivalent to x(t) = A1sin1(ω00t)

Question R2

Sketch the graph of the function

x(t) = A1cos1(ω00t + φ)

over the time interval from t = 0 to t = 4π/ω (i.e. t = 2T) for the situations where (a) φ = 0, (b) φ = π/2, and(c) φ = −π/2.


Question R3

Using the function x(t) given in Question R2

x(t) = A1cos1(ω00t + φ)

for the case φ = 0, sketch dx(t)/dt, its derivative with respect to t, as a function of t.

(dx(t)/dt denotes the rate of change of x with respect to t.)

Question R4

A body of mass 0.401kg is suspended from the end of a spring of spring constant 1601N1m−1.Calculate the angular frequency, the frequency and the period of small oscillations of the system aboutequilibrium. If the body is pulled down to a position 0.051m below its equilibrium position, and released fromrest at time t = 0, write down an expression which gives the displacement in the subsequent oscillation as afunction of time.


2 Energy in oscillating systemsIn a mechanical oscillation the moving mass m has kinetic energy by virtue of its motion. In the case where theobject is in one-dimensional motion along the x-axis, with velocity of magnitude vx, the kinetic energy Ekin is:

Ekin = 12 mvx

2 (1)

In addition to this, the mass has potential energy by virtue of its position. Potential energy might arise from avariety of sources. For example, the mass may change its height above the Earth’s surface (as for a swingingpendulum bob), in which case there will be corresponding changes in the gravitational potential energy.Alternatively, the mass may be attached to a spring or to some other elastic body and there will then be changesin the stored strain potential energy during the motion. In this latter context we use the term elastic to describea body which will deform when subjected to appropriate forces but which fully recovers its original form whenthe forces are removed; the deformation does not produce any permanent distortion. The fact that energy can bestored in a compressed or stretched elastic body is well known to any archer or child with a catapult!


In physics, it is generally true that the calculation of potential energy poses far greater difficulties than thecalculation of kinetic energy does . Kinetic energy is simply determined from the mass and speed whereaspotential energy calculations require some knowledge of the forces acting 1—1and these are often not simple.In the case of strain potential energy in particular, the internal forces involved in a deformation may be verycomplicated and the energy difficult to calculate. A second but less serious problem is that while kinetic energyhas an obvious zero, when the particle is stationary, potential energy often has no obvious zero. For example,when dealing with gravitational potential energy we might choose the position of zero potential energy to be atthe Earth’s surface, or at the height of our laboratory floor or maybe even at the centre of the Earth, or totallyremote from any gravitating body such as the Earth. This problem may be simplified because it is only changesin potential energy that have physical significance, so the zero point can be chosen arbitrarily. When dealingwith strain potential energy there is an obvious zero point1— 1at the position where the elastic body is notdeformed. Nonetheless, the choice of zero point remains arbitrary and we may always choose some other zeropoint in order to simplify the problem in hand.


If the position of zero potential energy of an oscillator is chosen in such a way that it is also the position at whichthe oscillator has its maximum kinetic energy, then the total energy itself will ‘oscillate’ between being totallykinetic and totally potential. It is the purpose of this section to provide mathematical expressions for the totalenergy of a mechanical oscillator that show how this energy oscillation occurs. As a first step towards this goalwe will investigate the behaviour of the potential energy in three particular oscillators; a simple pendulum, amass attached to a spring and sliding on a table, and a mass hanging from a spring. We will then combine thegeneral expression for potential energy that emerges from these investigations with the general expression forkinetic energy to find a general expression for the total energy of any one-dimensional simple harmonicoscillator.


2.1 Gravitational potential energy for the simple pendulumWe can approach the problem of potential energy in vibrational motion by considering a simple example inwhich the potential energy involves only gravitational energy1— 1the swinging of a simple pendulum. As thependulum bob rises and falls during its oscillation, its gravitational potential energy changes in proportion to itsheight above a chosen reference level, such as the surface of the Earth. If the height of the bob (mass m) abovethe surface is h and the magnitude of the acceleration due to gravity is g then the gravitational potential energyEpot is:

Epot = mgh (2)

This expression may be interpreted as a special case of the formula for the energy transferred (or work done) bya force Fx when it acts through a displacement sx:

∆Epot = −Fx0sx (3)

In our case Fx is the gravitational force downwards of magnitude mg and sx is h, measured upwards as positive.If we take the point where h = 0 to be a point of zero gravitational potential energy, it follows that the energytransferred when Fx acts over a displacement sx is equal to mgh the gravitational potential energy at height h.


s

L

h

θ

L cos θ

Figure 13The simple pendulum.

We may apply Equation 2

Epot = mgh (Eqn 2)

to the simple pendulum, as shown in Figure 1. When the bob is displaced adistance s along the arc from the equilibrium position the additional height his given by

h = L − L cos θ = L(1 − cos θ )

If we take the equilibrium position of the bob (h = 0) as the point of zeropotential energy

Epot = mgh = mgL(1 − cos θ )

During the oscillation the angle θ varies and with it the potential energy.We could also express θ in terms of the displacement along the arc s bynoting that in radians θ = s/L and by using the approximation for small θ thatcos1θ ≈ 1 − θ12/2 thus:


The stored gravitational potential energy, when a pendulum is displaced by a small amount θ = s/L from itsequilibrium, is given by

Epot = mgL 1 − 1 − θ 2

2

= mgL

s2

2 L2

= 12

mg

L

s2 (4)

where L is the length of the pendulum.

Notice that this motion can still be described as one-dimensional motion along the arc, even though the motionis not along a straight line. Restoring forces and stored energy can be written in terms of a single variable s, thedisplacement along the arc from equilibrium.


s0−s0 s

potentialenergy

1–2

Epot = mgL

s2

Figure 23See Question T1.

Question T1

Figure 2 shows the potentialenergy of the simple pendulum as afunct ion of the (small)displacement along the arc, asgiven by Equation 4.

Epot = mgL 1 − 1 − θ 2

2

= mgLs2

2L2

= 12

mg

L

s2

(Eqn 4)

The potential energy is taken as zero at the equilibrium position. What is the shape of the graph?3


x

Fx

(a)

x

Fx

frictionless surface

Figure 3a3Motion of a mass attachedto a spring and oscillating along ahorizontal frictionless surface.

2.2 Strain potential energy in a stretched or compressedspringPotential energy may also appear as stored strain potential energy in astretched or compressed spring. An illustration of an oscillator based onthis idea is shown in Figure 3a. Here a mass is attached to a spring andrests on a horizontal frictionless surface. If the mass is initially displacedby stretching the spring and is then released, it will oscillate — producingalternately, compressions and extensions in the spring.

