Before starting our quantitative analysis, let’s qualitatively consider what to expect. If the frequency of the drive is very low, then the mass should simply move up and down together with the support point, that is,x =xc =Adcosωdt. On the other hand, ifωdis very high, then the sign of the drive force keeps changing at a high frequency. So, the drive force doesn’t have much time to accelerate the mass before the sign of the drive force changes. So, for largeωd, we expect a small motion of the mass.
Somewhere between low ωd and high ωd we expect to find a resonant response, that is, a special frequency of drive force for which the mass oscillates with large amplitude.
The damped oscillator without a drive force is just a special case of the damped driven oscillator, with zero drive amplitude, that is, F0=0. Thus, thefull solution for the oscillator with both driving and damping should include the possibility of an oscillation at angular frequency ωv that decays away exponentially and has an amplitude and phase that depend on the initial position and velocity. However, for times well beyond t =0, it is reasonable to expect that the effect of these initial conditions will decay away, and that we will see a “steady-state” oscillation in which the energy per cycle delivered to the system by the drive equals the energy per cycle that leaks away from the system because of the damping. This “steady-state solution”
is actually the one of most interest, since it persists indefinitely.
We consider a general sinusoidal drive force
Fdrive=F0cosωdt, (4.1.2)
which might be applied by a motion of the support point (equation (4.1.1), or might be applied in some other way.
To find the behavior of the system,x(t), we again follow our three-step procedure:
steady-state asωs. To be as general as possible, we include a phase factor, and of course an amplitude. Thus, our guess is
z=? Aei(ωst−δ).
Here, the phase factor is written as −δ, rather than, for example, +ϕ. We do this because it will turn out thatδas defined this way is always positive, though this is not yet obvious. The above guess can be rewritten
z=?Ae−iδeiωst. (4.1.5) 3. Substitute the guess into the DEQ to see if it is a solution, and if there are restrictions on the parameters
Your turn:Plug our guess (4.1.5) into (4.1.4) to show that (−ωs2A+iγ ωsA+ω20A)e−iδeiωst=? F0
meiωdt. (4.1.6)
We see that the left side oscillates at an angular frequencyωs, while the right side oscillates atωd. So, if they are to be equal, we must haveωs =ωd. In other words (assuming other aspects of our guess turn out to be correct):
In the steady-state, the oscillator moves with the same angular frequency as the drive.
This is perhaps an unexpected result. In the steady-state, the system doesnot move with its “natural” angular frequencyω0or atωv, but rather it moves atωd.
So, our guess now becomes
z=? Ae−iδeiωdt, (4.1.7) and is shown graphically in figure 4.1.4. But we haven’t yet shown that our guess really works. Sinceωs=ωd, equation (4.1.6) becomes
(−ωd2A+iγ ωdA+ω02A)e−iδeiωdt=? F0 meiωdt.
⇒(−ω2dA+iγ ωdA+ω02A)e−iδ=? F0 m.
⇔(ω02−ωd2)A+iγ ωdA=? F0 meiδ
Figure 4.1.4 Our guess represented in the complex plane. Note that, since the units ofzand F0are different, the length of the two vectors cannot be compared.
For this equation to hold, the real part of the left side must equal the real part of the right side, and also the imaginary part of the left side must equal the imaginary part of the right side:
Real: (ω20−ω2d)A=? F0
m cosδ. (4.1.8)
Imaginary:γ ωdA=? F0
m sinδ. (4.1.9)
To make these equations true, we will need particular values ofAandδ. To isolateA, we square these two equations and add them, giving
A=Aωd
= F0/m
(ω20−ωd2)2+(γ ωd)2
, (4.1.10)
in which we emphasize thatA is a function ofωd. To isolateδ, we instead divide equation (4.1.9) by (4.18), giving
tanδ ωd
= γ ωd
ω20−ω2d, (4.1.11)
again emphasizing thatδis a function ofωd.1
So, our guess (4.1.7) is indeed a solution of the differential equation (4.1.4)! (But only ifAandδare as given above.) Of course,
x=Re (z)=Acos (ωdt−δ). (4.1.12) We can see from equation (4.1.10) that the amplitude depends on the angular frequency of the drive,ωd. It might appear from the equation that the maximum amplitude occurs
1. It is correct to writeδ ωd
=tan−1 γ ωd
ω20−ω2d, however recall that one must be careful with the tan−1function. For a negative argument, your calculator (or Mathematica or other equivalent program) returns a negative value for the tan−1. Since we wantδto be positive, we must addπ to such a result.