We can find the stored strain potential energy by using Equation 3

∆Epot = −Fx0sx (Eqn 3)

to calculate the energy transferred by the tension force Fx in compressingor extending the spring from its unstretched length. The calculation is alittle more complicated than for the previous example because themagnitude of the tension force increases as the extension increases, andso it is not constant throughout the energy transfer. Extensions andcompressions take place along a single axis, the x-axis, so the problem isone-dimensional with the x-components of displacement and force givenby x and Fx, respectively.


Provided the spring is not stretched too far, this tension force is linearly proportional to x, but has the oppositesign since it is directed towards the origin.

Thus, Fx = −ksx (5)

Equation 5 is a statement of Hooke’s law.

Suppose the spring is stretched to some maximum extension x = xmax. Equation 3


then gives the potential energy of the system when the mass m is at the position x = xmax as

(Epot0)max = −0⟨ 1Fx1⟩xmax = ⟨ 1−Fx1⟩xmax (6)

where we have replaced the varying tension force Fx over the range from x = 0 to x = xmax by its average value⟨ 1Fx1⟩ over this range. Since Fx varies linearly with x the average force ⟨ 1Fx1⟩ is just half the final force(Fx)max = −ksxmax.

Equation 6 then gives

(Epot0)max = 12 ksxmax

2 (7)


ksxmax

(b)

x

ksxmax

−Fx

= area 1–2

xmax

Fx = ksx

2

Figure 3b3Motion of a mass attached to a springand oscillating along a horizontal frictionlesssurface; minus the tension force, plotted againstdisplacement for this oscillator.

Aside If you are familiar with integral calculus you will perhapsrecognize Equation 7


2 (Eqn 7)

as the integral of Equation 3


over the extension x, i.e. Epot = − Fx0

x

∫ dx = ks x0

x

∫ dx = 12 ks x2

If you are unfamiliar with this approach it is not essential here − itamounts to finding the area under the graph in Figure 3b; byintegration we have done this here directly from the graph. Details ofintegration can be found in the maths strand of FLAP.

A related point is that there is a general relationship between theconservative force acting on an object and the way its potential energy

changes with position. The general relationship is: Fx = −dEpot

dx. If

we differentiate the general form of Epot ( i.e. 12 ks x2 ) with respect to x

we obtain the tension Fx = −ksx as in Equation 5.


ksxmax

(b)

x

ksxmax

−Fx

= area 1–2

xmax

Fx = ksx

2

Figure 3b3Motion of a mass attached to a springand oscillating along a horizontal frictionlesssurface; minus the tension force, plotted againstdisplacement for this oscillator.

It can be seen that the quantity in Equation 7


2 (Eqn 7)

is equal to the area under the graph of Figure 3b between x = 0and x = xmax. It is straightforward to generalize Equation 7 forany extension x as:


The stored strain potential energy when a spring is stretched (x > 0) or compressed (x < 0) by an amount xfrom its original length is given by

Epot = 12 ksx2 (8)

where ks is the spring constant.

Notice that since x appears as a squared quantity in Equation 8 the magnitude but not the sign of x is important;the energy stored is the same for an extension or a compression of the same magnitude.


x

potentialenergy

−x0

(c)

1–2

Epot =

x0

ksx2

Figure 3c3 Motion of a mass attached to a spring andoscillating along a horizontal frictionless surface; graphshowing the potential energy of this oscillator as a functionof its displacement from the equilibrium position (Equation7).

The potential energy of the horizontal spring oscillatoras a function of the displacement, as given by Equation8,

Epot = 12 ksx2 (Eqn 8)

is shown in Figure 3c. The similarity between Figure3c for the spring oscillator and Figure 2 for thependulum is obvious1—1both relationships describeparabolic curves and for the pendulum the quantitymg/L plays the same role as the spring constant ks. Thissimilarity in the mathematical expressions for potentialenergy of the pendulum and the spring oscillator willbe useful at the end of this section where we set up amathematical model for the general energy oscillationsin any SHM.


Question T2

A spring-powered gun fires a pellet of mass 0.11g using a spring of spring constant 5 × 1031N1m−1.

If the maximum compression of the spring is 0.11m and 30% of the strain potential energy is given to the pellet,calculate its speed on leaving the gun.3


(a) mg

reaction

(b) mg

xeq

(c) mg

Fx1=1−ksxeq Fx1=1−ks(xeq1+1x)

xeq1+1x

Figure 43A light coil spring shown (a) unloaded (with a supported mass held at theunstretched spring position), (b) at the position of equilibrium under the load, and(c) loaded and displaced below its position of equilibrium. Note that x is taken to be positive in the downward direction.

2.3 Total potential energyfor a mass suspended on aspringA vertical spring oscillator, with amass hanging from a light spring , is shown in Figure 4.Figure 4a shows the unstretchedspring, with the mass attached butsupported externally, rather thanby the spring; Figure 4b showsthe mass hanging freely at rest atits equilibrium position xeq andFigure 4c shows the massdisplaced to some position xeq + xbelow its equilibrium position, asit might be at some point duringan oscillation of the spring.


Calculating the potential energy of this oscillator requires some care, since the gravitational energy of the massmust be added to the strain energy of the spring. Moreover, we have to choose zero points for both the strainenergy and the gravitational energy, and they need not be the same. Should we choose the zero point of thestrain energy so that it corresponds to the unstretched length or to the equilibrium position? What position of themass should correspond to the zero point of gravitational energy? From what was said at the start of Section 2we can choose these zero points arbitrarily, but we would be wise to choose them in such a way that they makethe calculations as simple as possible. Let us investigate some possible choices to see what is best.


(a) mg

reaction

(b) mg

xeq

(c) mg


xeq1+1x


If we take the unstretched spring(Figure 4a) as having zero strainpotential energy, as we did inSubsection 2.2, the strainpotential energy at theequilibrium position (Figure 4b)is ksxeq

2/2 and at displacement xbelow the equilibrium position(Figure 4c) it is k s(xeq + x )2/2.If we take the zero ofgravitational potential energywith the mass at the equilibriumposition (Figure 4b) then atdisplacement x below theequilibrium position (Figure 4c) itis −mgx. When the mass is at theequilibrium position, the net forceon it is zero and its weight, mg downwards, is balanced by the spring tension, ksxeq, upwards.

So, mg = ksxeq


We may therefore write the gravitational energy at displacement x as −mgx = −ks 0xeq0x.

Table 1 summarizes the potential energy situation as shown in Figure 4, taking these potential energy zeros.

Table 13Potential energies for Figure 4 with the gravitational potential energy zero atthe equilibrium position and the strain potential energy zero for the unstretched spring.

Position Gravitationalpotential energy

Egrav

Strainpotential energy

Estrain

Totalpotential energy

Epot

As in Figure 4a mgxeq = ks(xeq)2 0 ks(xeq)2

As in Figure 4b 0 ks(xeq)2/2 ks(xeq)2/2

As in Figure 4c −mgx = −ksxeqx ks(xeq + x)2/2 ks(xeq)2/2 + ksx2/2


(a) mg

reaction

(b) mg

xeq

(c) mg


xeq1+1x


Question T3

For the system shown in Figure 4,construct a table similar toTable 1 but with both the strainpotential energy zero and thegravitational potential energy zerofor the unstretched spring, asshown in Figure 4a.3

In the next question you will needto think carefully about how todeal with a potential energy zerodefined where the spring isalready stretched. You mustremember that it is the r e a lextension from the unstretchedlength which controls the storedenergy.


(a) mg

reaction

(b) mg

xeq

(c) mg


xeq1+1x

Figure 43A light coil spring shown (a) unloaded (with a supported mass held at theunstretched spring position), (b) at the position of equilibrium under the load, and(c) loaded and displaced below its position of equilibrium. Note that x is taken to bepositive in the downward direction.

Question T4

For the system shown in Figure 4,construct a table similar toTable 1 but with the gravitationalpotential energy zero for theunstretched spring and the strainpotential energy zero at theequilibrium position.3

There are two lessons to belearned from Table 1 andAnswers T3 and T4. First, it isclear that although any choice forthe zero point of the strain energyis equally valid, the most naturalchoice is that in which the springis neither stretched norcompressed.


The reason for this is that the strain potential energy is a quadratic function of displacement, so the graph ofEstrain against x is a parabola with a unique minimum at x = −xeq.

It makes good sense to let this minimum value of Estrain correspond to the zero point of the strain energy to avoidassigning a negative value to the strain energy of the unstretched spring.

The second lesson from Table 1 and Answers T3 and T4 is that when the mass has been displaced by an amountx from the equilibrium position the total potential energy is always a quadratic function of the form

Epot = 12 ksx2 + constant

where the value of the constant depends on the choice of zero points for the gravitational and strain energies.Now you may find this surprising since the gravitational potential is a linear function of x, yet there is no termproportional to x in the expression for Epot. It is worth noting how this comes about.


If we take the unstretched spring as having zero strain energy, then

Estrain = 12 ks(x + xeq)2

while the gravitational potential energy is given by

Egrav = −mgx + E0

where E0 is a constant determined by the location of the zero point of the gravitational potential energy. Hence

Epot = Estrain + Egrav = 12 ks(x + xeq)2 − mgx + E0

i.e. Epot = 12 ksx2 + E0 + 1

2 mgxeq

Note that the linear terms, proportional to x alone, have cancelled. Also note that Epot becomes especially simpleif E0 = − 1

2 mgxeq. In other words the expression for the total potential energy becomes as simple as possible ifwe require that

Egrav = −mgx − 12 mgxeq

So Egrav = 0 corresponds to x = −xeq/2.


If the zero point of the strain energy corresponds to the unstretched spring, and the zero point ofgravitational energy corresponds to x = −xeq/2, what value of x corresponds to the zero point of Epot = Estrain +Egrav?

The stored potential energy when a mass suspended from a spring is displaced by a small amount x from itsequilibrium position is given by

Epot = 12 ksx2 (Eqn 8)

where ks is the spring constant and the equilibrium position (x = 0) is the position of zero point of potentialenergy.

Question T5

The introduction to Section 2 claimed that we were free to choose any position of zero of potential energy sinceit was only the differences in potential energy which were significant. Look at the data in Tables 1, 3 and 4,which encompass two different choices of the zero for each of the two types of potential energy. What supportfor this claim can you find in these tables?3


Study comment This subsection draws on the results of the general mathematical description of one-dimensional SHM1—1in particular, the expressions for the displacement, velocity and restoring forces. These equations are developed elsewhere inFLAP and if you are unfamiliar with the results you should review them now through the Glossary entries under simpleharmonic motion.

2.4 Energy oscillations in SHMThe results of Subsections 2.1, 2.2 and 2.3 have all shown that the potential energy of a simple harmonicoscillator can be written in the form

Epot = 12 kx2 (9)


provided we interpret k and x appropriately (see Table 2), and provided we take due care over the choice of thezero point of potential energy.

Table 2

Oscillator k x

pendulum mg/L displacement, s along the arc from theequilibrium position

sliding mass on a spring ks displacement from the unstretchedposition

suspended mass on a spring ks displacement from the equilibriumposition

This common form for the potential energy of a simple harmonic oscillator given in Equation 9

Epot = 12 kx2 (Eqn 9)

is in fact quite general. It follows directly from the force law that characterizes one-dimensional SHM.


SHM is generally characterized by a restoring force and an acceleration that is linearly proportional to thedisplacement from some stable position of equilibrium. The implication of this is that for any one-dimensionalSHM we may write

Fx = −kx (10)

and that this is consistent with the general relation between a conservative force and potential energy

Fx =−dEpot

dx

which also gives

Epot = 12 kx2 (Eqn 9)

where k is the general SHM force constant and x is the displacement from the equilibrium position, provided wechoose Epot = 0 when x = 0.


Let us now move on to consider the kinetic energy and then the total energy of an oscillator in SHM.From Equation 1

Ekin = 12 mvx

2 (Eqn 1)

and we may write the general expression for the total energy of any one-dimensional simple harmonic oscillator

Etot = Ekin ( x ) + Epot ( x ) = 12 mvx

2 ( x ) + 12 kx2 4

Here we have recognized that both kinetic and potential energies vary with displacement x although the totalenergy should not vary with x, to be consistent with the principle of energy conservation. Since x itself is anoscillatory quantity it must be a function of time, we can write:

Etot = Ekin (t ) + Epot (t ) = 12 mvx

2 (t ) + 12 kx2 (t ) (11)


To take the discussion further we need to use the general expressions for x(t) and vx(t) in one-dimensional SHM.

These expressions are:

x(t) = A cos2πt

T+ φ

= A cos(ω t + φ ) (12)

vx (t) = dx(t)

dt= − Aω sin (ω t + φ ) (13)

where A is the amplitude, φ the initial phase or phase constant, ω the angular frequency, T the period orperiodic time and (ω1t + φ) the phase.