Figure 4.1.5 Amplitude and phase of a damped driven oscillator as a function of the angular frequency of the drive.
atωd=ω0; this is almost right, but the actual maximum is at a slightly lower value2of ωd. However, the difference is quite small, except for heavy damping, and is usually unimportant. (We shall discuss this in more detail in section 4.2.) This maximum amplitude is the resonance discussed earlier in this section.
We also see from equation (4.1.11) that the phase δ by which the oscillator’s response lags behind the drive force also depends onωd. When ωd=ω0, equation (4.1.11) becomes tanδ→ ∞, so thatδ=π/2. The dependencies ofAandδonωdare shown in figure 4.1.5.
We see from equation (4.1.10) that the response amplitude at high frequencies approaches zero, as anticipated in our initial qualitative discussion. In the opposite limit,ωd→0, equation (4.1.10) reduces to
Aωd→0=F0/m ω20 =F0
k .
If the drive force is applied by moving the support point, then, using equation (4.1.1):
F0=kAd, this becomes
Aωd→0=Ad, (4.1.13)
which is also as we anticipated.
2. To see this, we rewrite equation (4.1.10):
A= F0/m
(ω02−ω2d)2+(γ ωd)2
= 1 ωd
F0/m (ω20−ωd2)2
ωd2 +γ2
!
peaks atωd=ω0
.
The second part of this has a peak at exactlyωd=ω0. However, it is multiplied by the factor 1/ωd, which increases asωddecreases, shifting the peak to a slightly lower value ofωd.
4.2 Effects of damping
The shapes of the curves in figure 4.1.5 are profoundly important for applications of resonance, both those applications in which we want to maximize resonance effects (such as in radio receivers) and applications in which we want to minimize them (such as in building designs). These curves are strongly affected by the degree of damping, as we’ll explore in this section.
Often, it is revealing to re-express functions in terms of their dependence on dimensionless variables; this frequently reveals a universal behavior that was obscured in the original form of the function. In our case, we will try re-expressing the steady- state amplitude Aand the phase shift δ of the response relative to the drive force.
In equations (4.1.10) and (4.1.11), these two functions are expressed in terms of the angular frequency of the drive,ωdand the factorγ ≡b/mwhich characterizes the damping. We will now re-express them in terms of ωd/ω0andQ≡ω0/γ, both of which are dimensionless.
Your turn:Starting from equation (4.1.10), A= F0/m
(ω20−ωd2)2+(γ ωd)2
, and using
γ=ω0/Q, show thatA= F0/m ω0ωd ω0
ωd−ωd ω0
2
+ 1 Q2
.
We can rewrite this result asA= F0 mω20
ω0/ωd
ω0 ωd−ωd
ω0 2
+ 1 Q2
. Sinceω0≡ k
m, this becomes
A= F0 k
ω0/ωd ω0
ωd−ωωd
0
2
+Q12
. (4.2.1)
We can now see the universal behavior that we hoped would arise; the particular values ofωdandω0are not really central – what really matters is the ratioωd/ω0.
If the drive force is applied by moving the support point, then equation (4.1.1):
F0=kAd, so that
A=Ad ω0/ωd
ω0 ωd−ωωd
0
2
+Q12
. (4.2.2)
This relation is graphed in figure 4.2.1. Several different curves are shown for different values ofQ. The most important effect of increasing theQ(i.e., lowering the damping) is to make the peak higher and sharper.
A more subtle effect is that the peak, which is always at an angular frequency close toω0, moves even closer toω0asQincreases. You can show in problem 4.8 that the
Figure 4.2.1 Dependence of steady state amplitude onωd/ω0.
peak occurs at
ωd, peak=ω0 1− 1
2Q2. (4.2.3)
Recall from chapter 3 thatQ=2πcorresponds to reasonably heavy damping. (Without a driving force, the energy decays by a factor of 1/e per cycle, and the amplitude decays by a factor of 1/√
e∼=0.61 per cycle.) SubstitutingQ=2πinto equation (4.2.3) gives ωd,peak=0.994ω0. For light damping,ωd,peakis even closer toω0, so for most purposes the difference can be ignored.
For moderate to light damping, for which the peak amplitude occurs atωd∼=ω0, we can obtain a useful result for the height of the peak. Substitutingωd =ω0 into equation (4.2.2) gives
Aωd=ω0
=QAd. (4.2.4)
In other words,
For light to moderate damping:
(peak response amplitude)∼=Q×(drive amplitude).