Expressions for the acceleration and the force acting can be obtained by further differentiation and by Newton’ssecond law, respectively:

ax (t ) = dvx (t )

dt= d2 x (t )

dt2= − Aω 2 cos (ω t + φ ) = −ω 2 x (t ) (14)

and Fx (t) = max (t) = −mω 2 x(t) (15)


A comparison of Equation 15

and Fx (t) = max (t) = −mω 2 x(t) (Eqn 15)

with Equation 10

Fx = −kx (Eqn 10)

identifies the force constant k with mω12 and so the angular frequency ω is determined by the force constant kand the mass m of the oscillating object:

k = − Fx (t )x (t )

= mω 2

so ω = k

m(16)

and T = 2πω

= 2π m

k(17)


It is important to note that Equations 12 to 17

x(t) = A cos2πt

T+ φ

= A cos(ω t + φ ) (Eqn 12)

vx (t) = dx(t)

dt= − Aω sin (ω t + φ ) (Eqn 13)

ax (t ) = dvx (t )

dt= d2 x (t )

dt2= − Aω 2 cos (ω t + φ ) = −ω 2 x (t ) (Eqn 14)

Fx (t) = max (t) = −mω 2 x(t) (Eqn 15)

ω = k

m(Eqn 16)

T = 2πω

= 2π m

k(Eqn 17)

are valid for any one-dimensional SHM, irrespective of the specific system, providing k is interpreted as theappropriate force constant1—1this ‘universality’ is the great power of the mathematical model.


Now we can return to our discussion of energy in SHM by using Equations 12 and 13

x(t) = A cos2πt

T+ φ

= A cos(ω t + φ ) (Eqn 12)

vx (t) = dx(t)

dt= − Aω sin (ω t + φ ) (Eqn 13)

to eliminate x(t) and vx(t) from Equation 11.

Etot = Ekin (t ) + Epot (t ) = 12 mvx

2 (t ) + 12 kx2 (t ) (Eqn 11)

We find:

Etot = 12 m − Aω sin (ω t + φ )[ ]2 + 1

2 k A cos (ω t + φ )[ ]2

i.e. Etot = A2

2mω 2 sin2 (ω t + φ ) + A2

2k cos2 (ω t + φ )


in which we can identify the kinetic and potential energy expressions:

Ekin = A2

2mω 2 sin2 (ω t + φ ) (18)

Epot = A2

2k cos2 (ω t + φ ) (19)

The relationship between ω and k as given in Equation 16

ω = k

m(Eqn 16)

allows the total energy to be written in either of two equivalent forms:

Etot = A2

2mω 2 sin2 (ω t + φ ) + cos2 (ω t + φ )[ ] = A2

2mω 2 (20a)

or Etot = A2

2k sin2 (ω t + φ ) + cos2 (ω t + φ )[ ] = A2

2k (20b)

In Equations 20a and 20b we have used the trigonometric identity that sin21θ + cos2

1θ = 1 for any angle θ.


Study comment We can now draw some important conclusions from Equations 18, 19, 20a and 20b. This will be done viaQuestions T6 to T9. You should be sure you understand the answers to these questions before moving on.

Question T6

Are Equations 18, 19, 20a and 20b

Ekin = A2

2mω 2 sin2 (ω t + φ ) (Eqn 18)

Epot = A2

2k cos2 (ω t + φ ) (Eqn 19)

Etot = A2


2mω 2 (Eqn 20a)

Etot = A2

2k sin2 (ω t + φ ) + cos2 (ω t + φ )[ ] = A2

2k (Eqn 20b)

consistent with the principle of the conservation of energy in an isolated system?3


Question T7

For the special case where the phase constant is zero, draw graphs (on the same piece of graph paper) showingthe behaviour of the kinetic, potential and total energies over two periods of the displacement. What is the periodof any oscillations in these energies? Describe what is happening over the time interval.3

Question T8

By inspecting your answer to Question T7 determine the average values of each of Epot and Ekin over a fullperiod of the oscillation.3

Question T9

Explain, in your own words, what factors determine the total energy in the oscillation; are these dependencieslinear?3


3 Damped and driven harmonic oscillatorsOur discussions in Section 2 have assumed that in SHM the amplitude is constant. From Equations 20a and 20b

Etot = A2


2mω 2 (Eqn 20a)

Etot = A2

2k sin2 (ω t + φ ) + cos2 (ω t + φ )[ ] = A2

2k (Eqn 20b)

or from Answer T9 you will appreciate that it is the amplitude (together with the force constant) that determinesthe total energy in SHM. If the amplitude is constant the total energy is constant and vice versa. Now, the totalenergy is certainly constant if the oscillating system is completely isolated from its environment since no energycan then pass to or from the oscillator. In practice, however, this condition is never rigorously true for anyoscillator; there is always some exchange of energy with the environment. In this section we will examine theconsequences of such energy exchanges.


There are two situations in which changes in the energy of an oscillator are of particular importance.

The first arises when we deliberately put energy into the oscillator1—1this is the case of the driven oscillator orof forced vibrations, which will be discussed in Subsection 3.5. The second is where energy is transferred outof the oscillator and the total energy of the oscillator (and hence its amplitude) decreases with time. This isthe case of the damped oscillator and the process by which energy is transferred out of the oscillator is calleddamping; this will be discussed in Subsections 3.1 to 3.4. All natural oscillations are damped to some extent andthere are many cases where we wish to change this damping artificially. For both these reasons, damping is animportant topic.


3.1 The mechanisms of damping: frictionThere are several mechanisms by which energy may be transferred away from a moving object.These mechanisms frequently involve frictional forces or, more succinctly, simply friction. Frictional forcesact to prevent or reduce relative motion between two parts of a system and hence produce a reduction in kineticenergy. The origin of frictional forces is complex, involving intermolecular interactions between materials inclose contact. Fortunately for our purpose here we need not understand the details of these processes, providedwe appreciate the following characteristics of frictional forces:

o Frictional forces occur whenever there is relative motion between two surfaces or where there would besuch motion if it were not being prevented by friction. The frictional forces are generally different in thesetwo situations. When relative motion does occur it is said to be opposed by dynamic friction which isusually less than the static friction that acts to prevent relative motion. This difference explains why it isusually easier to keep something moving (thus overcoming dynamic friction) than it is to start the objectmoving in the first place (thus overcoming static friction). An example of dynamic friction would be theforce that opposes the motion of a book sliding down a steeply inclined plane. An example of static frictionwould be the force that stops a book from sliding down an inclined plane when the angle of inclination issmall.


o Frictional forces always act in a direction that opposes relative motion or the tendency to begin relativemotion.

o In dynamic friction the general mechanism of the force is that momentum is exchanged between the surfacesor particles at the molecular level so that their relative velocity is reduced.

o The magnitude F of the dynamic frictional force often (but not always) depends on the relative speed v.Sometimes this dependence can be approximated by a linear relationship:

F = bv (21)

where b is a positive constant (known as the damping coefficient).

o The effect of dynamic friction is to reduce the relative motion and to convert the kinetic energy of directedmotion into kinetic energy of random motion at the molecular level1—1that is, into heat energy.