Since the low-frequency response amplitude is equal to the drive amplitude, we could also say that the peak response amplitude is Q times the low-frequency response amplitude.
Forωdω0, the response becomes A=Ad ω0/ωd
ω0 ωd−ωωd
0
2
+Q12
limωdω0
−−−−−−→Ad ω0
ωd 2
, (4.2.5)
so that, at high frequencies, the response is universal, and does not even depend on the degree of damping.
Figure 4.2.2 Phase shiftδof the steady-state response.
Now, we will consider how the graph of the phase shiftδis affected by damping.
Again, we begin by expressingδ in terms of the dimensionless quantitiesωd/ω0andQ.
Your turn:Starting from equation (4.1.11), tanδ = γ ωd
ω20−ω2d, and using γ = ω0/Q, show that
tanδ= 1/Q ω0 ωd−ωd
ω0
. (4.2.6)
This relation is graphed in figure 4.2.2.3At lowωd, the phase shift is zero, meaning (as we anticipated) that the oscillator moves in phase with the drive. At highωd, the phase shift is 180◦; the oscillator is exactly out of phase with the drive. Atωd=ω0, the phase shift is exactly 90◦. The effect of increasing theQ(i.e., lowering the damping) is to sharpen the transition fromδ=0 at low drive frequency toδ=π at high-drive frequency.
Connection to current research: Tapping Mode Atomic Force Microscopy
The Atomic Force Microscope (AFM), invented in 1986 by Gerd Binnig, Calvin Quate, and Christoph Gerber, images a sample by touching it very lightly with a sharp tip mounted on the end of a cantilever (figure 4.2.3). The upward force exerted on the tip by the sample causes a bend in the cantilever, so that this force can be measured quantitatively. In the original mode of operation (“contact mode”), the tip is first lowered into contact with the sample until a pre-set force is achieved (typically about 100 nN). Then, the tip is moved laterally across the sample. As it moves, the force is monitored, and the base of the cantilever is moved up or down as needed to keep the force constant at the pre-set level. In this way, the microscope “feels” the shape continued
3. We can rewrite equation (4.2.6) asδ=tan−1 1/Q ω0 ωd−ωd
ω0
. Recall that, in this case if the argument of the tan−1function is negative, that is, ifωd> ω0, we must addπto the result given by a calculator or a program like Mathematica in order to get the correct value ofδ. (See footnote 1 after equation (4.1.11).)
metal samples, but the lateral motion of the tip is very destructive for the soft samples which are of most interest in biology and in molecular electronics.
Figure 4.2.3 The cantilever for an atomic force microscope (top image) must have a moderately highQ, so that a small amplitude vibration applied at the base of the cantilever results in a vibration amplitude at the tip of about 100 nm. The length of the cantilever is 125μm, about the same as the diameter of a human hair. The
pyramid-shaped tip located at the end of the cantilever (bottom image) must be very sharp to obtain high resolution images. (Image courtesy of Veeco Instruments Inc.)
In 1993, Zhong, Inniss, Kjoller, and Elings introduced the Tapping modeTMAFM. (The trademark belongs to Digital Instruments (now part of Veeco Corporation), which was founded by Virgil Elings and is the leading manufacturer of AFMs.) In this mode, the base of the cantilever is vibrated vertically by means of a piezoelectric crystal. The frequency of vibration is chosen to match the resonant frequency of the cantilever, so that the tip vibrates with an amplitude of about 100 nm. The vibration amplitude is measured, and the base of the cantilever is slowly lowered toward the sample. When the tip begins to tap the sample, this contact reduces the oscillation amplitude. The base is lowered further until the amplitude reaches a pre-set value. Then, as in contact mode, the base of the cantilever is moved laterally across the sample. As it moves, the amplitude of the tip’s oscillation is monitored, and the base of the cantilever is moved up or down as needed to keep the amplitude constant. Because the tip only touches the sample briefly during each cycle of oscillation, the lateral forces applied to soft features on the sample are minimized, so that they can be imaged without damage.
TheQof the cantilever is typically about 100. Therefore, sinceA
ωd=ω0
=QAd, the drive amplitudeAdapplied by the piezoelectric crystal to the base of the cantilever needs to be only about 1 nm in order to produce the desired tip vibration amplitude Aof 100 nm. This large ratio is essential, because if one had to vibrate the base by the full 100 nm, the entire AFM would vibrate, dramatically degrading image quality.
94
4.3 Energy flow
Energy is perhaps the most fundamental idea in physics. In many situations, including many parts of quantum mechanics, finding the energy is the central problem. We will find that, for damped driven oscillators, a careful consideration of the energy for the steady-state behavior provides insights that can be very broadly applied.
The power (energy per time) supplied to the oscillator by the drive force is Pdrive=Fdrivev.
In the steady-state,x=Acos (ωdt−δ), and so v= ˙x= −Aωdsinωdt−δ
=Aωdsinδ−ωdt=Aωdcosδ−π/2−ωdt
=Aωdcosωdt+π/2−δ Therefore,
Pdrive=
F0cosωdt !
Fdrive
Aωdcosωdt+π/2−δ
!
v
. (4.3.1)
Concept test (answer4below):For what value ofωd/ω0 are the oscillations ofFdrive in phase with the oscillations inv?Hint: review figure 4.2.2.
Since, for the value you just found, the oscillations are always in phase,Pdriveis always positive, that is, the drive force always supplies powertothe oscillator. For any other value ofωd/ω0, the oscillations ofFdriveare not in phase with the oscillations inv;
therefore, for some parts of the cyclePdriveis positive (the drive force supplies power to the oscillator), and for some parts of the cyclePdriveis negative (the oscillator supplies power to the entity providing the drive force). An example is shown in figure 4.3.1a.
We can definePdrive, av to be the power supplied by the drive averaged over a cycle.
For the example shown in figure 4.3.1a, Pdrive is positive for most of the cycle, so Pdrive, av >0.
Asωd→0,δ→0, so thatFdriveandvare out of phase byπ/2. For this condition, the net energy flowing from the drive to the oscillator is zero (i.e.,Pdrive, av =0), as suggested in figure 4.3.1b. (You can show rigorously that the net energy flow is zero in problem 4.11.) Asωd→ ∞,δ →π, so thatFdriveandvare out of phase byπ/2 in the other direction; again the net flow of energy from the drive over a cycle is zero.
Putting all this together, we expect that the graph ofPdrive, avas a function ofωd/ω0 must start at zero, reach a peak (probably atωd/ω0=1), and then go back to zero.
Let’s examine this quantitatively. The average power, Pdrive, av, can be computed using the average value theorem from calculus:
Pdrive,av= 1 T
T 0
Pdrivedt,
4. For the oscillations to be in phase, we needδ=π/2, which occurs whenωd/ω0=1.
Figure 4.3.1 a:Fdriveandvfor the caseδ=45◦. In the shaded regions, the two have opposite sign, soPdriveis negative, meaning that the oscillator supplies power to the drive. In the other regions,Fdriveandvhave the same sign, soPdriveis positive, and the drive supplies power to the oscillator. (SinceFdriveandvhave different units, the vertical scales cannot be compared.) b:Fdriveandvforωd→0; in this limitδ→0, soFdriveandvare out of phase byπ/2. In the shaded regions, the two have opposite sign, soPdriveis negative, meaning that the oscillator supplies power to the drive. In the other regions,Fdriveandvhave the same sign, soPdriveis positive, and the drive supplies power to the oscillator. (The velocity scale for this plot is greatly magnified compared to part a; for smallωd, the velocity is also small.)
whereT is the period of the steady-state motion. Plugging in our expression from equation (4.3.1), we get
Pdrive= 1 T
T 0
F0cosωdtAωdcosωdt+π/2−δ dt
= F0Aωd T
T 0
cosωdt cosωdt+π/2−δ dt.
This integral can be evaluated using Mathematica, or a similar symbolic algebra program or calculator; it can also be done “by hand” as shown in the footnote.5Bearing
5. First, we re-express the second term in the integral: cos
ωdt+π/2−δ
= cos δ − π/2 − ωdt
= sin δ−ωdt
. Therefore, Pdrive = F0Aωd T
$T 0
cosωdt sin δ−ωdt
dt.
Next, we use sin (A+B) = sinAcosB + cosAsinB, so that Pdrive =
Figure 4.3.2 Geometric representation of equation (4.2.6).
in mind thatT =2π/ωd, the result is
Pdrive, av =F0Aωd
2 sinδ. (4.3.2)
To evaluate sin δ, we can represent equation (4.2.6) geometrically, as shown in figure 4.3.2. From this, we see that
sinδ= 1/Q ω0
ωd −ωωd02
+Q12
.