In this section we are concerned with damping which is attributable to friction between the moving parts of asystem − that is to dynamic friction. Within the general principles outlined above it is possible to identify severaltypes of dynamic friction which may be involved in damped SHM:

1 Friction between two solid surfaces in relative motion. For example, a mass sliding on a surface as in ourhorizontal spring oscillator of Subsection 2.2. Equation 21 is not valid in such cases

F = bv (Eqn 21)

and the sliding friction is almost independent of the relative speed v.

2 Friction between a moving solid and a liquid (or vice versa). For example, a ship rolling in a rough sea.Friction involving liquids (and sometimes gases) is often called viscosity and the forces involved are termedviscous forces. Equation 21 works quite well here.

3 Friction affecting solids or liquids moving through gases. For example, a swinging pendulum in air or avehicle travelling through air. This is usually referred to as air friction or air resistance.

All of these processes result in the relative motion being reduced and heat being produced. Because the kineticenergy is dissipated by friction, frictional forces are often called dissipative forces.


3.2 Frictional forces as dissipative forces in mechanical SHMThe rate at which a damped oscillator loses energy depends on the nature of the damping force. As an example,let us consider the energy transferred from an object that oscillates along the x-axis while subject to a dampingforce that satisfies Equation 21,

F = bv (Eqn 21)

that is, one where its magnitude is proportional to the instantaneous speed of the oscillator.

What is the x-component of such a damping force at a time when the x-component of the instantaneousvelocity is vx?


If we consider a very short interval of time, ∆t, we can assume that the instantaneous velocity is effectivelyconstant throughout that interval, so the oscillator will change its position by an amount ∆x = vx1∆t and theenergy transferred to the oscillator by the damping force will be

∆E = Fx1∆x

If we substitute for Fx and ∆x, we find

∆E = −bvxvx∆t = −bvx2∆t (22)

In Equation 22 both b and vx2 are positive, so the energy transferred to the moving object must be negative.

Thus the effect of the damping force is to reduce the energy of the oscillator, irrespective of its direction ofmotion. (Reversing the sign of vx does not change the sign of ∆E.)

The instantaneous rate at which energy is transferred to the oscillator by the damping force is obtained byrearranging Equation 22 to find ∆E/∆t and then considering the limit in which ∆t tends to zero. In terms ofcalculus notation this instantaneous rate is

dE

dt= lim

∆t→0

∆E

∆t

= −bvx

2 (23)


dE

dt= lim

∆t→0

∆E

∆t

= −bvx

2 (Eqn 23)

Notice that dE/dt is also negative, as we would expect for a damping force that reduces the energy of theoscillator.

At what stage in the oscillatory cycle is energy transferred most rapidly according to Equation 23?

Of course, Equations 22 and 23 only apply to the particular case where the damping force is given by Fx = −bvx,but similar analyses can be applied to other cases.

∆E = −bvxvx∆t = −bvx2∆t (Eqn 22)


Question T10

Suppose that a one-dimensional oscillator is subject to a damping force of constant magnitude.

(a) Write down an expression for the damping force.

(b) Find an expression for the instantaneous rate of energy transfer to the oscillator by this force.

(c) At what stage in the oscillation is energy transferred most rapidly because of the damping?3


3.3 Lightly damped harmonic motionWe now return to the case of a one-dimensional oscillator subject to a damping force Fx = −bvx, and consider theinfluence of the damping on the motion itself. Equation 23

dE

dt= lim

∆t→0

∆E

∆t

= −bvx

2 (Eqn 23)

has already shown us that the energy of such an oscillator decreases continuously, and it follows from this thatthe amplitude of the oscillation must also decrease with time. What we now want to do is to develop amathematical expression for x(t), the displacement of the oscillator as a function of time, which will (amongother things) show us exactly how the amplitude diminishes as times passes.

The general solution to this problem involves some quite challenging mathematics. Instead of undertaking thistask we will follow a simpler approach by assuming that the energy is transferred slowly, so that the total energyof the oscillator is almost constant over any one cycle even though it will change appreciably over many cycles.This amounts to assuming that the damping force is weak, so that the oscillator is certain to undergo manyoscillations before the damping brings it to rest. Because of this limitation we can only claim to be dealing withthe motion of the oscillator in the case of light damping. More will be said about the general case where thedamping force may be strong or weak in the next subsection.


Since Equation 23

dE

dt= lim

∆t→0

∆E

∆t

= −bvx

2 (Eqn 23)

is true moment by moment, it must be the case that the average value of the right-hand side of Equation 23,taken over a full period of oscillation, is equal to the average value of the left-hand side taken over the sameperiod. Thus, using angular brackets to indicate the mean:

dE

dt= −bvx

2

but b is a constant and vx2 = 2Ekin1/m, so

dE

dt= −b

2 Ekin

m= − 2b

mEkin


Now, Subsection 2.4 and Question T8 have shown us that in SHM the average kinetic energy and the averagepotential energy over a full oscillation are each equal to half the total energy so, over a full oscillation it muststill be the case that

Ekin = 12

E

ThusdE

dt= − b

mE

Now, if we introduce the light damping assumption so that the total energy only changes slowly with time, theaverage rate of change of energy will be equal to the rate of change of the average energy, i.e.

dE

dt= d E

dt

Moreover, if we represent the slowly varying average energy ⟨ 1E1⟩ by E(t) then we have

dE(t )dt

= − bE(t )m

= − γ E(t ) (24)

where the positive quantity b/m has been written as γ.


Study comment Equation 24 is an example of a differential equation. Such equations are the subject of a block of modulesin the maths strand of FLAP. If you are already familiar with differential equations you will probably know how to solveEquation 24 and you will be able to consider the following question. If not, you should simply read the question and itsanswer, and pursue the topic of differential equations on some other occasion.

dE(t )dt

= − bE(t )m

= − γ E(t ) (Eqn 24)

Equation 24 tells us that the function E(t) is such that its rate of change is proportional to the instantaneousvalue of E(t) at any time t. What is the form of such a function?


γ1 > γ2

t

E(t)

E0 E0 exp (–γ1t)

E0 exp (–γ2t)

Figure 53Graphs of E(t) = E0e−γ1t fordifferent values of the damping constantγ. (γ01 > γ02)

Graphs of E(t) = E0e−γ1t for two different values of γ are shown in Figure 5. It is apparent that the larger value of γ corresponds to a morerapidly decaying average energy for the oscillator. The coefficientγ = b/m is called the damping constant of the oscillator. Over any timeinterval of duration 1/γ the value of E(t) changes by a factor of e−1

(from e−γ1t1 to e−γ11(t1+11/γ11) = e−1e−γ11t0).

Thus, γ determines the rate of exponential decay of the average energy.The time required for any exponentially decaying quantity to decreaseby a factor of e−1 is known as the time constant of the decay, so in ourexample the time constant is 1/γ and we may write

E(t) = E0e−γ1t = E0e−1t1/τ (25)

where τ = 1/γ represents the time constant.


γ1 > γ2

t

E(t)

E0 E0 exp (–γ1t)

E0 exp (–γ2t)

Figure 53Graphs of E(t) = E0e−γ1t fordifferent values of the damping constantγ. (γ01 > γ02)

Question T11Confirm that the function E(t) = E0e−γ1t satisfies Equation 24,

dE(t )dt

= − bE(t )m

= − γ E(t ) (Eqn 24)

and explain the physical significance of the constant E0.

(If you are familiar with the process of differentiation you can use it toshow that Equation 25

E(t) = E0e−γ1t = E0e−1t1/τ (Eqn 25)

provides a solution to Equation 24. If not, you should add arbitraryscales to Figure 5, select a point on one of the curves, determine thegradient of the curve at that point (this corresponds to dE/dt at thatpoint) and then confirm that at that particular point dE/dt = −γ1E(t).)3


Question T12For damped one-dimensional harmonic motion, write down expressions for (a) the damping constant for theenergy and (b) the time constant of the energy decay, in terms of the damping coefficient b and the oscillatormass m.3

Question T13

Use the SHM relationship between total energy and amplitude to derive expressions for (a) the amplitude A(t)and (b) the time constant of the amplitude decay, in terms of the damping coefficient b and the oscillatormass m.3


Question T14

In Question T13 you derived an expression for A(t). Use this result as the amplitude in Equation 12

x(t) = A cos2πt

T+ φ

= A cos(ω t + φ ) (Eqn 12)

to write down an expression for x(t) for damped one-dimensional SHM. Sketch the graph of this function.4

Study comment

The rigorous mathematical treatment of damped harmonic motion shows that the introduction of damping, even lightdamping, modifies the angular frequency of the motion as well as its amplitude. The changes for light damping are verysmall so we will continue to use the values of angular frequency and period that pertain in the undamped situation, but inrecognition of this we will henceforth denote the undamped angular frequency by ω0 and use T0 to represent thecorresponding period (ω002 = k/m and T0 = 2π/ω00). You should not forget that in the case of damped harmonic motion, ω0and T0 are only approximations to the true angular frequency and period.


In the case of damped harmonic motion, the natural logarithm of the ratio of two successive maximumdisplacements (i.e. one period apart) is known as the logarithmic decrement of the decay. Question T13 showsthat the damping constant for the amplitude is half that for the energy, since E(t) ∝ A2(t). This gives the dampingconstant for the amplitude as γ/2 and it follows that

logarithmic decrement = loge expγ T0

2

= 12 γ T0 = πγ

ω0(26)

The case of light damping applies well to the important class of applications where engineers seek to minimizethe natural damping of an oscillator in order to prolong its natural oscillation or to minimize the energy inputneeded to maintain the oscillations1—1a pendulum clock, for example. In such cases, the quality of the oscillatoris determined by the fraction of its energy which is lost per cycle of the displacement (remember fromQuestion T7 that one displacement cycle corresponds to two cycles of the energy oscillation). The larger thisfraction, the poorer the performance of the oscillator, so its reciprocal can be taken as a ‘figure of merit’ for theoscillator. This inverted fraction, when multiplied by 2π, is known as the quality factor or Q-factor of theoscillator.


So, Q = 2πE(t)

∆E( )T

(27)

where E(t) is the average energy at time t and |1(∆E)T1| is the energy transferred from the oscillator in onecomplete oscillation at time t.

A very lightly damped oscillator has a Q >> 1, while a more severely damped oscillator has a Q of order 1.An oscillator with a large Q -factor damps slowly and oscillations are maintained for many periods.A high Q-factor mechanical oscillator, such as a tuning fork, might have a Q-factor of several thousand.

Now, according to Equation 24

dE(t )dt

= − bE(t )m

= − γ E(t ) (Eqn 24)

the average rate of energy transfer from the lightly damped oscillator is dE(t)/dt = −γ1E(t), so the energytransferred from the lightly damped oscillator over one period of duration T0 is

(∆E)T = T0dE(t)

dt= T0 γ E(t) = 2πγ

ω0E(t)


so Q = 2πE(t)(∆E)T

= 2πγ T0

= ω0

γ(28)

We can now summarize our results as follows:

In lightly damped harmonic motion

E(t) = E0 exp (−γ t) = E0 exp −ω0t Q( ) (29)

A(t) = A0 exp −γ t 2( ) = A0 exp −ω0t (2Q)( ) (30)

and

x(t) = A0 exp −γ t 2( )cos(ω t + φ ) = A0 exp −ω0t (2Q)( )cos(ω t + φ ) (31)

Question T15

By what factor does the energy of a damped oscillator decrease in Q cycles?3


We see from Answer T15 that when a high Q-factor oscillator is excited it completes Q/2π cycles before itsenergy has fallen by a factor of 1/e. If we call this time the ringing time then Q is 2π × the ringing time of theoscillator, expressed in periods.

3.4 Qualitative discussion of general dampingWhen the damping can no longer be considered to be light, the mathematics becomes quite complicated.This topic is dealt with elsewhere in FLAP. Here, we will describe the results qualitatively, so that you canidentify and describe the types of behaviour that damped oscillators exhibit.

We have already covered light damping quantitatively in this module. In that discussion we were able to use theundamped angular frequency ω00 as an approximation to the angular frequency of the damped motion.The full mathematical treatment shows that the angular frequency ω of the damped motion is slightly differentfrom ω00, but not significantly so in the case of light damping. However, as the damping is increased thisapproximation begins to break down, and we need to distinguish between ω and ω00.


3π/ω

x

0

(a)

(c)

(b)

t

A0

−A0

π/ω2π/ω

4π/ω5π/ω

(d)

As the damping is increased progressively, fewerand fewer oscillations are completed before theoscillator comes to rest; the various forms ofbehaviour that might arise are shown in Figure 6.The most familiar is illustrated by curve (a) and isgenerally known as underdamping; light dampingis an extreme case of this. If we increase thedamping beyond a certain level we eventually findthe behaviour indicated by curve (b) in which theoscillator fails to complete even a single fulloscillation. It may overshoot the equilibriumposition once only, after which its displacementdecays exponentially as it returns to theequilibrium position. This situation is generally

known as overdamping.

Figure 63The effect of different damping levels on theone-dimensional oscillator: (a) underdamped, (b) overdamped, (c) heavily overdamped, (d) critically damped.