Your turn:Substitute the above, and also equation (4.2.1):A= F0 k
ω0/ωd
ω0 ωd−ωd
ω0 2
+ 1 Q2 into equation (4.3.2), and show that the result is
Pdrive, av= F02ω0 2kQ
1 ω0 ωd −ωd
ω0 2
+ 1 Q2
. (4.3.3)
F0Aωd T
+ sinδ
$T 0
cosωdt cos
−ωdt
dt + cosδ
$T 0
cosωdt sin
−ωdt dt
,
. This simplifies to Pdrive = F0Aωd
T +
sinδ
$T 0
cos2ωdtdt − cosδ
$T 0
cosωdt sinωdtdt ,
. These are both standard integrals which can be looked up in a table. The first one is especially important:
$T 0
cos2ωdtdt =
"
t 2+ 1
4ωdsin 2ωdt
#T 0
. Since T = 2π/ωd, the second term evaluates to zero at both limits, so that the integral is just T /2. Since we integrated over a whole period, we can see that the average value of a squared sinusoid over one period is 1/2.
The second integral can also be looked up in a table (or you can integrate it by parts):
$T 0
cosωdt sinωdtdt=
"
1
2ωdsin2ωdt
#T 0
. SinceT=2π/ωd, this evaluates to zero at both limits.
Putting this all together givesPdrive=F0Aωd T
T
2sinδ=F0Aωd 2 sinδ.
This relation is plotted in figure 4.3.3a. Inspection of the equation shows that the peak is at exactlyωd/ω0=1. As for the graph ofA versusωd(figure 4.2.1), largerQ(less damping) leads to a higher and sharper peak.
The width of the peak is quite important for a variety of applications. One common way to define the width is the “full width at half maximum,” or FWHM, which is the width at half of the peak height, as shown in figure 4.3.3b. Let us calculate the FWHM exactly. Referring to equation (4.3.3), we see that the values ofωdcorresponding to half the maximum height are determined by
ω0 ω± −ω±
ω0 2
= 1
Q2, (4.3.4)
whereω±is eitherω+orω−. First, we considerω−. Sinceω0> ω−, equation (4.3.4) gives
ω20−ω2−= ω0ω−
Q ⇔ω
2−+ω0ω−
Q −ω
2
0=0⇒ω−= −ωQ0 ±
ω20 Q2+4ω20
2 .
Sinceω−must be greater than 0, we choose the positive square root, giving
ω−= −ω0 2Q+
1 2
ω02 Q2 +4ω20.
Figure 4.3.3 a: Power delivered by drive force to oscillator (averaged over a cycle) as a function ofωd. This is called the power resonance curve. b: Definition of FWHM.
Next, we considerω+. Sinceω+> ω0, equation (4.3.4) gives ω+2 −ω20=ω0Qω+ ⇔ω+2 −ω0Qω+ −ω02=0
⇒ω+= ω0
Q ± ω20 Q2+4ω02
2 .
Sinceω+must be greater than 0, we choose the positive square root, giving ω+= ω0
2Q+ 1 2
ω20
Q2 +4ω40. The FWHM is given byω+−ω−, so that
FWHM=ω0
Q =γ . (4.3.5)
Full Width at Half Maximum for the graph ofPdrive, avversusωd.
This important equation tells us that systems with more damping have a broader power resonance peak. This is a very universal behavior, which is observed even for systems with turbulence or frictional damping.
4.4 Linear differential equations, the superposition principle for driven systems, and the response to multiple drive forces
What happens if we apply a nonsinusoidal drive force, or if we apply two or three different sinusoids at different frequencies and amplitudes? We will show in chapter 8 thatanyfunction can be represented as a sum of sinusoids. Therefore, if we can figure out how a damped driven oscillator responds to multiple sinusoidal drives, it will be relatively easy, after understanding the contents of chapter 8, to determine the response to any driving force.
We will make use of an important property of the differential equation governing a damped driven oscillator:
mx¨= −kx−bx˙+F0cosωdt ⇒ ¨x+γx˙+ω20x= F0
m cosωdt.
This differential equation is linear, meaning that it has no terms proportional tox2, or tox˙2, or toxx, etc. That means, for example, that if we multiply˙ xby two, then the entire left side of the above equation becomes twice as big. (This would not be true, for example, if the left side werex2.) It also means, as we’re about to show, that we can simply add the steady-state solutions for each of the individual driving forces to get the steady-state solution for the combined driving force. The term on the right side, which does not depend onx, and which comes from the drive (also called the excitation) is called the “source term.”