3π/ω

x

0

(a)

(c)

(b)

t

A0

−A0

π/ω2π/ω

4π/ω5π/ω

(d)

If we continue to increase the damping further wereach an extreme case of heavy overdamping(curve (c)) in which the system does not passthrough the equilibrium position at all, but decaysexponentially and very slowly back to theequilibrium position. The situation betweenunderdamping (curve (a)) and overdamping (curve(b)) is called critical damping and is shown incurve (d). In the case of critical damping, thedamping force has the minimum value that willavoid an oscillation and, after one overshoot, thedisplacement decays exponentially to reach thestable equilibrium position in the shortest overalltime. This condition of critical damping is the onesought by designers of electrical meters and shock-absorbers.Figure 63The effect of different damping levels on theone-dimensional oscillator: (a) underdamped, (b) overdamped, (c) heavily overdamped, (d) critically damped.


3.5 Qualitative discussion of forced vibrations: resonanceIf the oscillations of a damped oscillator are to be maintained at a constant level then we must apply an externalforce which will, on average, transfer energy to the oscillator at the same rate that energy is transferred into heatby friction. This external force is called a driving force and the oscillator is then a driven oscillator.In this subsection we present a qualitative treatment of this problem; the quantitative treatment is coveredelsewhere in FLAP.

A familiar example of a driving force is the force that every child learns to apply to a swing to establish andmaintain its motion. If we ask how best to apply this driving force in order to minimize effort and maximize theamplitude then, as every child discovers, the answer is that the force should be applied periodically and shouldbe timed to coincide closely with the natural motion of the swing. In other words, a driven oscillator respondsmost strongly when driven by a periodically varying force, the frequency of which is closely matched to thefrequency with which the system would freely oscillate if left to itself. This frequency is called thenatural frequency of the oscillator.


A one-dimensional oscillator of mass m and force constant k , has a natural frequency ω0 = k m .It can be shown that this system, when subject to a damping force Fx = −bvx and a periodic driving forceFx = F01sin1(Ω0t), will eventually exhibit displacement oscillations described by

x(t) = A01sin1(Ω0t + φ) (32)

where A0 = F0 m

(ω02 − Ω 2 )2 + (γ Ω )2

3and3φ = arctan1− γ Ω

ω02 − Ω 2

Note that the angular frequency of these steady oscillations is determined by the angular frequency of the drivingforce and not by the natural frequency of the oscillator. Also note that the amplitude of the driven oscillations A0is independent of time but does depend on the amplitude and angular frequency of the driving force.In particular, the amplitude will be large when Ω ≈ ω00.

(It can be shown that the maximum value of A0 actually occurs when Ω12 = ω002 − γ12/2.)


(a)

(b)

A0

Ωω0

Figure 73The resonant response of adriven damped oscillator under twolevels of damping. (a) The steady stateamplitude established in a high Q-factoroscillator with very small γ, (b) thesteady state response in a low Q-factoroscillator with larger γ. The effect of γon the resonant frequency has beenignored in this graph.

The eventual steady amplitude of the driven oscillator for a range ofdriving frequencies is shown in Figure 7. There are two curves in thefigure, corresponding to different values of the damping constant γ.As you can see, although the location of the peak is mainly determinedby ω0, the height and width of the peak both depend on γ. The role of γin determining the amplitude is not surprising. Most damping lossesincrease with the speed of the oscillator and so the dissipative losses riseas the oscillation amplitude grows. With any given driving forceapplied, the oscillation amplitude will grow until the energy transferredfrom the oscillator by the damping force exactly balances the energytransferred to the oscillator by the driving force. Steady oscillationconditions are then established.


The condition in which an oscillator responds with maximum amplitude to a periodic driving force is calledresonance. For a lightly damped oscillator this condition occurs when the angular frequency of the drivingforce is close to but slightly less than the natural frequency of the oscillator and corresponds to the state ofaffairs in which the energy transferred to the oscillator by the driving force is a maximum.

If we refer back to Equation 32,

x(t) = A01sin1(Ω0t + φ) (Eqn 32)

you will see that there is a phase difference φ between the displacement oscillation and the driving force.

What is the value of φ when the resonance condition is satisfied?

So, in resonance, in the steady condition, the displacement of a lightly damped oscillator lags 90° behind thedriving force. You may find it surprising that the displacement is not in phase with the driving force, but yousaw in Subsection 3.2 that the power transferred to an oscillator depends on the velocity of the oscillator, not itsdisplacement. A phase lag of 90° between the driving force and the displacement ensures that the driving force isin phase with the oscillator’s velocity rather than displacement.


Thus a given force gives the oscillating mass its greatest rate of transfer of energy (i.e. power) just when itsspeed is greatest. This is exactly the principle most children learn to apply when pushing a swing.

There are many important illustrations and applications of resonance. Mechanical resonance plays a crucial rolein many pieces of machinery, such as when one part of a machine vibrates and drives another part into vibration.If the driven part is underdamped, the resonant response can build up a very large amplitude, which might end inthe destruction of the machine. The suspension bridge across the Tacoma Narrows on the western coast of theUSA was driven into resonance by unusually violent winds in 1940; it was destroyed. A similar incident tookplace in 1850 at Angers (in France) where the marching of about 500 soldiers across a bridge resonantly excitedthe structure, with the resultant death of more than 200 soldiers.


Perhaps some of the most important applications of resonance occur in electrical oscillations. Electrons arecharged particles and electrical forces can be exerted on them by electric fields. In an electrical conductor someelectrons are free to move and this motion can constitute an electric current. In an electrical circuit,electrical energy can be stored as potential energy, by electrons in electric fields, or associated with the kineticenergy of the moving electrons, through the magnetic field caused by the electric current. Electrical energyoscillations can be produced by the action of electric or magnetic fields on the electrons in a conductor. A goodexample of this process is when the oscillating electric and magnetic fields associated with anelectromagnetic wave (such as a radio wave or a light wave) interact with electrons in a circuit. An electricaloscillator has a natural frequency and a resonant response when driven at this frequency. The driving force foran electrical oscillator can be provided by the currents from an aerial. At the circuit’s resonant frequency largeoscillatory currents may be produced, and these can easily be detected. We have described here a radio or TVreceiver, which must be set or tuned to the required resonant frequency.

A surprising example of electrical resonance is the response of the electrons in an atom to an incoming lightwave. At some characteristic resonant frequencies (or wavelengths of light) these electrons respond strongly,absorbing the light. This is why materials absorb some colours of light but not others.

Resonance is an extremely widespread and important topic in physics. Although we have been concerned withmechanical oscillators in this module it is important to remember the wide applicability of all the ideas discussedhere.

Mike Tinker


4 Closing items

4.1 Module summary1 In simple harmonic motion, the displacement of an oscillator may be described by the sinusoidal function

x(t) = A1cos1(ω1t + φ) where A is the amplitude, φ is the phase constant, and the angular frequency ω is givenby ω = k m , where k is the force constant and m the mass. The period of such an oscillator is

T = 2πω

= 2π m

k(Eqn 17)

2 Kinetic energy and potential energy are associated with SHM. In one-dimensional mechanical oscillators

the kinetic energy 12 mvx

2 is associated with a moving mass and the potential energy 12 kx2 with

gravitational or strain energy. The zero of strain energy is best taken where the elastic body isun-deformed, the zero of gravitational energy should be chosen to simplify the problem.

3 In undamped SHM the kinetic energy and potential energy each oscillate with time at a frequency which istwice that of the displacement oscillation, but the sum of kinetic and potential energy, the total energy,remains constant.


4 The average values for kinetic energy and potential energy over one cycle of undamped SHM are eachequal to half the total energy1— 1which is itself determined by the square of the amplitude through theexpression:

Etot = A2

2k = A2

2mω 2 (Eqn 20)

5 When dissipative forces, such as friction, are present the energy of an oscillator is gradually transferred tothe environment, often as heat, and the amplitude falls with time. The oscillator is then said to be damped.

6 Frictional forces often depend on the relative motion and oppose this motion; in some cases their magnitudeis linearly proportional to the relative speed and then the power loss from the damped oscillator isproportional to the square of this speed.


7 In light damping the oscillation takes many cycles to damp away and, to a good approximation, the motionremains periodic with the undamped frequency. The energy decays exponentially with a damping constant γgiven by

E(t) = E0 exp (−γ t) (Eqn 29)

where γ = b/m (and b is a positive constant). The amplitude also decays exponentially, as given by

A(t ) = A0 exp (− γ t 2) (Eqn 30)

and the displacement is given by

x (t ) = A0 exp (− γ t 2) cos ω t + φ( ) (Eqn 31)

8 The degree of damping is indicated by the Q-factor of the oscillator, which is 2π times the reciprocal of thefractional energy loss per cycle

Q = 2πE(t)

∆E( )T

(Eqn 27)

For a lightly damped oscillator Q = ω0/γ and oscillations take many cycles to decay away.


9 In general, three levels of damping can be identified. Underdamping, when the oscillator completes severaloscillations before being damped to rest. Overdamping when the displacement is no longer periodic butdecays exponentially, or where it may overshoot the equilibrium position once, before decayingexponentially to the equilibrium position. Critical damping, when the displacement decays to rest withoutoscillatory behaviour but does so in the minimum time, passing through the equilibrium position no morethan once.

10 When a damped oscillator is subject to a periodic external driving force at or near to its natural undampedfrequency it exhibits resonance. In resonant behaviour the amplitude builds up to a high value, until thepower lost through dissipative forces equals the power input from the driving force. The amplitude atresonance and the sharpness of the response near ω0 are each determined by the damping constant γ (or theQ-factor), and with low damping (high Q-factor) this amplitude can be very large and the response verysharp. Resonance is an important and widespread process in physics and engineering.


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 Account for and describe, qualitatively, the oscillations of kinetic and potential energy in undamped one-dimensional SHM, and explain why the total energy is constant.

A3 Derive and use expressions for the kinetic, potential and total energies in undamped one-dimensional SHM,from an expression for the displacement as a function of time. Show how the total energy depends on theamplitude.

A4 Sketch the variations in kinetic and potential energy in undamped one-dimensional SHM and identify theiraverage values over one oscillation.

A5 Describe the operation of frictional forces and their qualitative effect on one-dimensional harmonic motion.

A6 Describe the motion of lightly damped one-dimensional harmonic motion in terms of an energy–time graph,an amplitude–time graph and a displacement–time graph.

A7 Use the approximation of light damping, with a damping force which is proportional to the velocity, toderive expressions for the energy and displacement in one-dimensional harmonic motion.


A8 Describe and identify the general behaviour of damped one-dimensional harmonic motion when conditionsare no longer those of light damping.

A9 Describe the process of driving and resonance in harmonic motion, explaining its significance and itsdependence on the damping constant or the Q-factor.

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.

(a) mg

reaction

(b) mg

xeq

(c) mg


xeq1+1x

Question E1

(A2)3For the system shown inFigure 4,

Figure 43A light coil spring shown(a) unloaded (with a supported massheld at the unstretched springposition), (b) at the position ofequilibrium under the load, and(c) loaded and displaced below itsposition of equilibrium. Note that x istaken to be positive in the downwarddirection.


Table 13Potential energies for Figure 4 with the gravitational potential energy zero atthe equilibrium position and the strain potential energy zero for the unstretched spring.

Position Gravitationalpotential energy

Egrav

Strainpotential energy

Estrain

Totalpotential energy

Epot

As in Figure 4a mgxeq = ks(xeq)2 0 ks(xeq)2

As in Figure 4b 0 ks(xeq)2/2 ks(xeq)2/2

As in Figure 4c −mgx = −ksxeqx ks(xeq + x)2/2 ks(xeq)2/2 + ksx2/2

construct a table such as Table 1 but with the gravitational potential energy and the strain potential energy bothzero at the equilibrium position.


Question E2

(A3 and A4)3Show that in a one-dimensional SHM the kinetic energy, potential energy and the total energy alldepend on the square of the amplitude. Sketch each of these energies over one cycle of the displacementoscillation and give their average values over this interval.

Question E3

(A5 and A6)3Describe the motion of a lightly damped oscillator, and explain why its total energy is notconserved.

Question E4

(A5)3A body of mass 0.301kg is suspended from the end of a spring with a spring constant of 1201N1m−1.When oscillating, it is subject to a damping force that varies in proportion to the velocity, with constant ofproportionality −b = −0.601N1s1m−1.

Calculate the damping constant, logarithmic decrement and Q-factor of the system.


Question E5

(A7)3A lightly damped oscillator with undamped angular frequency 161s−1 is set in motion by releasing it fromrest at a displacement of 0.101m from its equilibrium position. It is found that the turning point at the end of thefirst cycle has a displacement of 0.081m. What are the logarithmic decrement, damping constant and Q-factor ofthe system? Estimate the amplitude of the oscillation after five cycles.

Question E6

(A8)3For one-dimensional damped harmonic motion explain what is meant by (a) underdamped, (b)overdamped, and (c) critically damped, explaining how these conditions might be established.


Question E7

(A9)3Describe the process of resonance in SHM and its significance. Describe qualitatively how resonantbehaviour depends on the level of damping.

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.

enrgy, damping and resonance

Documents

energy oscillations

total energy

thekinetic energy

total potential energy

gravitational potential

strain potential energy

module p5

isolated system